Bài tập 23 trang 46 SGK Toán 8 Tập 1

Câu a:

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

\( =\dfrac{y}{x(2x-y)}+\dfrac{-4x}{y(2x-y)}\) (Áp dụng quy tắc đổi dấu ở phân thức thứ hai)

\(=\dfrac{y^{2}}{xy(2x-y)}+\dfrac{-4x^{2}}{xy(2x-y)}\) (Quy đồng hai phân thức với \(MTC = xy(2x – y))\)

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

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

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

Câu b:

Xét các mẫu thức:

\(\eqalign{
& {x^2} - 4 = \left( {x - 2} \right)\left( {x + 2} \right) \cr 
& \left( {{x^2} + 4x + 4} \right)\left( {x - 2} \right) \cr&= \left( {{x^2} + 2.x.2 + {2^2}} \right)\left( {x - 2} \right) \cr&= {\left( {x + 2} \right)^2}\left( {x - 2} \right) \cr} \)

MTC \(={\left( {x + 2} \right)^2}\left( {x - 2} \right)\)

Ta có: 

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

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

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

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

\( =\dfrac{x^{2}-4+3x+6+x-14}{(x+2)^{2}(x-2)}= \dfrac{x^{2}+4x-12}{(x+2)^{2}(x-2)}\)

\( =\dfrac{x^{2}-2x+6x-12}{(x+2)^{2}(x-2)}= \dfrac{x(x-2)+6(x-2)}{(x+2)^{2}(x-2)}\)

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

Câu c:

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

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

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

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

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

Câu d:

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

\( =\left(\dfrac{1}{x+3}+\dfrac{1}{(x+3)(x+2)}\right)+\dfrac{1}{(x+2)(4x+7)}\)

\( =\left( \dfrac{x+2}{(x+3)(x+2)}+\dfrac{1}{(x+3)(x+2)}\right)\)\(+\dfrac{1}{(x+2)(4x+7)}\)

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

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

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

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

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

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

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

-- Mod Toán 8 HỌC247