تمارين في القوى والجذور التربيعية

1) $a = 5^{-1} \times \sqrt{2}^2 - \sqrt{3}^2$

$a = \frac{1}{5} \times 2 - 3$

$a = \frac{2}{5} - 3 = \frac{2}{5} - \frac{15}{5}$

$a = -\frac{13}{5}$

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2) $b = (\frac{1}{\sqrt{5}-2})^{2012} \times (\sqrt{5}+2)^{2012}$

$b = (\frac{1}{\sqrt{5}-2} \times (\sqrt{5}+2))^{2012}$

$b = (\frac{\sqrt{5}+2}{(\sqrt{5}-2)(\sqrt{5}+2)})^{2012}$

$(a-b)(a+b) = a^2-b^2$ قاعدة رياضية

$b = (\frac{\sqrt{5}+2}{5-4})^{2012} = (\frac{\sqrt{5}+2}{1})^{2012}$

$b = (\sqrt{5}+2)^{2012}$

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3) $c = \frac{2\sqrt{2}-1}{2\sqrt{2}+1} + \frac{2\sqrt{2}+1}{2\sqrt{2}-1}$

$c = \frac{(2\sqrt{2}-1)^2 + (2\sqrt{2}+1)^2}{(2\sqrt{2}+1)(2\sqrt{2}-1)}$

$c = \frac{(8 - 4\sqrt{2} + 1) + (8 + 4\sqrt{2} + 1)}{8 - 1}$

$c = \frac{9 - 4\sqrt{2} + 9 + 4\sqrt{2}}{7}$

$c = \frac{18}{7}$

1) $(\frac{\sqrt{2}}{3})^{-7} \times (\frac{2}{9})^2$

$= (\frac{3}{\sqrt{2}})^7 \times (\frac{(\sqrt{2})^2}{3^2})^2$

$= \frac{3^7}{(\sqrt{2})^7} \times \frac{(\sqrt{2})^{4}}{3^4}$

$= \frac{3^{7-4}}{(\sqrt{2})^{7-4}} = \frac{3^3}{(\sqrt{2})^3} = (\frac{3}{\sqrt{2}})^3$

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2) $\frac{0,00001^3 \times \frac{3}{100}}{0,0003 \times 10^{10}}$

$= \frac{(10^{-5})^3 \times 3 \times 10^{-2}}{3 \times 10^{-4} \times 10^{10}}$

$= \frac{3 \times 10^{-15} \times 10^{-2}}{3 \times 10^{6}}$

$= 10^{-15-2-6} = 10^{-23}$

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3) $\frac{-5\sqrt{5} \times \sqrt{5}^4}{25^3}$

$= \frac{-5^1 \times 5^{0.5} \times 5^{4/2}}{(5^2)^3}$

$= \frac{-5^{1+0.5+2}}{5^6} = \frac{-5^{3.5}}{5^6} = -5^{3.5-6}$

$= -5^{-2.5}$

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4) $\frac{(\frac{\sqrt{2}}{3})^{-9}}{(\frac{3}{\sqrt{2}})^{-9}}$

$= (\frac{\frac{\sqrt{2}}{3}}{\frac{3}{\sqrt{2}}})^{-9} = (\frac{\sqrt{2}}{3} \times \frac{\sqrt{2}}{3})^{-9}$

$= (\frac{2}{9})^{-9} = ((\frac{3}{\sqrt{2}})^2)^{-9}$

$= ((\frac{2}{9})^2)^{-9}$

1) $F = \frac{(a^{-2}b)^{-3} a^{-3}b^2}{a^{2}b^{-5}(ab^{-6})^{-1}}$

$F = \frac{a^{6}b^{-3} \times a^{-3}b^2}{a^{2}b^{-5} \times a^{-1}b^{6}}$

$F = \frac{a^{6-3} b^{-3+2}}{a^{2-1} b^{-5+6}}$

$F = \frac{a^{3} b^{-1}}{a^{1} b^{1}}$

$F = a^{3-1} b^{-1-1}$

$F = a^{2} b^{-2}$

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2) لدينا $F = a^2b^{-2}$

نعيد كتابة $F$ بدلالة $\frac{a}{b}$

$F = a^2b^{-2} = \frac{a^2}{b^2} = (\frac{a}{b})^2$

لدينا $\frac{a}{b} = 3\sqrt{2}-5$

إذن $F = (3\sqrt{2}-5)^2$

$F = (3\sqrt{2})^2 - 2(3\sqrt{2})(5) + 5^2$

$F = (9 \times 2) - 30\sqrt{2} + 25$

$F = 18 - 30\sqrt{2} + 25$

$F = 43 - 30\sqrt{2}$

1) $A = \frac{(3^{2}a^{-3}b^{-6})^{-1} \times 27 \times (ab^{-2})^{-5}}{ (3a^{-2}b^5)^{4}}$

$A = \frac{(3^{-2}a^{3}b^{6}) \times 3^3 \times (a^{-5}b^{10})}{3^4 a^{-8}b^{20}}$

$A = \frac{3^{-2+3}a^{3-5}b^{6+10}}{3^4 a^{-8}b^{20}}$

$A = \frac{3^{1}a^{-2}b^{16}}{3^4 a^{-8}b^{20}}$

$A = 3^{1-4}a^{-2-(-8)}b^{16-20}$

$A = 3^{-3}a^{-2+8}b^{-4}$

$A = 3^{-3}a^{6}b^{-4}$

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2) لدينا $A = 3^{-3}a^{6}b^{-4}$

لدينا $a=\sqrt{7}$ و $\frac{a}{b}=3$

من $\frac{a}{b}=3$ نستنتج أن $b = \frac{a}{3}$

$A = 3^{-3}a^{6}( \frac{a}{3})^{-4}$

$A = 3^{-3}a^{6}( \frac{3}{a})^{4}$

$A = 3^{-3}a^{6}( \frac{3^4}{a^4})$

$A = 3^{-3+4} a^{6-4}$

$A = 3^{1} a^{2}$

بما أن $a=\sqrt{7}$ فإن $a^2 = (\sqrt{7})^2 = 7$

$A = 3 \times 7$

$A = 21$

1) $a = (-\sqrt{3})^5 \times 3^7 = -(\sqrt{3})^5 \times 3^7$

$a = -(3^{1/2})^5 \times 3^7 = -3^{5/2} \times 3^7$

$a = -3^{5/2 + 7} = -3^{5/2 + 14/2} = -3^{19/2}$

$a = -\sqrt{3^{19}}$

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2) $b = 36 \times 15^{-2} = 6^2 \times (3 \times 5)^{-2}$

$b = (2 \times 3)^2 \times 3^{-2} \times 5^{-2}$

$b = 2^2 \times 3^2 \times 3^{-2} \times 5^{-2}$

$b = 2^2 \times 3^{2-2} \times 5^{-2} = 2^2 \times 3^0 \times 5^{-2}$

$b = 4 \times 1 \times \frac{1}{25} = \frac{4}{25} = (\frac{2}{5})^2$

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3) $c = \sqrt{\frac{8^{10} + 4^{10}}{2^{10} + 1}}$

$c = \sqrt{\frac{(2^3)^{10} + (2^2)^{10}}{2^{10} + 1}}$

$c = \sqrt{\frac{2^{30} + 2^{20}}{2^{10} + 1}}$

$c = \sqrt{\frac{2^{20}(2^{10} + 1)}{2^{10} + 1}}$

$c = \sqrt{2^{20}} = 2^{10}$