Toán Lớp 8: Rút gọn phân thức `A` `=` ($\frac{x}{x^2 + 4}$ `+` $\frac{2}{2 – x }$ `+` $\frac{1}{x – 2}$ ) `.` $\frac{x+2}{2}$

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Toán Lớp 8: Rút gọn phân thức `A` `=` ($\frac{x}{x^2 + 4}$ `+` $\frac{2}{2 – x }$ `+` $\frac{1}{x – 2}$ ) `.` $\frac{x+2}{2}$, hướng dẫn giải giúp em bài này ạ, em cảm ơn thầy cô và các bạn nhiều.

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Dạ Nguyệt 1 ngày 2022-06-20T22:23:38+00:00 2 Answers 0 views 0

TRẢ LỜI ( 2 )

  1. landinhngoc97
    0
    2022-06-20T22:24:40+00:00
    Reply
    Gửi bạn:
    Đề đúng: $A=(\dfrac{x}{x^2-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}).\dfrac{x+2}{2}$
    $(ĐKXĐ:x\neq±2)$
    $A=(\dfrac{x}{x^2-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}).\dfrac{x+2}{2}$
    $A=(\dfrac{x}{(x-2)(x+2)}-\dfrac{2}{x-2}+\dfrac{1}{x+2}).\dfrac{x+2}{2}$
    $A=(\dfrac{x}{(x-2)(x+2)}-\dfrac{2(x+2)}{(x-2)(x+2)}+\dfrac{x-2}{(x+2(x-2)}).\dfrac{x+2}{2}$
    $A=(\dfrac{x}{(x-2)(x+2)}-\dfrac{2x+4}{(x-2)(x+2)}+\dfrac{x-2}{(x+2(x-2)}).\dfrac{x+2}{2}$
    $A=(\dfrac{x-2x-4+x-2}{(x-2)(x+2)}).\dfrac{x+2}{2}$
    $A=\dfrac{-6}{(x-2)(x+2)}.\dfrac{x+2}{2}$
    $A=\dfrac{-6}{x-2}.\dfrac{1}{2}$
    $A=\dfrac{-3}{x-2}$

  2. ngoclandiep
    0
    2022-06-20T22:25:23+00:00
    Reply
    Giải đáp:
    A=(x/(x^(2)-4)+2/(2-x)+1/(x+2)).(x+2)/2    (ĐK : x \ne +-2)
    =(x/(x^(2)-4)-2/(x-2)+1/(x+2)).(x+2)/2
    =(x/[(x-2)(x+2)]-[2(x+2)]/[(x-2)(x+2)]+(x-2)/[(x-2)(x+2)]).(x+2)/2
    =((x-2x-4+x-2)/[(x-2)(x+2)]).(x+2)/2
    =(-6/[(x-2)(x+2)]).(x+2)/2
    =(-6.(x+2))/[(x-2)(x+2).2]
    =-3/(x-2)
     

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