مراجعة الرياضيات

تمرين عدد 1

(I نعتبر العبارة: $A = (4 - x)^2 - 9$ حيث $x \in \mathbb{R}$[span_7](start_span)[span_7](end_span).
1) فكك إلى جذاء عوامل العبارة $A$[span_8](start_span)[span_8](end_span).
2) أوجد العدد الحقيقي $x$ حيث $A = 0$[span_9](start_span)[span_9](end_span).
3) بين أن $A + 5 = (6 - x)(2 - x)$[span_10](start_span)[span_10](end_span).

(II يُمثل الرسم التالي مثلثا $OAB$ قائم الزاوية في $O$ حيث $OA = 4$ و $OB = 2$ و $K$ منتصف $[AB]$ و $M$ نقطة من $[OA]$ بحيث $OM = x$[span_11](start_span)[span_11](end_span).
المستقيم المار من $M$ والعمودي على $(OA)$ يقطع $(AB)$ في $N$[span_12](start_span)[span_12](end_span).

1) بين أن $MN = \frac{4 - x}{2}$[span_13](start_span)[span_13](end_span).
2) لتكن $P$ المسقط العمودي للنقطة $K$ على $(OA)$ و $S$ مساحة شبه المنحرف $MNKP$[span_14](start_span)[span_14](end_span).
أ- بين أن $KP = 1$[span_15](start_span)[span_15](end_span).
ب- بين أن $S = \frac{1}{4}(6 - x)(2 - x)$[span_16](start_span)[span_16](end_span).
3) أوجد العدد الحقيقي $x$ حيث $4S = 5$[span_17](start_span)[span_17](end_span).

إصلاح التمرين عدد 1:
I)

1) $A = (4 - x)^2 - 9$[span_18](start_span)[span_18](end_span)

$A = (4 - x - 3)(4 - x + 3)$[span_19](start_span)[span_19](end_span)

$A = (1 - x)(7 - x)$[span_20](start_span)[span_20](end_span)

2) $A = 0 \implies (1 - x)(7 - x) = 0$[span_21](start_span)[span_21](end_span)

يعني $(1 - x) = 0$ أو $(7 - x) = 0$[span_22](start_span)[span_22](end_span)

يعني $x = 1$ أو $x = 7$[span_23](start_span)[span_23](end_span)

3) $A + 5 = (4 - x)^2 - 9 + 5$[span_24](start_span)[span_24](end_span)

$A + 5 = (4 - x)^2 - 4$[span_25](start_span)[span_25](end_span)

$A + 5 = (4 - x - 2)(4 - x + 2)$[span_26](start_span)[span_26](end_span)

$A + 5 = (2 - x)(6 - x)$[span_27](start_span)[span_27](end_span)

II)

1) في المثلث $AOB$ لنا $M \in [OA]$ و $N \in [AB]$[span_28](start_span)[span_28](end_span)

بما أن $(MN) \perp (OA)$ و $(OB) \perp (OA)$ فإن $(MN) // (OB)$[span_29](start_span)[span_29](end_span)

حسب مبرهنة طالس: $\frac{AM}{AO} = \frac{MN}{OB}$[span_30](start_span)[span_30](end_span)

$\frac{4 - x}{4} = \frac{MN}{2}$[span_31](start_span)[span_31](end_span)

$MN = \frac{4 - x}{2}$[span_32](start_span)[span_32](end_span)

2) أ- في المثلث $AOB$ لنا $K$ منتصف $[AB]$ و $(PK) // (OB)$ فإن $P$ هي منتصف $[OA]$[span_33](start_span)[span_33](end_span)

إذن $KP = \frac{OB}{2} = \frac{2}{2} = 1$[span_34](start_span)[span_34](end_span)

ب- $S_{MNKP} = \frac{(PK + MN) \times MP}{2}$[span_35](start_span)[span_35](end_span)

$S = \frac{\left(1 + \frac{4 - x}{2}\right)(2 - x)}{2}$[span_36](start_span)[span_36](end_span)

$S = \frac{\left(\frac{6 - x}{2}\right)(2 - x)}{2}$[span_37](start_span)[span_37](end_span)

$S = \frac{(6 - x)(2 - x)}{4}$[span_38](start_span)[span_38](end_span)

3) $4S = 5 \implies 4 \times \frac{1}{4}(6 - x)(2 - x) = 5$[span_39](start_span)[span_39](end_span)

$(6 - x)(2 - x) = 5$[span_40](start_span)[span_40](end_span)

$A + 5 = 5$[span_41](start_span)[span_41](end_span)

$A = 0$[span_42](start_span)[span_42](end_span)

وبما أن $M \in [OA]$ فإن $0 \le x \le 4$[span_43](start_span)[span_43](end_span)

إذن $x = 1$[span_44](start_span)[span_44](end_span)

تمرين عدد 2

1) لتكن العبارة $A = \frac{3}{2}x^2 - \frac{15}{2}x + 9$ حيث $x \in \mathbb{R}$[span_45](start_span)[span_45](end_span).
أ) احسب $A$ إذا علمت أن $x = 3$[span_46](start_span)[span_46](end_span).
ب) بين بالنشر أن $A = (3 - x)^2 - \frac{1}{2}x(3 - x)$[span_47](start_span)[span_47](end_span).
ج) استنتج تفكيكا إلى جذاء عوامل للعبارة $A$[span_48](start_span)[span_48](end_span).

2) نعتبر الشكل التالي حيث $ABCD$ مربع و $EAD$ مثلث قائم الزاوية في $A$[span_49](start_span)[span_49](end_span).
و $EB = 3$ و $EA = x$ ($x$ عدد حقيقي موجب أصغر من 3)[span_50](start_span)[span_50](end_span).

ابحث عن العدد $x$ ليكون قيس مساحة المربع $ABCD$ يساوي قيس مساحة المثلث $EAD$[span_51](start_span)[span_51](end_span).

إصلاح التمرين عدد 2:

1) أ) إذا كان $x = 3$:

$A = \frac{3}{2}(3)^2 - \frac{15}{2}(3) + 9$[span_52](start_span)[span_52](end_span)

$A = \frac{27}{2} - \frac{45}{2} + \frac{18}{2} = 0$[span_53](start_span)[span_53](end_span)

ب) $(3 - x)^2 - \frac{1}{2}x(3 - x) = 9 - 6x + x^2 - \frac{3}{2}x + \frac{1}{2}x^2$[span_54](start_span)[span_54](end_span)

$= \frac{3}{2}x^2 - \frac{15}{2}x + 9 = A$[span_55](start_span)[span_55](end_span)

ج) $A = (3 - x)^2 - \frac{1}{2}x(3 - x)$[span_56](start_span)[span_56](end_span)

$A = (3 - x)\left[(3 - x) - \frac{1}{2}x\right]$[span_57](start_span)[span_57](end_span)

$A = (3 - x)\left(3 - \frac{3}{2}x\right)$[span_58](start_span)[span_58](end_span)

2) $S_{ABCD} = S_{AED}$[span_59](start_span)[span_59](end_span)

بما أن $AB = EB - EA = 3 - x$ فإن ضلع المربع هو $(3 - x)$[span_60](start_span)[span_60](end_span)

و $AD = AB = 3 - x$[span_61](start_span)[span_61](end_span)

مساحة المربع: $S_{ABCD} = (3 - x)^2$[span_62](start_span)[span_62](end_span)

مساحة المثلث القائم: $S_{AED} = \frac{EA \times AD}{2} = \frac{1}{2}x(3 - x)$[span_63](start_span)[span_63](end_span)

$(3 - x)^2 = \frac{1}{2}x(3 - x)$[span_64](start_span)[span_64](end_span)

$(3 - x)^2 - \frac{1}{2}x(3 - x) = 0$[span_65](start_span)[span_65](end_span)

$A = 0$[span_66](start_span)[span_66](end_span)

$(3 - x)\left(3 - \frac{3}{2}x\right) = 0$[span_67](start_span)[span_67](end_span)

يعني $3 - x = 0$ أو $3 - \frac{3}{2}x = 0$[span_68](start_span)[span_68](end_span)

يعني $x = 3$ أو $x = 2$[span_69](start_span)[span_69](end_span)

بما أن $x < 3$ فإن الحل هو $x = 2$[span_70](start_span)[span_70](end_span)

تمرين عدد 3

(I لتكن العبارة $A = x^2 + 2x - 80$[span_71](start_span)[span_71](end_span).
1) بين أن $A = (x + 1)^2 - 81$[span_72](start_span)[span_72](end_span).
2) استنتج تفكيكا للعبارة $A$[span_73](start_span)[span_73](end_span).

(II $ABC$ مثلث قائم الزاوية في $A$ بحيث $AC = x$ و $AB = x + 2$ و $x \in \mathbb{R}_+$[span_74](start_span)[span_74](end_span).
1) أوجد $x$ إذا علمت أن مساحة المثلث $ABC$ تساوي $40\text{ cm}^2$[span_75](start_span)[span_75](end_span).
2) لتكن $J$ نقطة من $[AB]$ حيث $BJ = 2\text{ cm}$ و $\Delta$ المستقيم المار من $J$ والعمودي على $(AB)$ يقطع $[BC]$ في $I$[span_76](start_span)[span_76](end_span).
أ) بين أن $\frac{IB}{IC} = \frac{2}{x}$[span_77](start_span)[span_77](end_span).
ب) أوجد $x$ ليكون $I$ منتصف $[BC]$[span_78](start_span)[span_78](end_span).

إصلاح التمرين عدد 3:
I)

1) $(x + 1)^2 - 81 = x^2 + 2x + 1 - 81 = x^2 + 2x - 80 = A$[span_79](start_span)[span_79](end_span)

2) $A = (x + 1)^2 - 81 = (x + 1 - 9)(x + 1 + 9)$[span_80](start_span)[span_80](end_span)

$A = (x - 8)(x + 10)$[span_81](start_span)[span_81](end_span)

II)

1) $S_{ABC} = 40 \implies \frac{AC \times AB}{2} = 40$[span_82](start_span)[span_82](end_span)

$\frac{x(x + 2)}{2} = 40 \implies x(x + 2) = 80$[span_83](start_span)[span_83](end_span)

$x^2 + 2x - 80 = 0 \implies A = 0$[span_84](start_span)[span_84](end_span)

$(x - 8)(x + 10) = 0$[span_85](start_span)[span_85](end_span)

يعني $x = 8$ أو $x = -10$[span_86](start_span)[span_86](end_span)

بما أن $x$ عدد موجب فإن $x = 8$[span_87](start_span)[span_87](end_span)

2) أ) لنا $(IJ) \perp (AB)$ و $(AC) \perp (AB)$ إذن $(IJ) // (AC)$[span_88](start_span)[span_88](end_span)

حسب مبرهنة طالس في المثلث $ABC$:[span_89](start_span)[span_89](end_span)

$\frac{IB}{IC} = \frac{JB}{JA}$[span_90](start_span)[span_90](end_span)

بما أن $BJ = 2$ و $JA = AB - BJ = (x + 2) - 2 = x$[span_91](start_span)[span_91](end_span)

فإن $\frac{IB}{IC} = \frac{2}{x}$[span_92](start_span)[span_92](end_span)

ب) $I$ منتصف $[BC]$ يعني $IB = IC \implies \frac{IB}{IC} = 1$[span_93](start_span)[span_93](end_span)

يعني $\frac{2}{x} = 1 \implies x = 2$[span_94](start_span)[span_94](end_span)

تمرين عدد 5

1) نعتبر العدد الحقيقي $a = \sqrt{3}^{1001} \times \sqrt{3}^{-1000} + \sqrt{(\sqrt{2} - \sqrt{3})^2}$[span_95](start_span)[span_95](end_span).
بين أن $a = 2\sqrt{3} - \sqrt{2}$[span_96](start_span)[span_96](end_span).

2) أوجد كتابة مقامها عدد صحيح للعدد $\frac{1}{\sqrt{2} + \sqrt{3}}$[span_97](start_span)[span_97](end_span).

3) نعتبر العدد الحقيقي $b = \frac{1}{\sqrt{3} + \sqrt{2}} - \sqrt{48} + \sqrt{8}$[span_98](start_span)[span_98](end_span).
بين أن $b = \sqrt{2} - 3\sqrt{3}$[span_99](start_span)[span_99](end_span).
أ- أوجد حصرا مداه $10^{-1}$ لكل من $\sqrt{2}$ و $\sqrt{3}$[span_100](start_span)[span_100](end_span).
ب- بين أن $a \in ]1.9, 2.2[$ و $b \in ]-4, -3.6[$[span_101](start_span)[span_101](end_span).
ج- استنتج حصرا للعدد $ab$[span_102](start_span)[span_102](end_span).

4) نعتبر المتراجحة $I : ax \le \sqrt{3}a - bx$[span_103](start_span)[span_103](end_span).
أ- حل في $\mathbb{R}$ المتراجحة $I$[span_104](start_span)[span_104](end_span).
ب- حدد مجموعة حلول المتراجحة $I$ في المجال $[-3, -2[$[span_105](start_span)[span_105](end_span).

إصلاح التمرين عدد 5:

1) $a = \sqrt{3}^{1001 - 1000} + |\sqrt{2} - \sqrt{3}|$[span_106](start_span)[span_106](end_span)

بما أن $\sqrt{2} < \sqrt{3}$ فإن $|\sqrt{2} - \sqrt{3}| = \sqrt{3} - \sqrt{2}$[span_107](start_span)[span_107](end_span)

$a = \sqrt{3} + \sqrt{3} - \sqrt{2} = 2\sqrt{3} - \sqrt{2}$[span_108](start_span)[span_108](end_span)

2) $\frac{1}{\sqrt{2} + \sqrt{3}} = \frac{\sqrt{3} - \sqrt{2}}{(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2})} = \frac{\sqrt{3} - \sqrt{2}}{3 - 2} = \sqrt{3} - \sqrt{2}$[span_109](start_span)[span_109](end_span)

3) $b = (\sqrt{3} - \sqrt{2}) - \sqrt{16 \times 3} + \sqrt{4 \times 2}$[span_110](start_span)[span_110](end_span)

$b = \sqrt{3} - \sqrt{2} - 4\sqrt{3} + 2\sqrt{2} = -3\sqrt{3} + \sqrt{2}$[span_111](start_span)[span_111](end_span)

أ- $1.4 < \sqrt{2} < 1.5$ و $1.7 < \sqrt{3} < 1.8$[span_112](start_span)[span_112](end_span)

ب- حصر $a$:

$1.7 < \sqrt{3} < 1.8 \implies 3.4 < 2\sqrt{3} < 3.6$[span_113](start_span)[span_113](end_span)

$1.4 < \sqrt{2} < 1.5 \implies -1.5 < -\sqrt{2} < -1.4$[span_114](start_span)[span_114](end_span)

$3.4 - 1.5 < 2\sqrt{3} - \sqrt{2} < 3.6 - 1.4 \implies 1.9 < a < 2.2$[span_115](start_span)[span_115](end_span)

إذن $a \in ]1.9, 2.2[$[span_116](start_span)[span_116](end_span)

حصر $b$:

$1.7 < \sqrt{3} < 1.8 \implies -5.4 < -3\sqrt{3} < -5.1$[span_117](start_span)[span_117](end_span)

$1.4 - 5.4 < \sqrt{2} - 3\sqrt{3} < 1.5 - 5.1 \implies -4 < b < -3.6$[span_118](start_span)[span_118](end_span)

إذن $b \in ]-4, -3.6[$[span_119](start_span)[span_119](end_span)

ج- لدينا $3.6 < -b < 4$ و $1.9 < a < 2.2$[span_120](start_span)[span_120](end_span)

$1.9 \times 3.6 < a(-b) < 2.2 \times 4$[span_121](start_span)[span_121](end_span)

$6.84 < -ab < 8.8 \implies -8.8 < ab < -6.84$[span_122](start_span)[span_122](end_span)

4) أ- $ax \le \sqrt{3}a - bx \implies ax + bx \le \sqrt{3}a$[span_123](start_span)[span_123](end_span)

$x(a + b) \le \sqrt{3}a$[span_124](start_span)[span_124](end_span)

نحسب $a + b = (2\sqrt{3} - \sqrt{2}) + (\sqrt{2} - 3\sqrt{3}) = -\sqrt{3} < 0$[span_125](start_span)[span_125](end_span)

$- \sqrt{3}x \le \sqrt{3}a \implies x \ge \frac{\sqrt{3}a}{-\sqrt{3}}$[span_126](start_span)[span_126](end_span)

$x \ge -a \implies S_{\mathbb{R}} = [-a, +\infty[$[span_127](start_span)[span_127](end_span)

ب- بما أن $1.9 < a < 2.2$ فإن $-2.2 < -a < -1.9$[span_128](start_span)[span_128](end_span)

المجال $[-a, +\infty[$ لا يتقاطع مع $[-3, -2[$[span_129](start_span)[span_129](end_span)

$S = [-a, +\infty[ \cap [-3, -2[ = \emptyset$[span_130](start_span)[span_130](end_span)

تمرين عدد 6

نعتبر المجالين التاليين $I = [-2, 3]$ و $J = [-\frac{5}{2}, -1]$[span_131](start_span)[span_131](end_span).
1) ليكن $x \in I$ و $y \in J$. أوجد حصرا لـ $x + y$ و $y - x$[span_132](start_span)[span_132](end_span).
2) أ- بين أن $0 \le (2x - 1)^2 \le 25$[span_133](start_span)[span_133](end_span).
ب- استنتج أن $-\frac{1}{4} \le x^2 - x \le 6$[span_134](start_span)[span_134](end_span).
3) أ- مثل على المستقيم العددي المجالين $I$ و $J$ ثم حدد $I \cap J$ و $I \cap \mathbb{N}$[span_135](start_span)[span_135](end_span).
ب- بين أن $-\frac{\sqrt{7}}{2} \in I \cap J$[span_136](start_span)[span_136](end_span).
4) لتكن $x$ عددا حقيقيا حيث $-5 < 3x - 5 < -2$[span_137](start_span)[span_137](end_span).
أ- بين أن $x \in ]0, 1[$[span_138](start_span)[span_138](end_span).
ب- لتكن العبارة $A = |3x - 5| - |x - 4|$. اختصر العبارة $A$ ثم استنتج أن $|A| < 1$[span_139](start_span)[span_139](end_span).

إصلاح التمرين عدد 6:

1) $-2 \le x \le 3$ و $-\frac{5}{2} \le y \le -1$[span_140](start_span)[span_140](end_span)

$-2 + \left(-\frac{5}{2}\right) \le x + y \le 3 + (-1) \implies -\frac{9}{2} \le x + y \le 2$[span_141](start_span)[span_141](end_span)

$-3 \le -x \le 2 \implies -\frac{5}{2} + (-3) \le y - x \le -1 + 2 \implies -\frac{11}{2} \le y - x \le 1$[span_142](start_span)[span_142](end_span)

2) أ- $-2 \le x \le 3 \implies -4 \le 2x \le 6 \implies -5 \le 2x - 1 \le 5$[span_143](start_span)[span_143](end_span)

إذن $|2x - 1| \le 5 \implies 0 \le (2x - 1)^2 \le 25$[span_144](start_span)[span_144](end_span)

ب- $(2x - 1)^2 = 4x^2 - 4x + 1$[span_145](start_span)[span_145](end_span)

$0 \le 4x^2 - 4x + 1 \le 25 \implies -1 \le 4(x^2 - x) \le 24$[span_146](start_span)[span_146](end_span)

$-\frac{1}{4} \le x^2 - x \le 6$[span_147](start_span)[span_147](end_span)

3) أ- $I \cap J = [-2, -1]$[span_148](start_span)[span_148](end_span)

$I \cap \mathbb{N} = \{0, 1, 2, 3\}$[span_149](start_span)[span_149](end_span)

ب- $\left(-\frac{\sqrt{7}}{2}\right)^2 = \frac{7}{4} = 1.75$[span_150](start_span)[span_150](end_span)

بما أن $1 < 1.75 < 4$ فإن $1 < -\frac{\sqrt{7}}{2} < 2 \implies -2 < -\frac{\sqrt{7}}{2} < -1$[span_151](start_span)[span_151](end_span)

إذن $-\frac{\sqrt{7}}{2} \in [-2, -1] = I \cap J$[span_152](start_span)[span_152](end_span)

4) أ- $-5 < 3x - 5 < -2 \implies 0 < 3x < 3 \implies 0 < x < 1 \implies x \in ]0, 1[$[span_153](start_span)[span_153](end_span)

ب- بما أن $3x - 5 < 0$ فإن $|3x - 5| = -3x + 5$[span_154](start_span)[span_154](end_span)

بما أن $x \in ]0, 1[$ فإن $x - 4 < 0 \implies |x - 4| = -x + 4$[span_155](start_span)[span_155](end_span)

$A = (-3x + 5) - (-x + 4) = -2x + 1$[span_156](start_span)[span_156](end_span)

$0 < x < 1 \implies 0 < 2x < 2 \implies -1 < -2x + 1 < 1 \implies |A| < 1$[span_157](start_span)[span_157](end_span)

تمرين عدد 7

(I نعتبر العبارة $A = 2x^2 - 8$ حيث $x \in \mathbb{R}$[span_158](start_span)[span_158](end_span).
1) أ- أحسب القيمة العددية للعبارة $A$ إذا كان $x = -1 + \sqrt{2}$[span_159](start_span)[span_159](end_span).
ب- بين أن $\frac{A}{2} = (x - 2)(x + 2)$[span_160](start_span)[span_160](end_span).
2) لتكن العبارة $B = 2(x - 1)^2 - 4(x - \frac{3}{2})$[span_161](start_span)[span_161](end_span).
أ- انشر واختصر العبارة $B$[span_162](start_span)[span_162](end_span).
ب- استنتج أن $B = 2(x - 2)^2$[span_163](start_span)[span_163](end_span).
3) بين أن $\frac{A}{2} - B = (x - 2)(6 - x)$[span_164](start_span)[span_164](end_span).
4) حل في $\mathbb{R}$ المعادلة $\frac{A}{2} = B$[span_165](start_span)[span_165](end_span).

(II نعتبر الرسم التالي حيث $BCE$ مثلث قائم في $C$ و $M$ منتصف $[BE]$[span_166](start_span)[span_166](end_span).
$ABCD$ مربع مساحته $x^2 - 4x + 4$ و $CE = 8x + 16$ و $x > 2$[span_167](start_span)[span_167](end_span).

1) بين أن $BC = x - 2$[span_168](start_span)[span_168](end_span).
2) بين أن مساحة المثلث $BCE$ هي $4(x + 2)(x - 2)$[span_169](start_span)[span_169](end_span).
3) أحسب $S_{MBC}$[span_170](start_span)[span_170](end_span).
4) أوجد $x$ في حالة ربع مساحة $MBC$ تساوي مساحة المربع $ABCD$[span_171](start_span)[span_171](end_span).

إصلاح التمرين عدد 7:
I)

1) أ- $A = 2(-1 + \sqrt{2})^2 - 8 = 2(1 - 2\sqrt{2} + 2) - 8 = 6 - 4\sqrt{2} - 8 = -2 - 4\sqrt{2}$[span_172](start_span)[span_172](end_span)

ب- $\frac{A}{2} = \frac{2(x^2 - 4)}{2} = (x - 2)(x + 2)$[span_173](start_span)[span_173](end_span)

2) أ- $B = 2(x^2 - 2x + 1) - 4x + 6 = 2x^2 - 4x + 2 - 4x + 6 = 2x^2 - 8x + 8$[span_174](start_span)[span_174](end_span)

ب- $B = 2(x^2 - 4x + 4) = 2(x - 2)^2$[span_175](start_span)[span_175](end_span)

3) $\frac{A}{2} - B = (x - 2)(x + 2) - 2(x - 2)^2 = (x - 2)[(x + 2) - 2(x - 2)]$[span_176](start_span)[span_176](end_span)

$= (x - 2)(x + 2 - 2x + 4) = (x - 2)(6 - x)$[span_177](start_span)[span_177](end_span)

4) $\frac{A}{2} = B \implies \frac{A}{2} - B = 0 \implies (x - 2)(6 - x) = 0$[span_178](start_span)[span_178](end_span)

يعني $x = 2$ أو $x = 6$[span_179](start_span)[span_179](end_span)

II)

1) $S_{ABCD} = x^2 - 4x + 4 = (x - 2)^2$[span_180](start_span)[span_180](end_span)

بما أن $x > 2$ فإن طول الضلع $BC = \sqrt{(x - 2)^2} = x - 2$[span_181](start_span)[span_181](end_span)

2) $S_{BCE} = \frac{CE \times BC}{2} = \frac{(8x + 16)(x - 2)}{2} = \frac{8(x + 2)(x - 2)}{2} = 4(x + 2)(x - 2)$[span_182](start_span)[span_182](end_span)

3) بما أن $M$ منتصف $[BE]$ فإن $[CM]$ هو الموسط الموافق لـ $[BE]$ في المثلث $BCE$[span_183](start_span)[span_183](end_span)

$S_{MBC} = \frac{S_{BCE}}{2} = 2(x + 2)(x - 2)$[span_184](start_span)[span_184](end_span)

4) $\frac{1}{4} S_{MBC} = S_{ABCD} \implies \frac{2(x + 2)(x - 2)}{4} = (x - 2)^2$[span_185](start_span)[span_185](end_span)

$(x + 2)(x - 2) = 2(x - 2)^2 \implies \frac{A}{2} = B$[span_186](start_span)[span_186](end_span)

وبما أن $x > 2$ فإن الحل هو $x = 6$[span_187](start_span)[span_187](end_span)

المدرسة الإعدادية النموذجية ضفاف البحيرة

سلسلة تمارين مراجعة - 9 أساسي | الأستاذ: فوزي الغربي