Câu 1:

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

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

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

Câu 2: thay x vào A có :

\(A=\frac{-\frac{1}{2}}{\frac{1}{4}-1}=\frac{2}{3}\)

Câu c :

2A=1 => \(\frac{x}{x^4-1}=\frac{1}{2}\)ĐK \(\hept{\begin{cases}x\ne1\\x\ne-1\end{cases}}\)

\(\Leftrightarrow x^4-2x-1=0\Leftrightarrow\left(x+1\right)\left(x^3-x^2+x-1\right)=0\)

\(\left(x+1\right)\left(x^2+1\right)\left(x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)loại do điều kiện  vậy ko có giá trị nào của x thỏa mãn

\(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{8}{x^2-1}\right):\left(\frac{1}{x-1}-\frac{7x+3}{1-x^2}\right)\)

\(A=\left[\frac{x^2+2x+1}{\left(x-1\right)\left(x+1\right)}-\frac{x^2-2x+1}{\left(x+1\right)\left(x-1\right)}+\frac{8}{\left(x+1\right)\left(x-1\right)}\right]:\left[\frac{x+1}{\left(x+1\right)\left(x-1\right)}-\frac{3-7x}{\left(x+1\right)\left(x-1\right)}\right]\)

\(A=\left[\frac{x^2+2x+1-x^2+2x-1+8}{\left(x+1\right)\left(x-1\right)}\right]:\frac{x+1-3+7x}{\left(x+1\right)\left(x-1\right)}\)

\(A=\frac{4x+8}{\left(x+1\right)\left(x-1\right)}.\frac{\left(x+1\right)\left(x-1\right)}{8x-2}\)

...................... 

tìm giá trị x nguyên để A nguyên đi

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)\(P=\left[\frac{2}{\left(x+1\right)^3}.\left(\frac{1}{x}+1\right)+\frac{1}{x^2+2x+1}.\left(\frac{1}{x^2}+1\right)\right]:\frac{x-1}{x^3}\left(ĐKXĐ:x\ne0;-1\right)\)

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

\(P=\left[\frac{2}{\left(x+1\right)^2x}+\frac{x^2+1}{\left[x\left(x+1\right)\right]^2}\right]:\frac{x-1}{x^3}\)

\(P=\left[\frac{x^2+2x+1}{\left[x\left(x+1\right)\right]^2}\right]:\frac{x-1}{3}\)

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

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

b)Bài này liên quan đến dấu lớn nên mk ko làm đc

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