`đk:x ne 0,-2`

`a)D=(x/(x+2)+(8x+8)/(x^2+2x)-(x+2)/x):((x^2-x-3)/(x^2+2x)+1/x)`

`=((x^2+8x+8-x^2-4x-4)/(x(x+2))):((x^2-x-3+x+2)/(x(x+2)))`

`=(4x+4)/(x(x+2)):(x^2-1)/(x(x+2))`

`=(4x+4)/(x^2-1)(x ne +-1)`

`=4/(x-1)`

`b)x(x-2)-(x-2)=0`

`<=>(x-2)(x-1)=0`

Vì `x ne 1=>x-1 ne 0`

`=>x-2=0<=>x=2`

`=>D=4/(2-1)=4`

`c)D<0`

Mà `4>0`

`=>x-1<0`

`=>x<1`

Kết hợp đkxđ:

`=>x<1,x ne 0,x ne -2`

`d)D=2`

`<=>4/(x-1)=2`

`<=>2/(x-1)=1`

`<=>x-1=2`

`<=>x=3(tm)`

\(a,\)

\(B=\dfrac{x-1}{x+1}-\dfrac{x+1}{x-1}-\dfrac{4}{1-x^2}\) (Điều kiện xác định: \(x\ne\pm1\))

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

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

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

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

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

\(=-\dfrac{4}{x+1}\)

\(b,\)

\(x^2-x=0\)

\(\Leftrightarrow x\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

Với \(x=0\Rightarrow B=-\dfrac{4}{0+1}=-4\)

Với \(x=1\Rightarrow B=-\dfrac{4}{1+1}=-2\)

\(c,\)

\(B=-3\Rightarrow-\dfrac{4}{x+1}=-3\)

\(\Leftrightarrow-3\left(x+1\right)=-4\)

\(\Leftrightarrow-3x-3+4=0\)

\(\Leftrightarrow-3x+1=0\)

\(\Leftrightarrow-3x=-1\)

\(\Leftrightarrow x=\dfrac{1}{3}\)

\(d,\)

\(B< 0\Rightarrow-\dfrac{4}{x+1}< 0\)

\(\Leftrightarrow x+1>0\)

\(\Leftrightarrow x>-1\)

Kết hợp điều kiện \(x\ne\pm1\) 

\(\Rightarrow-1< x< 1\)

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

\(=\dfrac{x+4x+8+x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x+2\right)}{3\left(x+1\right)}\)

\(=\dfrac{6\left(x+1\right)\cdot x\left(x+2\right)}{3\left(x+1\right)\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{2x}{x-2}\)

cho mình sửa lại đề là cho biểu thức 

Câu 1 :

a) ĐKXĐ : \(\hept{\begin{cases}x+1\ne0\\2x-6\ne0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x\ne-1\\x\ne3\end{cases}}\)

b) Để \(P=1\Leftrightarrow\frac{4x^2+4x}{\left(x+1\right)\left(2x-6\right)}=1\)

\(\Leftrightarrow\frac{4x^2+4x-\left(x+1\right)\left(2x-6\right)}{\left(x+1\right)\left(2x-6\right)}=0\)

\(\Rightarrow4x^2+4x-2x^2+4x+6=0\)

\(\Leftrightarrow2x^2+8x+6=0\)

\(\Leftrightarrow x^2+4x+4-1=0\)

\(\Leftrightarrow\left(x+2-1\right)\left(x+2+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+3=0\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=-1\left(KTMĐKXĐ\right)\\x=-3\left(TMĐKXĐ\right)\end{cases}}\)

Vậy : \(x=-3\) thì P = 1.

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

a) ĐK: \(x\ne1,x\ge0\)

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

\(B=\left[\dfrac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right]\cdot\dfrac{\left(x-1\right)^2}{2}\)

\(B=\left[\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right]\cdot\dfrac{\left(x-1\right)^2}{2}\)

\(B=\left[\dfrac{x+\sqrt{x}-2\sqrt{x}-2-x+\sqrt{x}-2\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right]\cdot\dfrac{\left(x-1\right)^2}{2}\)

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

\(B=-\sqrt{x}\left(\sqrt{x}-1\right)\) 

a) ĐKXĐ: \(x\notin\left\{3;-3;-2\right\}\)

Ta có: \(P=\left(\dfrac{2x-1}{x+3}-\dfrac{x}{3-x}-\dfrac{3-10x}{x^2-9}\right):\dfrac{x+2}{x-3}\)

\(=\left(\dfrac{\left(2x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{3-10x}{\left(x-3\right)\left(x+3\right)}\right):\dfrac{x+2}{x-3}\)

\(=\dfrac{2x^2-6x-x+3+x^2+3x-3+10x}{\left(x-3\right)\left(x+3\right)}:\dfrac{x+2}{x-3}\)

\(=\dfrac{3x^2+6x}{\left(x-3\right)\left(x+3\right)}:\dfrac{x+2}{x-3}\)

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

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

b) Ta có: \(x^2-7x+12=0\)

\(\Leftrightarrow x^2-3x-4x+12=0\)

\(\Leftrightarrow x\left(x-3\right)-4\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(loại\right)\\x=4\left(nhận\right)\end{matrix}\right.\)

Thay x=4 vào biểu thức \(P=\dfrac{3x}{x+3}\), ta được: 

\(P=\dfrac{3\cdot4}{4+3}=\dfrac{12}{7}\)

Vậy: Khi \(x^2-7x+12=0\) thì \(P=\dfrac{12}{7}\)