toán 8

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pinkylun

a) ĐKXĐ x khác -y và x khác -1 y khác 1 :D


$A=\dfrac{x^2}{(x+y)(1-y)}-\dfrac{y^2}{(x+y)(1+x)}-\dfrac{x^2y^2}{(1+x)(1-y)}$

$A=\dfrac{x^2(x+1)-y^2(1-y)-x^2y^2(x+y)}{(x+y)(1+x)(1-y)}$

$A=\dfrac{x^3+x^2-y^2+y^3-x^3y^2-x^2y^3}{(x+y)(1+x)(1-y)}$

$A=\dfrac{x^3(1-y)(1+y)-y^2(1-y)+x^2(1-y)(1+xy+y^2)}{(x+y)(1+x)(1-y)}$

$A=\dfrac{x^3+x^3y-y^2+x^2+x^3y+x^2y^2}{(x+y)(1+x)}$

???
 
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T

thaotran19

Mik lấy 3 bước đầu của pinkylun nhé!! :D

a) ĐKXĐ: $x \not= -y ; x \not=-1$ và $y\not= 1 $

$A=\dfrac{x^2}{(x+y)(1-y)}-\dfrac{y^2}{(x+y)(1+x)}-\dfrac{x^2y^2}{(1+x)(1-y)}$

$A=\dfrac{x^2(x+1)-y^2(1-y)-x^2y^2(x+y)}{(x+y)(1+x)(1-y)}$

$A=\dfrac{x^3+x^2-y^2+y^3-x^3y^2-x^2y^3}{(x+y)(1+x)(1-y)}$

$A=\dfrac{(x^3+y^3)+(x^2-y^2)-(x^3y^2+x^2y^3)}{(x+y)(1+x)(1-y)}$

$A=\dfrac{x^3(x^2-xy+y^2)+(x-y)(x+y)-x^2y^2(x+y)}{(x+y)(1+x)(1-y)}$

$A=\dfrac{(x+y)(x^2-xy+y^2+x-y-x^2y^2)}{(x+y)(1+x)(1-y)}$

$A=\dfrac{(x+y)[(x^2+x)-(xy+y)+(y^2-x^2y^2)}{(x+y)(1+x)(1-y)}$

$A=\dfrac{(x+y)[x(x+1)-y(x+1)+y^2(1-x)(1+x)]}{(x+y)(1+x)(1-y)}$

$A=\dfrac{(x+y)(1+x)(x-y+y^2-xy^2}{(x+y)(1+x)(1-y)}$

$A=\dfrac{(x+y)(1+x)[x(1-y)(1+y)-y(1-y)}{(x+y)(1+x)(1-y)}$

$A=\dfrac{(x+y)(1+x)(1-y)(x+xy-y)}{(x+y)(1+x)(1-y)}$

$A=x+xy-y$
 
P

phamhuy20011801

$A=-7 \leftrightarrow x+xy-y=-7 \leftrightarrow x(y+1)-(y+1)=-8 \leftrightarrow (x-1)(y+1)=-8$
Xét các ước của $-8$...
 
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transformers123

Nốt câu b:

$A=-7$

$\Longrightarrow x+xy-y=-7$

$\iff x(1+y)-(y+1)=-8$

$\iff (1+y)(x-1)=-8$

$\iff \left[\begin{matrix}\begin{cases}1+y=-1\\x-1=8\end{cases}\\ \begin{cases}1+y=-2\\x-1=4\end{cases}\\ \begin{cases}1+y=-4\\x-1=2\end{cases}\\ \begin{cases}1+y=-8\\x-1=1\end{cases}\\\begin{cases}1+y=1\\x-1=-8\end{cases}\\ \begin{cases}1+y=2\\x-1=-4\end{cases}\\ \begin{cases}1+y=4\\x-1=-2\end{cases}\\ \begin{cases}1+y=8\\x-1=-1\end{cases}\end{matrix}\right. \iff ......................$
 
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