Giải bài 6 trang 17 vở thực hành Toán 8 tập 2

Đề bài

Rút gọn biểu thức $P=\frac{3}{\left(x+1\right)\left(x+4\right)}-\frac{1}{\left(x+2\right)\left(x+3\right)}-\frac{1}{\left(x+2\right)\left(x+1\right)}$.

Lời giải chi tiết

$\begin{array}{c}P=\frac{3}{\left(x+1\right)\left(x+4\right)}-\frac{1}{\left(x+2\right)\left(x+3\right)}-\frac{1}{\left(x+2\right)\left(x+1\right)}\\ =\frac{3}{\left(x+1\right)\left(x+4\right)}-\left(\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+2\right)\left(x+1\right)}\right)\\ =\frac{3}{\left(x+1\right)\left(x+4\right)}-\frac{x+1-(x+3)}{(x+1)(x+3)}\\ =\frac{3}{\left(x+1\right)\left(x+4\right)}-\frac{2}{\left(x+1\right)\left(x+3\right)}\\ =\frac{3\left(x+3\right)}{\left(x+1\right)\left(x+3\right)\left(x+4\right)}-\frac{2\left(x+4\right)}{\left(x+1\right)\left(x+3\right)\left(x+4\right)}\\ =\frac{x+1}{\left(x+1\right)\left(x+3\right)\left(x+4\right)}=\frac{1}{\left(x+3\right)\left(x+4\right)}\end{array}$