a) \(A=\left[\frac{\left(x+1\right)^2-\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}\right]:\left[\frac{1}{x+1}+\frac{x}{x-1}+\frac{2}{\left(x-1\right)\left(x+1\right)}\right]\)

\(A=\left[\frac{\left(x+1-x+1\right)\left(x-1+x-1\right)}{\left(x-1\right)\left(x+1\right)}\right]:\left[\frac{x-1}{\left(x-1\right)\left(x+1\right)}+\frac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{2}{\left(x-1\right)\left(x+1\right)}\right]\)

\(A=\left[\frac{4x}{\left(x-1\right)\left(x+1\right)}\right]:\left[\frac{x-1+x^2+x+2}{\left(x-1\right)\left(x+1\right)}\right]\)

\(A=\left[\frac{4x}{\left(x-1\right)\left(x+1\right)}\right]:\left[\frac{x^2+2x+1}{\left(x-1\right)\left(x+1\right)}\right]\)

\(A=\left[\frac{4x}{\left(x-1\right)\left(x+1\right)}\right]:\left[\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}\right]\)

\(A=\left[\frac{4x}{\left(x-1\right)\left(x+1\right)}\right]:\left(\frac{x+1}{x-1}\right)\)

\(A=\frac{4x}{\left(x-1\right)\left(x+1\right)}\cdot\frac{x-1}{x+1}\)

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

\(A=\frac{4x}{2\left(x+1\right)}\)

\(A=\frac{2x}{x+1}\)

b) Thay A = -3 vào biểu thức a ta được:

\(\frac{2x}{x+1}=-3\)

\(\Rightarrow\)\(2x=-3\left(x+1\right)\)

\(\Rightarrow\)\(2x=-3x-3\)

\(\Rightarrow\)\(2x+3x=-3\)

\(\Rightarrow\)\(5x=-3\)

\(\Rightarrow\)\(x=-\frac{3}{5}\)

Vậy khi \(x=-\frac{3}{5}\)thì biểu thức A có giá trị là -3

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

ĐK: \(\hept{\begin{cases}x\ne0\\x\ne1\end{cases}}\)

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

    = \(\frac{\left(x-1\right)^3-1+2x^2-4x+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}.\frac{x^2+1}{x+1}\)

    = \(\frac{x^3-3x^2+3x-1+3x^2-3x}{\left(x-1\right)\left(x^2+x+1\right)}.\frac{x^2+1}{x+1}\)

    = \(\frac{x^3-1}{\left(x-1\right)\left(x^2+x+1\right)}.\frac{x^2+1}{x+1}=1.\frac{x^2+1}{x+1}=\frac{x^2+1}{x+1}\)

b) Để A > - 1 <=> \(\frac{x^2+1}{x+1}>-1\)

                       <=> \(\frac{x^2+1}{x+1}+1>0\)

                        <=> \(\frac{x^2+x+2}{x+1}>0\)

Vì x² + x + 2 >0 \(\forall x\)

=> A > 0 <=> x + 1 > 0 <=> x > -1

a.

\(ĐKXĐ:x\ne\pm1;\)

Ta có:

\(P=\left(\frac{x^4+x^2-4x+1}{x^2-1}-\frac{x-1}{x+1}+\frac{x+1}{x-1}\right)\cdot\frac{x\left(x+1\right)-\left(1+x\right)}{x^3-1}\)

\(\Rightarrow P=\left(\frac{x^4+x^2-4x+1}{x^2-1}-\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}+\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}\right)\cdot\frac{\left(x+1\right)\left(x-1\right)}{x^3-1}\)

\(\Rightarrow P=\left(\frac{x^4+x^2-4x+1}{x^2-1}-\frac{x^2-2x+1}{x^2-1}+\frac{x^2+2x+1}{x^2-1}\right)\cdot\frac{x^2-1}{x^3-1}\)

\(\Rightarrow P=\frac{x^4+x^2+1}{x^2-1}\cdot\frac{x^2-1}{x^3-1}\)

\(\Rightarrow P=\frac{x^4+x^2+1}{x^3-1}\)

b.

Để P là số nguyên thì  \(x^4+x^2+1⋮x^3-1\)

\(\Rightarrow\left(x^4-x\right)+\left(x^2+x+1\right)⋮\left(x-1\right)\left(x^2+x+1\right)\)

\(\Rightarrow x\left(x^3-1\right)+\left(x^2+x+1\right)⋮\left(x-1\right)\left(x^2+x+1\right)\)

\(\Rightarrow x\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)⋮\left(x-1\right)\left(x^2+x+1\right)\)

\(\Rightarrow\left(x^2+x+1\right)\left(x^2-x+1\right)⋮\left(x-1\right)\left(x^2+x+1\right)\)

\(\Rightarrow x^2-x+1⋮x-1\)

\(\Rightarrow x\left(x-1\right)+1⋮x-1\)

\(\Rightarrow1⋮x-1\)

\(\Rightarrow x-1\in\left\{1;-1\right\}\)

\(\Rightarrow x=1\left(KTMĐK\right);x=0\)

Vậy x=0.

P/S:Không chắc chắn lắm đâu nha mn,nếu có j sai thì ib vs em ah.

a) \(B=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1-\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\left(x\ge0,x\ne1\right)\)

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

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

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

a) Ta có: \(B=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1-\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)

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

\(=\dfrac{1}{x+\sqrt{x}+1}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)

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