Giải bài 1 trang 36 – SGK Toán lớp 8 tập 1
Dùng định nghĩa hai phân thức bằng nhau chứng tỏ rằng:
a) \(\dfrac{5y}{7} = \dfrac{20xy}{28x};\)
b) \(\dfrac{3x(x + 5)}{2(x + 5)} = \dfrac{3x}{2};\)
c) \(\dfrac{x + 2}{x - 1} = \dfrac{(x + 2)(x + 1)}{x^2 - 1};\)
d) \(\dfrac{x^2 - x - 2}{x + 1} = \dfrac{x^2 - 3x + 2}{x - 1};\)
e) \(\dfrac{x^3 + 8}{x^2 - 2x + 4} = x + 2\)
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 \(\dfrac{x^3 + 8}{x^2 - 2x + 4} = x + 2\)