Question

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

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Rút gọn B

Answer 1

ĐK: x khác 1 ; -1

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

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

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

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