Giải bài 23 trang 46 - Sách giáo khoa Toán 8 tập 1

Đề bài

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Giải bài 23 trang 46 Toán 8 Tập 1 | Giải bài tập Toán 8

Hướng dẫn giải

a) Ta có : \(2x^2 - xy = x(2x - y)\)

                \(y^2 - 2 xy = y(y-2x) = -y(2x - y )\)

MTC = \(xy(2x - y )\)

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

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

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

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

b) Ta có : \(x^2 -4 = (x-2)(x+2)\)

               \((x^2 + 4x + 4 )(x-2) = ( x + 2)^2(x-2)\)

MTC \(= ( x + 2)^2(x-2)\)

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

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

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

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

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

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

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

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

d) MTC = \((x +2 ) ( x+ 3 ) (4x+7)\)

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

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

\(= \dfrac{4x^2 + 7x + 8 x +14 + 4x + 7 + x + 3}{(x+3)(x+2)(4x+7)}\)

\(= \dfrac{4x^2 + 20x+ 24}{(x+3)(x+2)(4x+7)} = = \dfrac{4(x^2 + 5x +6 )}{(x+3)(x+2)(4x+7)}\)

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

 

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