Question

ghi kq thôi cũng được:

Rút gọn

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

Answer 1

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

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

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

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

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

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

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

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

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

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

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

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

Answer 2

Ta có :

\(\left(\dfrac{1-x^3}{1-x}+x\right)\left(\dfrac{1+x^3}{1+x}-x\right)=\left(\dfrac{\left(1-x\right)\left(1+x+x^2\right)}{1-x}+x\right)\left(\dfrac{\left(1+x\right)\left(1-x+x^2\right)}{1+x}-x\right)\)\(=\left(1+x+x^2+x\right)\left(1-x+x^2-x\right)=\left(1+x\right)^2\left(1-x\right)^2\)Thay vào A , ta được:

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