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\(\frac{2x+y}{2x^2-xy}+\frac{8y}{y^2-4x^2}+\frac{2x-y}{2x^2+xy}\)
\(=\frac{2x+y}{x\left(2x-y\right)}-\frac{8y}{\left(2x-y\right)\left(2x+y\right)}+\frac{2x-y}{x\left(2x+y\right)}\)
\(=\frac{\left(2x+y\right)^2-8xy+\left(2x-y\right)^2}{x\left(2x-y\right)\left(2x+y\right)}=\frac{4x^2+4xy+y^2-8xy+4x^2-4xy+y^2}{x\left(2x-y\right)\left(2x+y\right)}\)
\(=\frac{8x^2-8xy+2y^2}{x\left(2x-y\right)\left(2x+y\right)}=\frac{2\left(4x^2-4xy+y^2\right)}{x\left(2x-y\right)\left(2x+y\right)}\)
\(=\frac{2\left(2x-y\right)^2}{x\left(2x-y\right)\left(2x+y\right)}=\frac{2\left(2x-y\right)}{x\left(2x+y\right)}\)
\(\frac{y}{2x^2-xy}+\frac{4x}{y^2-2xy}=0\)
<=>\(\frac{y}{x\left(2x-y\right)}-\frac{4x}{y\left(2x-y\right)}=0\)
<=>\(\frac{y^2}{xy\left(2x-y\right)}-\frac{4x^2}{xy\left(2x-y\right)}=0\)
=>y2-(2x)2=0
<=>(y-2x)(y+2x)=0
<=>y-2x=0 hoặc y+2x=0
M chỉ làm đc đến đó thôi!!!!!
1,\(\frac{3}{2x+6}-\frac{x-6}{x\left(2x+6\right)}\)
=\(\frac{3x}{x\left(2x+6\right)}+\frac{x-6}{x\left(2x+6\right)}\)
=\(\frac{3x+x-6}{x\left(2x+6\right)}\)=\(\frac{4x-6}{x\left(2x+6\right)}=\frac{2\left(2x-3\right)}{x\left(2x+6\right)}\)
\(\frac{1}{x}-\frac{1}{x+1}=\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{x^2+x}\)
b, \(\frac{1}{xy-x^2}-\frac{1}{y^2-xy}=\frac{y^2-xy-xy+x^2}{\left(xy-x^2\right)\left(y^2-xy\right)}=\frac{x^2+y^2}{xy^3-xyxy-xyxy+x^3y}\)Tu rut gon tiep
c, tt
d, cx r
a) \(\frac{1}{x}-\frac{1}{x+1}=\frac{x+1}{x\left(x+1\right)}-\frac{x}{x\left(x+1\right)}\)
\(=\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{x\left(x+1\right)}\)
b) \(\frac{1}{xy-x^2}-\frac{1}{y^2-xy}=\frac{1}{x\left(y-x\right)}-\frac{1}{y\left(y-x\right)}\)
\(=\frac{y}{xy\left(y-x\right)}-\frac{x}{xy\left(y-x\right)}=\frac{y-x}{xy\left(y-x\right)}=\frac{1}{xy}\)
c) \(\frac{9x-3}{4x-1}-\frac{3x}{1-4x}=\frac{9x-3}{4x-1}+\frac{3x}{4x-1}\)
\(=\frac{9x-3+3x}{4x-1}=\frac{6x-3}{4x-1}\)
1) \(\frac{x+1}{2x-2}+\frac{x^2+3}{2-2x^2}\)
\(=\frac{-4x^2+8x-4}{-4x^3+4x^2+4x-4}\)
\(=\frac{-x^2+2x-1}{-x^3+x^2+x-1}\)
\(=\frac{\left(-x+1\right)\left(x-1\right)}{\left(-x-1\right)\left(x-1\right)\left(x-1\right)}\)
\(=\frac{1}{x+1}\)
2) \(\frac{1-2x}{2x}+\frac{2x}{2x-1}+\frac{1}{2x-4x^2}\)
\(=\frac{-16x^3+16x^2-4x}{-16x^4+16x^3-4x^2}\)
\(=\frac{-16x^2+16x-4}{-16x^3+16x^2-4x}\)
\(=\frac{-4x^2+4x-1}{-4x^3+4x^2-x}\)
\(=\frac{\left(-2x+1\right)\left(2x-1\right)}{x\left(-2x+1\right)\left(2x-1\right)}\)
\(=\frac{1}{x}\)