Question

B= \(\left[\frac{x}{2x+1}+\frac{4x+1}{4x^2-1}\right]:\frac{x^3+1}{2x-1}\)

a. tìm dk của x dể gia trị phan thưc dược xac dịnh

b. rut gọn phan thưc

Answer 1

ĐKXĐ: \(x\ne\left\{-\frac{1}{2};\frac{1}{2};-1\right\}\)

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

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

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