Bài tập về biến đổi các biểu thức hữu tỉ.
1.
a) $\left ( \frac{2x+1}{2x-1}-\frac{2x-1}{2x+1} \right ):\frac{4x}{10x+5}$
= $\frac{(2x+1)^{2}-(2x-1)^{2}}{(2x-1)(2x+1)}.\frac{5(2x+1)}{4x}$
= $\frac{8x}{(2x-1)(2x+1)}.\frac{5(2x+1)}{4x}$
= $\frac{10}{2x-1}$
b) $\left ( \frac{1}{x^{2}+x}-\frac{2-x}{x+1} \right ):\left ( \frac{1}{x}+x-2 \right )$
= $\left [ \frac{1}{x(x+1)}+\frac{x-2}{x+1} \right ]:\frac{1+x^{2}-2x}{x}$
= $\frac{1+x(x-2)}{x(x+1)}.\frac{x}{1+x^{2}-2x}$
= $\frac{1}{x+1}$
c) $\frac{1}{x-1}-\frac{x^{3}-x}{x^{2}+1}.\left ( \frac{1}{x^{2}-2x+1}+\frac{1}{1-x^{2}} \right )$
= $\frac{1}{x-1}-\frac{x^{3}-x}{x^{2}+1}.\left [\frac{1}{(x-1)^{2}}-\frac{1}{(x-1)(x+1)} \right ]$
= $\frac{1}{x-1}-\frac{x(x-1)(x+1)}{x^{2}+1}.\frac{(x+1)-(x-1)}{(x-1)^{2}(x+1)}$
= $\frac{1}{x-1}-\frac{2x}{(x^{2}+1)(x-1)}$
= $\frac{x^{2}+1}{(x^{2}+1)(x-1)}-\frac{2x}{(x^{2}+1)(x-1)}$
= $\frac{(x-1)^{2}}{(x^{2}+1)(x-1)}$
= $\frac{x-1}{x^{2}+1}$
d) $\left ( \frac{x}{x+1}-\frac{x^{3}-2x^{2}}{x^{3}+1} \right ):\frac{x}{x+1}+\frac{2x+1}{x^{3}+1}$
= $\frac{x(x^{2}-x+1)-x^{3}+2x^{2}}{(x+1)(x^{2}-x+1)}.\frac{x+1}{x}+\frac{2x+1}{(x+1)(x^{2}-x+1)}$
= $\frac{x^{3}-x^{2}+x-x^{3}+2x^{2}}{(x+1)(x^{2}-x+1)}.\frac{x+1}{x}+\frac{2x+1}{(x+1)(x^{2}-x+1)}$
= $\frac{x+1}{x^{2}-x+1}+\frac{2x+1}{(x+1)(x^{2}-x+1)}$
= $\frac{(x+1)^{2}+2x+1}{(x+1)(x^{2}-x+1)}$
= $\frac{x^{2}+4x+2}{x^{3}+1}$
2. $\left ( \frac{x-2}{x+2}+\frac{6x-4}{x^{2}-4} \right ):\frac{x+1}{x-2}$
= $\frac{(x-2)^{2}+6x-4}{(x+2)(x-2)}.\frac{x-2}{x+1}$
= $\frac{x^{2}-4x+4+6x-4}{(x+2)(x-2)}.\frac{x-2}{x+1}$
= $\frac{x^{2}+2x}{(x+2)(x+1)}$
= $\frac{x(x+2)}{(x+2)(x+1)}$
= $\frac{x}{x+1}$