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SNata97 @SNata97

August 2022 1 4 Report

Сократив дробь [tex] \frac{3a^{2} +2ab-b^{2} }{9a^{2}+10ab+b^{2} } [/tex] , вычислите ее значение при [tex] \frac{a}{b} = \frac{3}{5} [/tex]

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dtnth

Verified answer

$3a^2+2ab-b^2=3a^2+3ab-ab-b^2=(3a^2+3ab)-(ab+b^2)=\\\\3a(a+b)-b(a+b)=(3a-b)(a+b)$
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$9a^2+10ab+b^2=9a^2+9ab+ab+b^2=\\\\(9a^2+9ab)+(ab+b^2)=\\\\9a(a+b)+b(a+b)=(9a+b)(a+b)$
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$\frac{3a^2+2ab-b^2}{9a^2+10ab+b^2}=\\\\\frac{(3a-b)(a+b)}{(9a+b)(a+b)}=\\\\\frac{3a-b}{9a+b}=\\\\\frac{3\frac{a}{b}-1}{9\frac{a}{b}+1}=\\\\\frac{3*\frac{3}{5}-1}{9*\frac{3}{5}+1}=\\\\\frac{9-5}{27+5}=\\\\\frac{4}{32}=\frac{1}{8}$

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