Question

rút gọn và tính giá trị biểu thức 

A=\(\dfrac{2x}{1-x^3}+\dfrac{1}{x^2-x}+\dfrac{1}{x^2+x+1}\) khi x=10

Answer 1

Ta có: \(A=\dfrac{2x}{1-x^3}+\dfrac{1}{x^2-x}+\dfrac{1}{x^2+x+1}\)

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

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

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

Thay x=10 vào A, ta được:

\(A=\dfrac{1}{10\cdot\left(10^3-1\right)}=\dfrac{1}{10\cdot999}=\dfrac{1}{9990}\)