a) ${\displaystyle \frac{5y}{7}}={\displaystyle \frac{20xy}{28x}}$;

b) ${\displaystyle \frac{3x(x+5)}{2(x+5)}}={\displaystyle \frac{3x}{2}}$;

c) ${\displaystyle \frac{x+2}{x-1}}={\displaystyle \frac{(x+2)(x+1)}{x^2-1}}$;

d) ${\displaystyle \frac{x^2-x-2}{x+1}}={\displaystyle \frac{x^2-3x+2}{x-1}}$;

e) ${\displaystyle \frac{x^3+8}{x^2-2x+4}}=x+2$;

Phương pháp giải: Áp dụng định nghĩa hai phân thức bằng nhau: ${\displaystyle \frac{A}{B}}={\displaystyle \frac{C}{D}}$ nếu $AD=BC$.

Lời giải:

a) ${\displaystyle \frac{5y}{7}}={\displaystyle \frac{20xy}{28x}}$

$\begin{array}{c}5y.28x=140xy\\ 7.20xy=140xy\end{array}\}$ $\Rightarrow 5y.28x=7.20xy$

nên ${\displaystyle \frac{5y}{7}}={\displaystyle \frac{20xy}{28x}}$

b) ${\displaystyle \frac{3x(x+5)}{2(x+5)}}={\displaystyle \frac{3x}{2}}$

Xét tích chéo:

$3x(x+5).2=6x(x+5)$

$3x.2(x+5)=6x(x+5)$

Suy ra $3x(x+5).2=3x.2(x+5)$

Do đó ${\displaystyle \frac{3x(x+5)}{2(x+5)}}={\displaystyle \frac{3x}{2}}$

c) ${\displaystyle \frac{x+2}{x-1}}={\displaystyle \frac{(x+2)(x+1)}{x^2-1}}$;

Xét tích chéo:

$(x+2)(x^2-1)$$=(x+2)(x+1)(x-1)$.

Nên ${\displaystyle \frac{x+2}{x-1}}={\displaystyle \frac{(x+2)(x+1)}{x^2-1}}$

d) ${\displaystyle \frac{x^2-x-2}{x+1}}={\displaystyle \frac{x^2-3x+2}{x-1}}$

$\begin{array}{l}\left(x^2-x-2\right)\left(x-1\right)\\ =\left(x^2-2x+x-2\right)\left(x-1\right)\\ =\left[x\left(x-2\right)+\left(x-2\right)\right]\left(x-1\right)\\ =\left(x-2\right)\left(x+1\right)\left(x-1\right)\\ \left(x+1\right)\left(x^2-3x+2\right)\\ =\left(x+1\right)\left(x^2-2x-x+2\right)\\ =\left(x+1\right)\left[x\left(x-2\right)-\left(x-2\right)\right]\\ =\left(x+1\right)\left(x-2\right)\left(x-1\right)\end{array}$

$\Rightarrow \left(x^2-x-2\right)\left(x-1\right)=$$\left(x+1\right)\left(x^2-3x+2\right)$

Vậy  ${\displaystyle \frac{x^2-x-2}{x+1}}={\displaystyle \frac{x^2-3x+2}{x-1}}$

Cách khác:

$\left(x^2-x-2\right)\left(x-1\right)$$=x^2.x+x^2.(-1)+(-x).x$$+(-x).(-1)+(-2).x+(-2).(-1)$$=x^3-x^2-x^2+x-2x+2$$=x^3-2x^2-x+2$

$\left(x+1\right)\left(x^2-3x+2\right)$$=x.x^2+x.\left(-3x\right)+x.2+1.x^2$$+1.\left(-3x\right)+1.2$$=x^3-3x^2+2x+x^2-3x+2$$=x^3-2x^2-x+2$

$\Rightarrow \left(x^2-x-2\right)\left(x-1\right)=$$\left(x+1\right)\left(x^2-3x+2\right)$

Vậy  ${\displaystyle \frac{x^2-x-2}{x+1}}={\displaystyle \frac{x^2-3x+2}{x-1}}$

e) ${\displaystyle \frac{x^3+8}{x^2-2x+4}}=x+2$;

Ta có: ${\displaystyle \frac{x^3+8}{x^2-2x+4}}=x+2$

Suy ra ${\displaystyle \frac{x^3+8}{x^2-2x+4}}={\displaystyle \frac{x+2}{1}}$

Xét tích chéo:

$(x^3+8).1=x^3+2^3$$=(x+2)(x^2–2x+4)$

Do đó: ${\displaystyle \frac{x^3+8}{x^2-2x+4}}=x+2$

Xem thêm lời giải bài tập Toán lớp 8 hay, chi tiết khác: