toán 8

M

math012

C

congchuaanhsang

1, $(\dfrac{a+b-x}{c}+1)+(\dfrac{b+c-x}{a}+1)+(\dfrac{c+a-x}{b}+1)+(\dfrac{4x}{a+b+c}-4)=0$

\Leftrightarrow $(a+b+c-x)(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{4}{a+b+c})=0$

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R

ronaldover7

A=$\frac{2a-b}{3a-b}$+$\frac{5b-a}{3a+b}$
=$\frac{(2a-b)(3a+b)}{(3a-b)(3a+b)}$+$\frac{(5b-a)(3a-b)}{(3a+b)(3a-b)}$
=$\frac{(2a-b)(3a+b)+(5b-a)(3a-b)}{(3a-b)(3a+b)}$
=$\frac{(6a^2-3ab+2ab-b^2)+(15ab-3a^2-5b^2+ab)}{(3a-b)(3a+b)}$
=$\frac{3a^2-6b^2+15ab}{9a^2-b^2}$
10$a^2$-3$b^2$+5ab=0
\Rightarrow 10$a^2$-3$b^2$=-5ab
\Rightarrow -30$a^2$+9$b^2$=15ab
\Rightarrow A =$\frac{3a^2-6b^2-30a^2+9b^2}{9a^2-b^2}$
\Rightarrow A = $\frac{-9(9a^2-b^2)}{9a^2-b^2}$
\RightarrowA=-9
 
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