Question

Сократив дробь $frac{3a^{2} +2ab-b^{2} }{9a^{2}+10ab+b^{2} }$ , вычислите ее значение при $frac{a}{b} = frac{3}{5}$

Answer 1

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