Giải bài 1 trang 36 – SGK Toán lớp 8 tập 1

Hướng dẫn:

Áp dụng:  $\dfrac{A}{B} = \dfrac{C}{D}$ nếu $A.D = B.C.$

a) Ta có:
$5y.28x = 140xy$
$7.20xy = 140xy$
$\Rightarrow 5y.28x = 7.20xy$
Vậy $\dfrac{5y}{7} = \dfrac{20xy}{28x}$
b) Ta có:
$2(x + 5).3x = 6x^2 + 30x$
$3x.2(x + 5) = 6x^2 + 30x$
$\Rightarrow 2(x + 5).3x = 3x.2(x + 5)$
Vậy $\dfrac{3x(x + 5)}{2(x + 5)} = \dfrac{3x}{2}$
c) Ta có:
$(x - 1)(x + 2)(x + 1) = (x^2 - 1)(x + 2)$
Vậy $\dfrac{x + 2}{x - 1} = \dfrac{(x + 2)(x + 1)}{x^2 - 1}$
d) Ta có:
$(x^2 - x - 2)(x - 1) = x^3 - x^2 - x^2 + x - 2x + 2 = x^3 - 2x^2 - x + 2$
$(x + 1)(x^2 - 3x + 2) = x^3 - 3x^2 + 2x + x^2 - 3x + 2 = x^3 - 2x^2 - x + 2$
$\Rightarrow (x^2 - x - 2)(x - 1) = (x + 1)(x^2 - 3x + 2)$
 Vậy $\dfrac{x^2 - x - 2}{x + 1} = \dfrac{x^2 - 3x + 2}{x - 1}$
 e) Ta có:
$x^3 + 8 = x^3 + 2^3 = (x + 2)(x^2 - 2x + 4)$
 Vậy $\dfrac{x^3 + 8}{x^2 - 2x + 4} = x + 2$