Giải bài 1 trang 47

Đề bài

Thực hiện phép tính:

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

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

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

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

Lời giải chi tiết

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

$b)\frac{x^2-36}{2x+10}.\frac{x+5}{6-x}=\frac{\left(x-6\right)\left(x+6\right)\left(x+5\right)}{2\left(x+5\right).\left(-1\right)\left(x-6\right)}=\frac{x+6}{-2}=\frac{-x-6}{2}$

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

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