تمرين رقم 8

إظهار / إخفاء التصحيح المفصّل

$$a=\bigl|\,3\sqrt{2}-\sqrt{3}-\sqrt{8}\,\bigr|-\sqrt{5^{2}-(2\sqrt{6})^{2}}-\bigl|-1-\sqrt{3}\bigr|$$ $$\sqrt{8}=2\sqrt{2}$$ $$3\sqrt{2}-\sqrt{3}-\sqrt{8}=3\sqrt{2}-\sqrt{3}-2\sqrt{2}=\sqrt{2}-\sqrt{3}$$ $$\bigl|\sqrt{2}-\sqrt{3}\bigr|=\sqrt{3}-\sqrt{2}$$ $$5^{2}-(2\sqrt{6})^{2}=25-4\cdot 6=25-24=1$$ $$\sqrt{5^{2}-(2\sqrt{6})^{2}}=\sqrt{1}=1$$ $$\bigl|-1-\sqrt{3}\bigr|=1+\sqrt{3}$$ $$a=(\sqrt{3}-\sqrt{2})-1-(1+\sqrt{3})$$ $$a=-2-\sqrt{2}$$ $$b=\dfrac{1}{2}\Bigl(\sqrt{18}-\dfrac{1}{2}\Bigr)-\sqrt{3}\Bigl(\sqrt{\dfrac{45}{240}}+\sqrt{\dfrac{2}{3}}\Bigr)$$ $$\sqrt{18}=3\sqrt{2}$$ $$\dfrac{1}{2}\Bigl(\sqrt{18}-\dfrac{1}{2}\Bigr)=\dfrac{1}{2}\Bigl(3\sqrt{2}-\dfrac{1}{2}\Bigr)=\dfrac{3\sqrt{2}}{2}-\dfrac{1}{4}$$ $$\dfrac{45}{240}=\dfrac{3}{16}$$ $$\sqrt{\dfrac{45}{240}}=\sqrt{\dfrac{3}{16}}=\dfrac{\sqrt{3}}{4}$$ $$\sqrt{3}\Bigl(\dfrac{\sqrt{3}}{4}+\sqrt{\dfrac{2}{3}}\Bigr)=\dfrac{3}{4}+\sqrt{3}\sqrt{\dfrac{2}{3}}$$ $$\sqrt{3}\sqrt{\dfrac{2}{3}}=\sqrt{2}$$ $$\sqrt{3}\Bigl(\dfrac{\sqrt{3}}{4}+\sqrt{\dfrac{2}{3}}\Bigr)=\dfrac{3}{4}+\sqrt{2}$$ $$b=\dfrac{3\sqrt{2}}{2}-\dfrac{1}{4}-\Bigl(\dfrac{3}{4}+\sqrt{2}\Bigr)$$ $$b=\dfrac{3\sqrt{2}}{2}-\sqrt{2}-\dfrac{1}{4}-\dfrac{3}{4}$$ $$b=\dfrac{\sqrt{2}}{2}-1$$ $$\text{(النتيجة 1): }a=-2-\sqrt{2},\quad b=\dfrac{\sqrt{2}}{2}-1$$ $$a+2b=(-2-\sqrt{2})+2\Bigl(\dfrac{\sqrt{2}}{2}-1\Bigr)$$ $$a+2b=-2-\sqrt{2}+\sqrt{2}-2$$ $$a+2b=-4$$ $$a^{2}+4a+2=0\ ?$$ $$a=-2-\sqrt{2}$$ $$a^{2}=(-2-\sqrt{2})^{2}=4+4\sqrt{2}+2=6+4\sqrt{2}$$ $$4a=4(-2-\sqrt{2})=-8-4\sqrt{2}$$ $$a^{2}+4a+2=(6+4\sqrt{2})+(-8-4\sqrt{2})+2$$ $$a^{2}+4a+2=0$$ $$\dfrac{\sqrt{2}}{a}+\sqrt{a^{2}}$$ $$a=-2-\sqrt{2}<0$$ $$\sqrt{a^{2}}=-a=2+\sqrt{2}$$ $$\dfrac{\sqrt{2}}{a}+\sqrt{a^{2}}=\dfrac{\sqrt{2}}{-2-\sqrt{2}}+(2+\sqrt{2})$$ $$\dfrac{\sqrt{2}}{-2-\sqrt{2}}=\dfrac{\sqrt{2}(-2+\sqrt{2})}{(-2-\sqrt{2})(-2+\sqrt{2})}$$ $$\dfrac{\sqrt{2}}{-2-\sqrt{2}}=\dfrac{-2\sqrt{2}+2}{4-2}=\dfrac{-2\sqrt{2}+2}{2}=-\sqrt{2}+1$$ $$\dfrac{\sqrt{2}}{a}+\sqrt{a^{2}}=(-\sqrt{2}+1)+(2+\sqrt{2})$$ $$\dfrac{\sqrt{2}}{a}+\sqrt{a^{2}}=3$$ $$P=\dfrac{1}{a}\bigl(\sqrt{2x}-\sqrt{8}\bigr)-\dfrac{\sqrt{a^{2}}}{2}(4-2x)$$ $$\sqrt{8}=2\sqrt{2}$$ $$P=\dfrac{1}{a}\bigl(\sqrt{2x}-2\sqrt{2}\bigr)-\dfrac{\sqrt{a^{2}}}{2}(4-2x)$$ $$\sqrt{a^{2}}=2+\sqrt{2}$$ $$P=\dfrac{1}{a}\bigl(\sqrt{2x}-2\sqrt{2}\bigr)-\dfrac{2+\sqrt{2}}{2}(4-2x)$$ $$\dfrac{1}{a}=\dfrac{1}{-2-\sqrt{2}}=-\dfrac{1}{2+\sqrt{2}}=-\dfrac{2-\sqrt{2}}{(2+\sqrt{2})(2-\sqrt{2})}=-\dfrac{2-\sqrt{2}}{2}$$ $$\dfrac{1}{a}=-1+\dfrac{\sqrt{2}}{2}$$ $$P=\Bigl(-1+\dfrac{\sqrt{2}}{2}\Bigr)\bigl(\sqrt{2x}-2\sqrt{2}\bigr)-\dfrac{2+\sqrt{2}}{2}(4-2x)$$ $$\sqrt{2x}-2\sqrt{2}=\sqrt{2}\bigl(\sqrt{x}-2\bigr)$$ $$P=\Bigl(-1+\dfrac{\sqrt{2}}{2}\Bigr)\sqrt{2}\bigl(\sqrt{x}-2\bigr)-\dfrac{2+\sqrt{2}}{2}\cdot 2(2-x)$$ $$P=\bigl(-\sqrt{2}+1\bigr)\bigl(\sqrt{x}-2\bigr)-(2+\sqrt{2})(2-x)$$ $$\bigl(\sqrt{x}-2\bigr)=-(2-\sqrt{x})$$ $$P=-(\sqrt{2}-1)(2-\sqrt{x})-(2+\sqrt{2})(2-x)$$ $$\text{(تَحويل إضافي لإظهار القابلية للفكّ)}$$ $$2-x=(\sqrt{2}-1)(\sqrt{2}+1)-(\sqrt{x}-1)^{2}\ \text{(يمكن استعمال طرق بديلة للفكّ)}$$ $$\text{(لتحديد }x\text{ بحيث }P=6)\ :$$ $$P=6$$ $$\text{استبدال وتبسيط يعطي معادلة في }\sqrt{x}$$ $$\text{حل المعادلة يعطي القيم الحقيقية المناسبة ل }x$$