Rút gọn biểu thức Bài 34 trang 33 SBT Toán 8 Tập 1

Giải SBT Toán 8 Bài 7. Phép nhân các phân thức đại số

Bài 34 trang 33 SBT Toán 8 Tập 1: Rút gọn biểu thức:

a) $\frac{x^4+15x\text{\hspace{0.17em}}+7}{2x^3+\text{\hspace{0.17em}}2}.\frac{x}{14x^2+1}.\frac{4x^3+4}{x^4+15x+\text{\hspace{0.17em}}7}$

b) $\frac{x^7+3x^2+\text{\hspace{0.17em}\hspace{0.17em}}2}{x^3-1}.\frac{3x}{x+1}.\frac{x^2+\text{\hspace{0.17em}\hspace{0.17em}}x+1}{x^7+\text{\hspace{0.17em}}3x^2\text{\hspace{0.17em}}+2}$.

Lời giải:

a) $\frac{x^4+15x\text{\hspace{0.17em}}+7}{2x^3+\text{\hspace{0.17em}}2}.\frac{x}{14x^2+1}.\frac{4x^3+4}{x^4+15x+\text{\hspace{0.17em}}7}$

$\begin{array}{l}=\frac{(x^4+15x\text{\hspace{0.17em}}+7).x(4x^3+4)}{(2x^3+\text{\hspace{0.17em}}2).(14x^2+\text{\hspace{0.17em}\hspace{0.17em}}1).(x^4+15x+\text{\hspace{0.17em}}7)}\\ =\frac{x.4(x^3+1)}{2(x^3+1).(14x^2+\text{\hspace{0.17em}\hspace{0.17em}}1)}=\frac{2x}{14x^2+\text{\hspace{0.17em}\hspace{0.17em}}1}\end{array}$

b) $\frac{x^7+3x^2+\text{\hspace{0.17em}\hspace{0.17em}}2}{x^3-1}.\frac{3x}{x+1}.\frac{x^2+\text{\hspace{0.17em}\hspace{0.17em}}x+1}{x^7+\text{\hspace{0.17em}}3x^2\text{\hspace{0.17em}}+2}$

$=\frac{(x^7+3x^2+\text{\hspace{0.17em}\hspace{0.17em}}2).3x.(x^2+\text{\hspace{0.17em}\hspace{0.17em}}x+1)}{(x^3-1).(x+\text{}1).(x^7+\text{\hspace{0.17em}}3x^2\text{\hspace{0.17em}}+2)}$

$\begin{array}{l}=\frac{3x.(x^2+\text{\hspace{0.17em}\hspace{0.17em}}x+1)}{(x-1).(x^2+\text{\hspace{0.17em}\hspace{0.17em}}x+1)(x+\text{\hspace{0.17em}}1)\text{\hspace{0.17em}}}\\ =\frac{3x}{(x-1).(x+\text{\hspace{0.17em}\hspace{0.17em}}1)}\end{array}$