Cho biểu thức A=\(\dfrac{x^2}{\left(x+y\right)\left(1-y\right)}-\dfrac{y^2}{\left(x+y\right)\left(1+x\right)}-\dfrac{x^2y^2}{\left(1+x\right)\left(1-y\right)}\)
a) Rút gọn A
b) Tính các cặp gia trị nguyên (x.y)để A=-3
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TN
23 tháng 7 2017
\(A=\dfrac{x^2}{\left(x+y\right)\left(1-y\right)}-\dfrac{y^2}{\left(x+y\right)\left(1+x\right)}-\dfrac{x^2y^2}{\left(1+x\right)\left(1-y\right)}\)đkxđ: \(y\ne1;x\ne-1;x\ne-y\)\(=\dfrac{x^2\left(1+x\right)}{\left(x+y\right)\left(1-y\right)\left(1+x\right)}-\dfrac{y^2\left(1-y\right)}{\left(x+y\right)\left(1-y\right)\left(1+x\right)}-\dfrac{x^2y^2\left(x+y\right)}{\left(x+y\right)\left(1-y\right)\left(1+x\right)}\)\(=\dfrac{x^2+x^3-y^2+y^3-x^3y^2-x^2y^3}{\left(x+y\right)\left(1-y\right)\left(1+x\right)}\)
\(=\dfrac{\left(x^3+y^3\right)+\left(x^2-y^2\right)-\left(x^3y^2+x^2y^3\right)}{\left(x+y\right)\left(1-y\right)\left(1+x\right)}\)
\(=\dfrac{\left(x+y\right)\left(x^2-xy+y^2\right)+\left(x-y\right)\left(x+y\right)-x^2y^2\left(x+y\right)}{\left(x+y\right)\left(1-y\right)\left(1+x\right)}\)\(=\dfrac{\left(x+y\right)\left(x^2-xy+y^2+x-y-x^2y^2\right)}{\left(x+y\right)\left(1-y\right)\left(1+x\right)}\)\(=\dfrac{\left(x^2+x\right)-\left(xy+y\right)+\left(y^2-x^2y^2\right)}{\left(1-y\right)\left(x+1\right)}\)
\(=\dfrac{x\left(x+1\right)-y\left(x+1\right)-y^2\left(x-1\right)\left(x+1\right)}{\left(1-y\right)\left(x+1\right)}\) \(=\dfrac{\left(x+1\right)\left(x-y-y^2x+y^2\right)}{\left(1-y\right)\left(x+1\right)}\)
\(=\dfrac{-\left(y-y^2\right)+\left(x-y^2x\right)}{1-y}\)
\(=\dfrac{-y\left(1-y\right)+x\left(1-y\right)\left(1+y\right)}{1-y}\)
\(=\dfrac{\left(1-y\right)\left(x+xy-y\right)}{1-y}=x+xy-y\)
2 tháng 10 2021
a: \(\left(x-2y\right)^2+\left(x-\dfrac{1}{2}y\right)\left(x+\dfrac{1}{2}y\right)\)
\(=x^2-4xy+4y^2+x^2-\dfrac{1}{4}y^2\)
\(=2x^2-4xy+\dfrac{15}{4}y^2\)
b: \(\left(x-2\right)^2+\left(x+3\right)^2-2\left(x-1\right)\left(x+1\right)\)
\(=x^2-4x+4+x^2+6x+9-2\left(x^2-1\right)\)
\(=2x^2+2x+13-2x^2+2\)
=2x+15
2 tháng 10 2021
a) \(=x^2-4xy+4y^2+x^2-\dfrac{1}{4}y^2=2x^2-4xy+\dfrac{15}{4}y^2\)
b) \(=x^2-4x+4+x^2+6x+9-2x^2+2\)
\(=2x+15\)
1 tháng 1 2022
\(A=\dfrac{x^2-y^2+2y^2}{y\left(x-y\right)}\cdot\dfrac{-\left(x-y\right)}{x^2+y^2}+\dfrac{2x^2+2-2x^2+x}{2\left(2x-1\right)}\cdot\dfrac{-\left(2x-1\right)}{x+2}\)
\(=\dfrac{-1}{y}+\dfrac{-1}{2}=\dfrac{-2-y}{2y}\)
20 tháng 11 2023
1: \(C=\left(x-\dfrac{4xy}{x+y}+y\right):\left(\dfrac{x}{x+y}+\dfrac{y}{y-x}+\dfrac{2xy}{x^2-y^2}\right)\)
\(=\dfrac{\left(x+y\right)^2-4xy}{x+y}:\left(\dfrac{x}{x+y}-\dfrac{y}{x-y}+\dfrac{2xy}{\left(x-y\right)\left(x+y\right)}\right)\)
\(=\dfrac{x^2+2xy+y^2-4xy}{x+y}:\dfrac{x\left(x-y\right)-y\left(x+y\right)+2xy}{\left(x+y\right)\left(x-y\right)}\)
\(=\dfrac{x^2-2xy+y^2}{x+y}:\dfrac{x^2-xy-xy-y^2+2xy}{\left(x+y\right)\left(x-y\right)}\)
\(=\dfrac{\left(x-y\right)^2}{x+y}\cdot\dfrac{x^2-y^2}{x^2-y^2}=\dfrac{\left(x-y\right)^2}{x+y}\)
2: \(\left(x^2-y^2\right)\cdot C=-8\)
=>\(\left(x-y\right)\left(x+y\right)\cdot\dfrac{\left(x-y\right)^2}{x+y}=-8\)
=>\(\left(x-y\right)^3=-8\)
=>x-y=-2
=>x=y-2
\(M=x^2\left(x+1\right)-y^2\left(y-1\right)-3xy\left(x-y+1\right)+xy\)
\(=\left(y-2\right)^2\left(y-2+1\right)-y^2\left(y-1\right)-3xy\left(-2+1\right)+xy\)
\(=\left(y-1\right)\left[\left(y-2\right)^2-y^2\right]+3xy+xy\)
\(=\left(y-1\right)\left(-4y+4\right)+4xy\)
\(=-4\left(y-1\right)^2+4y\left(y-2\right)\)
\(=-4y^2+8y-4+4y^2-8y\)
=-4
H
14 tháng 12 2018
\(a,\frac{x}{xy-y^2}+\frac{2x-y}{xy-x^2}:\left(\frac{1}{x}+\frac{1}{y}\right)\)
\(=\left(\frac{x}{y\left(x-y\right)}+\frac{y-2x}{x\left(x-y\right)}\right):\left(\frac{y}{xy}+\frac{x}{xy}\right)\)
\(=\left(\frac{x-y}{x\left(x-y\right)}\right):\left(\frac{x+y}{xy}\right)\)
\(=\frac{1}{x}.\frac{xy}{x+y}=\frac{y}{x+y}\)
\(\dfrac{x^2}{\left(x+y\right)\left(1-y\right)}-\dfrac{y^2}{\left(x+y\right)\left(1+x\right)}-\dfrac{x^2y^2}{\left(1+x\right)\left(1-y\right)}\) MTC : (x+y)(1-y)(1+x)
A=
\(\dfrac{x^2\times\left(1+x\right)}{\left(x+y\right)\left(1-y\right)\left(1+x\right)}-\dfrac{y^2\times\left(1-y\right)}{\left(x+y\right)\left(1-y\right)\left(1+x\right)}-\dfrac{x^2y^2\times\left(x+y\right)}{\left(x+y\right)\left(1-y\right)\left(1+x\right)}\)
A= \(\dfrac{x^2+x^3}{\left(x+y\right)\left(1-y\right)\left(1+x\right)}-\dfrac{y^2}{\left(x+y\right)\left(1-y\right)\left(1+x\right)}-\dfrac{x^3y^2+x^2y^3}{\left(x+y\right)\left(1-y\right)\left(1+x\right)}\)
\(\dfrac{x^2+x^3-y^2-x^3y^2-x^2y^3}{\left(x+y\right)\left(1-y\right)\left(1+x\right)}\)