Tìm Q Bài 40 trang 34 SBT Toán 8 Tập 1

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

Bài 40 trang 34 SBT Toán 8 Tập 1:

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

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

Lời giải:

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

Ta có $Q=\frac{x^2-\text{}2xy+y^2}{x^2-\text{\hspace{0.17em}}xy+y^2}:\frac{x-y}{x^3+\text{\hspace{0.17em}\hspace{0.17em}}{y}^3}$

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

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

$=(x-y).(x+\text{\hspace{0.17em}}y)$ = x² – y²

Vậy Q = x² – y².                                                                

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

$\begin{array}{l}Q=\frac{3x^2+3xy}{x^2+\text{\hspace{0.17em}}xy+y^2}:\frac{x+y}{x^3-\text{\hspace{0.17em}\hspace{0.17em}}{y}^3}.\\ =\frac{3x^2+3xy}{x^2+\text{\hspace{0.17em}}xy+y^2}.\frac{x^3-y^3}{x+y}\end{array}$

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

 $=3x(x-y)$

Vậy Q =  3x(x – y).