Giải bài 23 trang 46 – SGK Toán lớp 8 tập 1

Hướng dẫn:

Bước 1: Quy đồng mẫu các phân thức

Bước 2: Thực hiện tính

Bài giải

a) $\dfrac{y}{2x^2 - xy} + \dfrac{4x}{y^2 - 2xy}$

$= \dfrac{y}{x(2x - y)} + \dfrac{-4x}{y(2x - y)}$

$= \dfrac{y^2}{xy(2x - y)} + \dfrac{-4x^2}{xy(2x - y)}$

$= \dfrac{y^2 - 4x^2}{xy(2x - y)}$

$= \dfrac{(y - 2x)(y + 2x)}{xy(2x - y)}$

$= \dfrac{-(2x - y)(y + 2x)}{xy(2x - y)}$

$= \dfrac{-(2x + y)}{xy}$

b) $\dfrac{1}{x + 2} + \dfrac{3}{x^2 - 4} + \dfrac{x - 14}{(x^2 + 4x + 4)(x - 2)}$

$= \dfrac{1}{x + 2} + \dfrac{3}{(x + 2)(x - 2)} + \dfrac{x - 14}{(x + 2)^2(x - 2)}$

$= \dfrac{(x + 2)(x - 2)}{(x + 2)(x + 2)(x - 2)} + \dfrac{3(x + 2)}{(x + 2)(x - 2(x + 2)} + \dfrac{x - 14}{(x + 2)^2(x - 2)}$

$= \dfrac{x^2 - 4}{(x + 2)^2(x - 2)} + \dfrac{3(x + 2)}{(x + 2)^2(x - 2)} + \dfrac{x - 14}{(x + 2)^2(x - 2)}$

$= \dfrac{x^2 - 4 + 3(x + 2) + x - 14}{(x + 2)^2(x - 2)}$

$= \dfrac{x^2 - 4 + 3x + 6 + x - 14}{(x + 2)^2(x - 2)}$

$= \dfrac{x^2 + 4x - 12}{(x + 2)^2(x - 2)}$

$= \dfrac{x^2 + 4x + 4 - 16}{(x + 2)^2(x - 2)}$

$= \dfrac{(x + 2)^2 - 16}{(x + 2)^2(x - 2)}$

$= \dfrac{(x + 2 + 4)(x + 2 - 4)}{(x + 2)^2(x - 2)}$

$= \dfrac{(x + 6)(x - 2)}{(x + 2)^2(x - 2)}$

$= \dfrac{x + 6}{(x + 2)^2}$

c) $\dfrac{1}{x + 2} + \dfrac{1}{(x + 2)(4x + 7)}$

$= \dfrac{4x + 7}{(x + 2)(4x + 7)} + \dfrac{1}{(x + 2)(4x + 7)}$

$= \dfrac{4x + 7 + 1}{(x + 2)(4x + 7)}$

$= \dfrac{4x + 8}{(x + 2)(4x + 7)}$

$= \dfrac{4(x + 2)}{(x + 2)(4x + 7)}$

$= \dfrac{4}{4x + 7}$

d) $\dfrac{1}{x + 3} + \dfrac{1}{(x + 3)(x + 2)} + \dfrac{1}{(x + 2)(4x + 7)}$ 

$= \dfrac{(x + 2)(4x + 7)}{(x + 2)(x + 3)(4x + 7)} + \dfrac{4x + 7}{(x + 3)(x + 2)(4x + 7)} + \dfrac{x + 3}{(x + 2)(x + 3)(4x + 7)}$

$= \dfrac{(x + 2)(4x + 7) + 4x + 7 + x + 3}{(x + 2)(x + 3)(4x + 7)}$

$= \dfrac{4x^2 + 15x + 14 + 4x + 7 + x + 3}{(x + 2)(x + 3)(4x + 7)}$

$= \dfrac{4x^2 + 20x + 24}{(x + 2)(x + 3)(4x + 7)}$

$= \dfrac{4(x^2 + 5x + 6)}{(x + 2)(x + 3)(4x + 7)}$

$= \dfrac{4(x^2 + 2x + 3x + 6)}{(x + 2)(x + 3)(4x + 7)}$

$= \dfrac{4[x(x + 2) + 3(x + 2)]}{(x + 2)(x + 3)(4x + 7)}$

$= \dfrac{4(x + 2)(x + 3)}{(x + 2)(x + 3)(4x + 7)}$

$= \dfrac{4}{4x + 7}$