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Question

$\frac{a^2+3ab}{a^2-9b^2}+\frac{2a^2-5ab-3b^2}{6ab-a^2-9b^2}=\frac{a^2+an+ab+bn}{3bn-a^2-an+3ab}$

★★.P.I.N.O.★★ · Answer 1 · 8/1/2015 8:55:18 AM

Ta có:

$\frac{a^2+3ab}{a^2-9b^2}+\frac{2a^2-5ab-3b^2}{6ab-a^2-9b^2}=\frac{a(a+3b)}{(a-3b)(a+3b)}-\frac{(a-3b)(2a+b)}{(a-3b)^2}$

$=\frac{a}{a-3b}-\frac{2a+b}{a-3b} =\frac{-a^2+2ab+3b^2}{a^2-9b^2}=-\frac{(a+b)(a-3b)}{(a-3b)(a+3b) }=-\frac{a+b}{a+3b}$

$\frac{a^2+an+ab+bn}{3bn-a^2-an+3ab}=-\frac{(a+b)(a+n)}{(a+n)(a-3b)}=-\frac{a+b}{a+3b}$

$\Rightarrow \frac{a^2+3ab}{a^2-9b^2}+\frac{2a^2-5ab-3b^2}{6ab-a^2-9b^2}=\frac{a^2+an+ab+bn}{3bn-a^2-an+3ab}.$

Trả lời 01-08-15 08:55 AM

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