Giải bài tập 1 trang 47 sgk Toán 8 tập 1 CD

a. $\frac{3x+6}{4x-8}.\frac{2x-4}{x+2}$ = $\frac{3(x+2)}{2(2x-4)}.\frac{2x-4}{x+2}=\frac{3}{2}$

b. $\frac{x^{2}-36}{2x+10}.\frac{x+5}{6-x}$ = $\frac{-(36-x^{2})}{2x+10}.\frac{x+5}{6-x}=\frac{-(6-x)(6+x)}{2(x+5)}.\frac{x+5}{6-x}=\frac{-(6+x)}{2}$

c. $\frac{1-y^{3}}{y+1}.\frac{5y+5}{y^{2}+y+1}$ = $\frac{(1-y)(1+y+y^{2}).5(y+1)}{(y+1).(y^{2}+y+1)}=(1-y).5=5-5y$

d. $\frac{x+2y}{4x^{2}-4xy+y^{2}}.(2x-y)$ = $\frac{(x+2y)(2x-y)}{(2x-y)^{2}}=\frac{x+2y}{2x-y}$