Помогите!1) 2b²+3ax/bx - ab+5bx/ax2) 1/a²b³ - 1/a²b³ 3) a/a²-b² - b/(a-b)². Created by harton228. algebra-ru

(a-b)²...

harton228 @harton228

August 2021 1 13 Report

Помогите!1) 2b²+3ax/bx - ab+5bx/ax

2) 1/a²b³ - 1/a²b³

3) a/a²-b² - b/(a-b)²

Answers & Comments

nafanya2014

Verified answer

1) 2b²+3ax/bx - ab+5bx/ax=
$=\frac{2 b^{2} +3ax}{bx} - \frac{ax+5bx}{ax}= \frac{2ab ^{2} +3a ^{2}x-abx-5b ^{2}x }{abx}$
2) 1/a²b³ - 1/a²b³ =0
но если в условии второй дроби a³b², то решение другое:

$\frac{1}{a ^{2}b ^{3} } - \frac{1}{a ^{3}b ^{2} }= \frac{a-b}{a ^{3}b ^{3} }$

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

4 votes Thanks 4

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Verified answer

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