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\(\frac{1}{x}-\frac{1}{x+1}=\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{x^2+x}\)
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) \(\frac{9x-3}{4x-1}-\frac{3x}{1-4x}=\frac{9x-3}{4x-1}+\frac{3x}{4x-1}\)
\(=\frac{9x-3+3x}{4x-1}=\frac{6x-3}{4x-1}\)
a. \(=\frac{x+1}{2.\left(x+3\right)}+\frac{2x+3}{x.\left(x+3\right)}=\frac{x^2+x+4x+6}{2x.\left(x+3\right)}=\frac{x^2+5x+6}{2x.\left(x+3\right)}=\frac{\left(x+2\right).\left(x+3\right)}{2x.\left(x+3\right)}=\frac{x+2}{2x}\)
b. =\(\frac{2.\left(x+3\right)}{x.\left(3x-1\right)}.\frac{-\left(3x-1\right)}{x.\left(x+3\right)}=\frac{-2}{x^2}\)
Chắc chắn đúng, mik nhaaaaaa
a. 2x(x + y) - y(y + 2x) = 2x2 + 2xy - y2 - 2xy = 2x2 - y2
b.\(\frac{4x+3y}{7x^2y}-\frac{3x+3y}{7x^2y}=\frac{4x+3y-3x-3y}{7x^2y}=\frac{x}{7x^2y}=\frac{1}{7xy}\)
Phần c nản quá.
a) 2x(x + y) - y(y + 2x)
= 2x2 + 2xy - y2 - 2xy
= 2x2 - y2
b) \(\frac{4x+3y}{7x^2y}-\frac{3x+3y}{7x^2y}=\frac{4x+3y-3x-3y}{7x^2y}=\frac{x}{7x^2y}=\frac{1}{7xy}\)
c) \(\frac{x^3-4x^2}{x^3-1}+\frac{2}{x^2+x+1}+\frac{1}{x-1}\)
= \(\frac{x^3-4x^2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2\left(x-1\right)}{\left(x^2+x+1\right)\left(x-1\right)}+\frac{x^2+x+1}{\left(x^2+x+1\right)\left(x-1\right)}\)
= \(\frac{x^3-4x^2+2x-2+x^2+x+1}{\left(x^2+x+1\right)\left(x-1\right)}=\frac{x^3-3x^2+3x-1}{\left(x^2+x+1\right)\left(x-1\right)}=\frac{\left(x-1\right)^3}{\left(x^2+x+1\right)\left(x-1\right)}\)
\(=\frac{\left(x-1\right)^2}{x^2+x+1}\)
a,ĐKXĐ: \(\hept{\begin{cases}x\ne0\\x\ne\\x\ne2\end{cases}\pm1}\)
b,\(P=\left(\frac{x+1}{3x\left(x+1\right)}+\frac{1-2x}{3x\left(2x-1\right)}-1\right).\frac{2x}{1-x}\)
\(=\left(\frac{1}{3x}+\frac{-1}{3x}-1\right).\frac{2x}{1-x}\)
\(=\frac{2x}{x-1}\)
Khi P\(\le\)1=> \(\frac{2x}{x-1}\le1\)
=> 2x\(\le\)x-1
=> \(x\le-1\)(với x#0,X#-1)
ĐKXĐ : \(\left\{{}\begin{matrix}3x^2-x\ne0\\1-3x\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}3x\ne1\\1\ne3x\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne\frac{1}{3}\\x\ne\frac{1}{3}\end{matrix}\right.\)
=> \(x\ne\frac{1}{3}.\)
Ta có : \(\frac{2016}{3x^2-x}:\frac{x^2+3x}{1-3x}\)
= \(\frac{2016}{3x^2-x}.\frac{1-3x}{x^2+3x}\)
= \(\frac{2016}{x\left(3x-1\right)}.\frac{1-3x}{x\left(x+3\right)}\)
= \(\frac{2016}{x\left(3x-1\right)}.\frac{3x-1}{-x\left(x+3\right)}\)
= \(\frac{2016\left(3x-1\right)}{x\left(3x-1\right)\left(-x\left(x+3\right)\right)}\)
= \(\frac{2016}{x\left(-x\left(x+3\right)\right)}\)
= \(\frac{2016}{x\left(-x^2-3x\right)}\)
= \(\frac{2016}{-x^3-3x^2}\)