Rút gọn biểu thức Bài 38 trang 34 SBT Toán 8 Tập 1

Giải SBT Toán 8 Bài 8: Phép chia các phân thức đại số

Bài 38 trang 34 SBT Toán 8 Tập 1: 

a) $\frac{x^4-x{y}^3}{2xy+y^2}:\frac{x^3\text{\hspace{0.17em}}+\text{\hspace{0.17em}}{x}^{2}y+\text{\hspace{0.17em}}x{y}^2}{2x+y}$;

b) $\frac{5x^2-10xy+\text{\hspace{0.17em}\hspace{0.17em}}5y^2}{2x^2-2xy+2y^2}:\frac{8x-8y}{10x^3\text{\hspace{0.17em}}+10y^3}$.

Lời giải:

a) $\frac{x^4-x{y}^3}{2xy+y^2}:\frac{x^3\text{\hspace{0.17em}}+\text{\hspace{0.17em}}{x}^{2}y+\text{\hspace{0.17em}}x{y}^2}{2x+y}$

$=\frac{x^4-x{y}^3}{2xy+y^2}.\frac{2x+y}{x^3\text{\hspace{0.17em}}+\text{\hspace{0.17em}}{x}^{2}y+\text{\hspace{0.17em}}x{y}^2}$

$=\frac{(x^4-x{y}^3)(2x+y)}{(2xy+y^2)(x^3\text{\hspace{0.17em}}+\text{\hspace{0.17em}}{x}^{2}y+\text{\hspace{0.17em}}x{y}^2)}$

$=\frac{x(x^3-y^3)(2x+y)}{y(2x+y)x(x^2\text{\hspace{0.17em}}+\text{\hspace{0.17em}}xy+\text{\hspace{0.17em}}{y}^2)}$

$=\frac{x(x-y)(x^2+xy+y^2)(2x+y)}{y(2x+y)x(x^2\text{\hspace{0.17em}}+\text{\hspace{0.17em}}xy+\text{\hspace{0.17em}}{y}^2)}=\frac{x-y}{y}$

b) $\frac{5x^2-10xy+\text{\hspace{0.17em}\hspace{0.17em}}5y^2}{2x^2-2xy+2y^2}:\frac{8x-8y}{10x^3\text{\hspace{0.17em}}+10y^3}$

$\begin{array}{l}=\frac{5x^2-10xy+\text{\hspace{0.17em}\hspace{0.17em}}5y^2}{2x^2-2xy+2y^2}.\frac{10x^3\text{\hspace{0.17em}}+10y^3}{8x-8y}\\ =\frac{(5x^2-10xy+\text{\hspace{0.17em}\hspace{0.17em}}5y^2).(10x^3+10y^3)}{(2x^2-2xy+2y^2)(8x-8y)}\end{array}$

$=\frac{5(x^2-2xy+y^2).10(x^3+y^3)}{2(x^2-xy+y^2)8(x-y)}$

$=\frac{50{(x-y)}^2(x+y)(x^2-xy+\text{}{y}^2)}{16(x^2-xy+y^2)(x-y)}$

$=\frac{25(x-y)(x+\text{}y)}{8}$

$=\frac{25}{8}\left(x^2-y^2\right)$