Toán Lớp 8: a) 1/xy-x^2 – 1/ y^2 -xy b)x/x-2 + 2x+4/ x^2-4 c)(2/x+1 – 1/1-x) : x^2 -3x/ x+1 @thupham2008k

Question

Toán Lớp 8: a) 1/xy-x^2 – 1/ y^2 -xy
b)x/x-2 + 2x+4/ x^2-4
c)(2/x+1 – 1/1-x) : x^2 -3x/ x+1
@thupham2008k, hướng dẫn giải giúp em bài này ạ, em cảm ơn thầy cô và các bạn nhiều.

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Việt Hòa 2 tháng 2022-03-03T03:14:16+00:00 2 Answers 0 views 0

TRẢ LỜI ( 2 )

  1. $a)$

    $\dfrac{1}{xy – x^2}$ – $\dfrac{1}{y^2 – xy}$

    = $\dfrac{1}{x( y – x)}$ – $\dfrac{1}{y ( y – x)}$

    = $\dfrac{1y}{xy( y – x)}$ – $\dfrac{1x}{xy( y – x)}$

    = $\dfrac{y – x}{xy( y – x)}$

    =  $\dfrac{1}{xy}$

    $b)$

    $\dfrac{x}{x-2}$ + $\dfrac{2x + 4}{x^2 – 4}$

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

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

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

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

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

    = $\dfrac{(x + 2)}{(x – 2)}$

    -> Áp dụng hđt số 2 và 3

    $c)$

    ($\dfrac{2}{x + 1}$ – $\dfrac{1}{1 – x}$ ) : $\dfrac{x^2 – 3x}{x + 1}$ 

    = ($\dfrac{2}{x + 1}$ – $\dfrac{1}{1 – x}$ ) : $\dfrac{x^2 – 3x}{x + 1}$

    = ($\dfrac{2( 1 – x)}{(x + 1).(1 – x)}$ – $\dfrac{1( x + 1)}{(1 – x).( x +1)}$ ) : $\dfrac{x^2 – 3x}{x + 1}$

    = $\dfrac{2 – 2x – x – 1}{(x + 1).(1 – x)}$ : $\dfrac{x^2 – 3x}{x + 1}$

    = $\dfrac{ – 3x + 1}{(x + 1).(1 – x)}$. $\dfrac{x + 1}{x( x – 3)}$

    = $\dfrac{ (- 3x + 1). ( x + 1)}{x( x -3).(x + 1).(1 – x)}$

    = $\dfrac{ (- 3x + 1)}{x( x -3).(1 – x)}$

    __________________________________________________

    $#Rosé$

    =)))) mệt quớ

  2. Giải đáp:

    a, 1/(xy – x^2) – 1/(y^2 – xy)

    = 1/[x.(y – x)] – 1/[y.(y – x)]

    = y/[xy.(y – x)] – x/[xy.(y – x)]

    = [y  -x]/[xy.(y – x)]

    = 1/(xy)

    ________________

    b, x/(x – 2) + (2x + 4)/(x^2 – 4)

    =  x/(x – 2) + (2x + 4)/[(x – 2).(x + 2)]

    = [x.(x + 2)]/[(x – 2).(x + 2)] + (2x + 4)/[(x – 2).(x + 2)]

    = [x^2  + 2x]/[(x – 2).(x + 2)] + (2x + 4)/[(x – 2).(x + 2)]

    = [x^2 + 2x + 2x + 4]/[(x – 2).(x + 2)]

    = [x^2 + 4x + 4]/[(x – 2).(x + 2)]

    = [(x + 2)^2]/[(x – 2).(x + 2)]

    = [x + 2]/[x – 2]

    ________________

    c, (2/[x+1] – 1/[1-x]) : (x^2 -3x)/(x+1)

    = ([2.(1 – x)]/[(x + 1).(1 – x)] – [x + 1]/[(x + 1).(1 – x)]) : (x.(x – 3))/(x+1)

    = ([2 – 2x]/[(x + 1).(1 – x)] – [x + 1]/[(x + 1).(1 – x)]) : (x.(x – 3))/(x+1)

    = [2 – 2x – x – 1]/[(x + 1).(1 – x)]) : (x.(x – 3))/(x+1)

    = [-3x +1]/[(x + 1).(1 – x)] . (x + 1)/(x.(x – 3))

    = [(-3x  +1).(x + 1)]/[(x + 1).(1 – x).x.(x – 3)]

    = (-3x  +1)/[(1 – x).x.(x – 3)]

    $#dariana$

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