Giải bài tập 1 trang 49 sgk Toán 8 tập 1 CD

a. $\frac{x}{xy+y^{2}}-\frac{y}{x^{2}+xy}$

= $\frac{x}{y(x+y)}-\frac{y}{x(x+y)}$

= $\frac{x.x}{y(x+y).x}-\frac{y.y}{x(x+y).y}$

= $\frac{x^{2}-y^{2}}{xy(x+y)}$

= $\frac{(x-y)(x+y)}{xy(x+y)}$

= $\frac{x-y}{xy}$

b. $\frac{x^{2}+4}{x^{2}-4}-\frac{x}{x+2}-\frac{x}{2-x}$

= $\frac{x^{2}+4}{(x-2)(x+2)}-\frac{x(x-2)}{(x+2)(x-2)}-\frac{x(x+2)}{(x-2)(x+2)}$

= $\frac{x^{2}+4-x^{2}+2x-x^{2}-2x}{(x-2)(x+2)}$

= $\frac{4-x^{2}}{(x-2)(x+2)}$

= $-\frac{(x-2)(x+2)}{(x-2)(x+2)}=-1$

c. $\frac{a^{2}+ab}{b-a}:\frac{a+b}{2a^{2}-2b^{2}}$

= $\frac{a(a+b)}{b-a}.\frac{2(a-b)(a+b)}{a+b}$

= $\frac{a(a+b).2(a-b)}{b-a}$

= $\frac{a(a+b).(-2)(b-a)}{b-a}$

= $-2a^{2}-2ab$

d. $\left ( \frac{2x+1}{2x-1}-\frac{2x-1}{2x+1} \right ):\frac{4x}{10x-5}$

= $\left ( \frac{(2x+1)(2x+1)}{(2x-1)(2x+1)}-\frac{(2x-1)(2x-1)}{(2x+1)(2x-1)} \right ):\frac{4x}{10x-5}$

= $\frac{(2x+1)^{2}-(2x-1)^{2}}{(2x+1)(2x-1)}:\frac{4x}{10x-5}$

= $\frac{((2x)^{2}+2.2x+1^{2})-((2x)^{2}-2.2x+1^{2})}{(2x+1)(2x-1)}:\frac{4x}{10x-5}$

= $\frac{4x^{2}+4x+1-4x^{2}+4x-1}{(2x+1)(2x-1)}:\frac{4x}{10x-5}$

= $\frac{8x}{(2x+1)(2x-1)}.\frac{5(2x-1)}{4x}$

= $\frac{10}{2x+1}$