c) \(2x-3+2xy=1\)
d)\(3-2x-xy=2y\)
g)\(\frac{2}{x}+\frac{3}{y}=\frac{5}{6}\)
h)\(x^2-2xy+y^2=4xy-3\)
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Làmmmm
1/ \(\frac{1-2x}{2x}+\frac{2x}{2x-1}+\frac{1}{2x-4x^2}\)(ĐKXĐ:x\(\ne0\), x\(\ne\frac{1}{2}\))
= \(\frac{\left(1-2x\right)\left(2x-1\right)}{2x\left(2x-1\right)}+\frac{4x^2}{\left(2x-1\right)2x}-\frac{1}{2x\left(2x-1\right)}\)
\(=\frac{2x-1-4x^2+2x+4x^2-1}{2x\left(2x-1\right)}\)
\(=\frac{4x-2}{2x\left(2x-1\right)}=\frac{2\left(2x-1\right)}{2x\left(2x-1\right)}=\frac{1}{x}\)
KL:..............
2/\(\frac{x^2+2}{x^3-1}+\frac{2}{x^2+x+1}+\frac{1}{1-x}\)(ĐKXĐ : x\(\ne1\))
\(=\frac{x^2+2}{x^3-1}+\frac{2x-2}{x^3-1}-\frac{x^2+x+1}{x^3-1}\)
\(=\frac{x^2+2+2x-2-x^2-x-1}{x^3-1}=\frac{x-1}{x^3-1}=\frac{1}{x^2+x+1}\)
Kl:....................
Ta có: \(\frac{x^2y+2xy^2+y^3}{2x^2+xy-y^2}\)
\(=\frac{x^2y+xy^2+xy^2+y^3}{2x^2+2xy-xy-y^2}\)
\(=\frac{xy\left(x+y\right)+y^2\left(x+y\right)}{2x\left(x+y\right)-y\left(x+y\right)}\)
\(=\frac{\left(x+y\right)\left(xy+y^2\right)}{\left(2x-y\right)\left(x+y\right)}=\frac{xy+y^2}{2x-y}\left(đpcm\right)\)
Ta có: \(\frac{x^2+3xy+2y^2}{x^3+2x^2y-xy^2-2y^3}\)
\(=\frac{x^2+xy+2xy+2y^2}{x^2\left(x+2y\right)-y^2\left(x+2y\right)}\)
\(=\frac{x\left(x+y\right)+2y\left(x+y\right)}{\left(x^2-y^2\right)\left(x+2y\right)}\)
\(=\frac{\left(x+2y\right)\left(x+y\right)}{\left(x+y\right)\left(x-y\right)\left(x+2y\right)}=\frac{1}{x-y}\left(đpcm\right)\)
\(\frac{x^4-xy^3}{2xy+y^2}:\frac{x^3+x^2y+xy^2}{2x+y}\)
\(=\frac{x\left(x^3-y^3\right)}{y\left(2x+y\right)}.\frac{2x+y}{x^3+x^2y+xy^2}\)
\(=\frac{x\left(x-y\right)\left(x^2+xy+y^2\right)\left(2x+y\right)}{xy\left(2x+y\right)\left(x^2+xy+y^2\right)}\)
\(=\frac{x-y}{y}\)
\(\frac{x^4-xy^3}{2xy+y^2}:\frac{x^3+x^2y+xy^2}{2x+y}\)
\(=\frac{x\left(x^3-y^3\right)}{y\left(2x+y\right)}:\frac{x\left(x^2+xy+y^2\right)}{2x+y}\)
\(=\frac{x\left(x-y\right)\left(x^2+xy+y^2\right)}{y\left(2x+y\right)}:\frac{x\left(x^2+xy+y^2\right)}{2x+y}\)
\(=\frac{x\left(x-y\right)\left(x^2+xy+y^2\right)}{y\left(2x+y\right)}.\frac{2x+y}{x\left(x^2+xy+y^2\right)}\)
\(=\frac{x-y}{y}\)
a) Ta có: \(-2xy^2\cdot\left(x^3y-2x^2y^2+5xy^3\right)\)
\(=-2x^4y^3+4x^3y^4-10x^2y^5\)
b) Ta có: \(\left(-2x\right)\cdot\left(x^3-3x^2-x+1\right)\)
\(=-2x^4+6x^3+2x^2-2x\)
c) Ta có: \(3x^2\left(2x^3-x+5\right)\)
\(=6x^5-3x^3+15x^2\)
d) Ta có: \(\left(-10x^3+\frac{2}{5}y-\frac{1}{3}z\right)\cdot\left(-\frac{1}{2}xy\right)\)
\(=5x^4y-\frac{1}{5}xy^2+\frac{1}{6}xyz\)
e) Ta có: \(\left(3x^2y-6xy+9x\right)\cdot\left(-\frac{4}{3}xy\right)\)
\(=-4x^3y^2+8x^2y^2-12x^2y\)
f) Ta có: \(\left(4xy+3y-5x\right)\cdot x^2y\)
\(=4x^3y^2+3x^2y^2-5x^3y\)
2x - 3 + 2xy = 1
=> 2x - 2 - 1 + 2xy = 1
=> 2(x - 1) + 2xy = 0
=> 2(x - 1 + xy) = 0
=> x - 1 + xy = 0
=> x + xy - 1 = 0
=> x(1 + y) = -1
=> x và 1 + y thuộc Ư(-1) = {-1; 1}
ta có bảng :
vậy_