Bài 1. Dùng định nghĩa hai phân thức bằng nhau chứng tỏ rằng:
a) \( \frac{5y}{7}= \frac{20xy}{28x}\); b) \( \frac{3x(x + 5))}{2(x + 5)}= \frac{3x}{2}\)
c) \( \frac{x + 2}{x - 1}= \frac{(x + 2)(x + 1)}{x^{2} - 1}\); d) \( \frac{x^{2} - x - 2}{x + 1}= \frac{x^{2}- 3x + 2}{x - 1}\)
e) \( \frac{x^{3}+ 8 }{x^{2}- 2x + 4}= x + 2\);
Hướng dẫn giải:
a) \( \left.\begin{matrix} 5y.28x = 140xy\\ 7.20xy = 140xy \end{matrix}\right\}\) \(\Rightarrow 5y.28x = 7.20xy\)
nên \( \frac{5y}{7}= \frac{20xy}{28x}\)
b) \(3x(x + 5).2 = 3x.2(x + 5) = 6x(x + 5)\)
nên \( \frac{3x(x + 5)}{2(x +5)}= \frac{3x}{2}\)
c) \( \frac{x + 2}{x - 1}= \frac{(x + 2)(x + 1)}{x^{2} - 1}\)
Vì \((x + 2)(x^2- 1) = (x + 2)(x + 1)(x - 1)\).
d) \( \frac{x^{2} - x - 2}{x + 1}= \frac{x^{2}- 3x + 2}{x - 1}\)
Vì \((x^2- x - 2)(x - 1) = x^3- 2x^2– x + 2 = (x + 1)(x^2– 3x + 2)\)
e) \( \frac{x^{3}+ 8 }{x^{2}- 2x + 4}= x + 2\)
Vì \(x^3+ 8 = x^3+ 2^3= (x + 2)(x^2– 2x + 4)\)