Question

1. Tính:

\(F=\frac{\frac{x^3-x}{x+1}+\frac{2x-2}{1+\frac{x}{2}}}{\frac{x^3-3x^2}{x-3}-\frac{2x^2+8}{x+2}}\)

2.

\(G=\frac{\frac{x^4+1}{x^3-1}-x}{\frac{x}{x^2+x+1}-\frac{2}{x-1}}\)

    Tìm giá trị của G. Khi x=2017

Answer 1

Câu 1:
\(F=\frac{\frac{x^3-x}{x+1}+\frac{2x-2}{1+\frac{x}{2}}}{\frac{x^3-3x^2}{x-3}-\frac{2x^2+8}{x+2}}\left(ĐKXĐ:x\ne3;-2;-1\right)\)

\(F=\frac{\frac{x\left(x-1\right)\left(x+1\right)}{x+1}+\frac{2x-2}{1+\frac{x}{2}}}{\frac{x^2\left(x-3\right)}{x-3}-\frac{2x^2+8}{x+2}}\)

\(F=\frac{\frac{\left(x^2-x\right)\left(1+\frac{x}{2}\right)+2x-2}{1+\frac{x}{2}}}{\frac{x^2\left(x+2\right)-2x^2-8}{x+2}}\)

\(F=\frac{\frac{x^2+\frac{x^3}{2}-x-\frac{x^2}{2}+2x-2}{1+\frac{x}{2}}}{\frac{x^3-8}{x+2}}\)

\(F=\frac{\frac{x^2}{2}+\frac{x^3}{2}+x-2}{1+\frac{x}{2}}.\frac{x+2}{x^3-8}\)

Answer 2

Câu 2:

\(G=\frac{\frac{x^4+1}{x^3-1}-x}{\frac{x}{x^2+x+1}-\frac{2}{x-1}}\left(ĐKXĐ:x\ne1\right)\)

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

\(G=\frac{x+1}{x^3-1}:\frac{x^2-x-2x^2-2x-2}{x^3+1}\)

\(G=\frac{x+1}{-x^2-3x-2}\)

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

\(G=-\frac{1}{x+2}\)Tại x=2017 ta đc:\(G=-\frac{1}{2+2017}=-\frac{1}{2019}\)