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
