\(A=\dfrac{x^2-y^2+2y^2}{y\left(x-y\right)}\cdot\dfrac{-\left(x-y\right)}{x^2+y^2}+\dfrac{2x^2+2-2x^2+x}{2\left(2x-1\right)}\cdot\dfrac{-\left(2x-1\right)}{x+2}\)

\(=\dfrac{-1}{y}+\dfrac{-1}{2}=\dfrac{-2-y}{2y}\)

`a,ĐKXĐ:x-4 ne 0,2x+2 ne 0`

`<=>x ne 4,x me -1`

`b,ĐKXĐ:4x^2-25 ne 0`

`<=>(2x-5)(2x+5) ne 0`

`<=>x ne +-5/2`

`c,ĐKXĐ:8x^3+27 ne 0`

`<=>8x^3 ne -27`

`<=>2x ne -3`

`<=>x ne -3/2`

`d,2x+2 ne 0,4y^2-9 ne 0`

`<=>2x ne -2,(2y-3)(2y+3) ne 0`

`<=>x ne -1,y ne +-3/2`

b) ĐKXĐ: \(x\notin\left\{\dfrac{5}{2};-\dfrac{5}{2}\right\}\)

c) ĐKXĐ: \(x\ne-\dfrac{3}{2}\)

d) ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-1\\y\notin\left\{\dfrac{3}{2};-\dfrac{3}{2}\right\}\end{matrix}\right.\)

a: \(M=\left(\dfrac{\left(x-1\right)\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x+3\right)\left(x-1\right)}{2\left(x+1\right)\left(x-1\right)}\right):\dfrac{x-1-x+3}{x-1}\)

\(=\dfrac{x^2-1-x^2-2x+3}{2\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x-1}{2}\)

\(=\dfrac{-2x+2}{2\left(x+1\right)}\cdot\dfrac{1}{2}=\dfrac{-x+1}{2}\)

b: Thay x=-1/2 vào M, ta được:

\(M=\dfrac{\dfrac{1}{2}+1}{2}=\dfrac{3}{2}:2=\dfrac{3}{4}\)

a, \(M=\left(\dfrac{x^2-1-\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\right):\left(\dfrac{x-1-x+3}{x-1}\right)\)

\(=\left(\dfrac{-1+x-3x-3}{2\left(x-1\right)\left(x+1\right)}\right):\dfrac{2}{x-1}=\dfrac{-2x-4}{2\left(x-1\right)\left(x+1\right)}:\dfrac{2}{x-1}=\dfrac{-\left(x+2\right)}{2\left(x+1\right)}\)

b, Thay x  =-1/2 vào ta đc 

\(-\dfrac{\left(\dfrac{-1}{2}+2\right)}{2\left(-\dfrac{1}{2}+1\right)}=\dfrac{-\dfrac{3}{2}}{2\left(\dfrac{1}{2}\right)}=\dfrac{-3}{2}\)

a, \(M=\left(\dfrac{x^2-1-\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\right):\left(\dfrac{x-1-x+3}{x-1}\right)\)

\(=\dfrac{-2x+2}{2\left(x-1\right)\left(x+1\right)}:\dfrac{2}{x-1}=\dfrac{-2\left(x-1\right)^2}{4\left(x-1\right)\left(x+1\right)}=\dfrac{-\left(x-1\right)}{2\left(x+1\right)}\)

b, Thay x = -1/2 vào ta đc 

\(M=\dfrac{\dfrac{1}{2}+1}{2\left(\dfrac{-1}{2}+1\right)}=\dfrac{\dfrac{3}{2}}{\dfrac{2.1}{2}}=\dfrac{3}{2}\)

1.

ĐK: \(x\ne7;x\ne-1;x\ne3\)

\(\dfrac{2x-5}{x^2-6x-7}\le\dfrac{1}{x-3}\left(1\right)\)

TH1: \(x< -1\)

\(\left(1\right)\Leftrightarrow\left(2x-5\right)\left(x-3\right)\ge x^2-6x-7\)

\(\Leftrightarrow2x^2-11x+15\ge x^2-6x-7\)

\(\Leftrightarrow x^2-5x+22\ge0\)

\(\Leftrightarrow\) Bất phương trình đúng với mọi \(x< -1\)

TH2: \(-1< x< 3\)

\(\left(1\right)\Leftrightarrow\left(3-x\right)\left(2x-5\right)\ge\left(7-x\right)\left(x+1\right)\)

\(\Leftrightarrow-2x^2+11x-15\ge-x^2+6x+7\)

\(\Leftrightarrow-x^2+5x-22\ge0\)

\(\Rightarrow\) vô nghiệm

TH3: \(3< x< 7\)

Khi đó \(\dfrac{2x-5}{x^2-6x-7}\le0\); \(\dfrac{1}{x-3}>0\)

\(\Rightarrow\) Bất phương trình đúng với mọi \(3< x< 7\)

TH4: \(x>7\)

\(\left(1\right)\Leftrightarrow\left(2x-5\right)\left(x-3\right)\le x^2-6x-7\)

\(\Leftrightarrow2x^2-11x+15\le x^2-6x-7\)

\(\Leftrightarrow x^2-5x+22\le0\)

\(\Rightarrow\) vô nghiệm

Vậy ...

Các bài kia tương tự, chứ giải ra mệt lắm.

a) \(\dfrac{x}{x-3}-\dfrac{x^2+3x}{2x+3}\left(\dfrac{x+3}{x^2-3x}-\dfrac{x}{x^2-9}\right)\)

ĐKXĐ:\(\left\{{}\begin{matrix}x-3\ne0\\2x +3\ne0\\x^2-3x\ne0\\x^2-9\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne3\\x\ne-\dfrac{3}{2}\\x\ne0\\x\ne\pm3\end{matrix}\right.\)

\(=\dfrac{x}{x-3}-\dfrac{x\left(x+3\right)}{2x+3}\left(\dfrac{x+3}{x\left(x-3\right)}-\dfrac{x}{\left(x-3\right)\left(x+3\right)}\right)\)

\(=\dfrac{x}{x-3}-\dfrac{x\left(x+3\right)}{2x+3}.\dfrac{\left(x+3\right)^2-x^2}{x\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{x}{x-3}-\dfrac{x\left(x+3\right)}{2x+3}.\dfrac{\left(x+3-x\right)\left(x+3+x\right)}{x\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{x}{x-3}-\dfrac{x\left(x+3\right).3\left(2x+3\right)}{\left(2x+3\right)x\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{x}{x-3}-\dfrac{3}{x-3}\)

\(=\dfrac{x-3}{x-3}\)

=1

\(\Rightarrow\) ĐPCM

+ Pt thứ nhất :

Ta có mẫu thức chung là : \(2\left(x-3\right)\left(x+1\right)\)

\(\Rightarrow\left[{}\begin{matrix}x\ne2\\x-3\ne0\\x+1\ne0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x\ne2\\x\ne3\\x\ne-1\end{matrix}\right.\)

Vậy \(ĐKXĐ\) là :\(x\ne2;3;-1\)

+ Pt thứ hai : 

Ta có mẫu thức chung là : \(\left(x-2\right)\left(x+3\right)\)

\(\Rightarrow\left[{}\begin{matrix}x-2\ne0\\x+3\ne0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x\ne2\\x\ne-3\end{matrix}\right.\)

Vậy \(DKXD:\) \(\) \(x\ne2;-3\)

\(a,A=\dfrac{x+1+2-2x+5-x}{\left(1-x\right)\left(x+1\right)}\cdot\dfrac{\left(1-x\right)\left(x+1\right)}{2x-1}\left(x\ne1;x\ne-1;x\ne\dfrac{1}{2}\right)\\ A=\dfrac{8-2x}{2x-1}\\ b,A>0\Leftrightarrow\dfrac{8-2x}{2x-1}>0\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}8-2x>0\\2x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}8-2x< 0\\2x-1< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 4\\x>\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x>4\\x< \dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}< x< 4\\x\in\varnothing\end{matrix}\right.\Leftrightarrow\dfrac{1}{2}< x< 4\)

\(A=\dfrac{x^2+x-2+x^2-x-2-4}{x\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x-3\right)}{2\left(x+2\right)}=\dfrac{2\left(x-2\right)\left(x+2\right)\left(x-3\right)}{2\left(x-2\right)\left(x+2\right)^2}=\dfrac{x-3}{x+2}\\ A\le0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3\ge0\\x+2< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x-3\le0\\x+2>0\end{matrix}\right.\end{matrix}\right.\Rightarrow-2< x< 3;x\ne0\left(ĐKXD\right)\)