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

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

TRẢ LỜI

1. landinhngoc97
Gửi bạn:
Đề đúng: $A=(\dfrac{x}{x^2-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}).\dfrac{x+2}{2}$
$(ĐKXĐ:x\neq±2)$
$A=(\dfrac{x}{x^2-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}).\dfrac{x+2}{2}$
$A=(\dfrac{x}{(x-2)(x+2)}-\dfrac{2}{x-2}+\dfrac{1}{x+2}).\dfrac{x+2}{2}$
$A=(\dfrac{x}{(x-2)(x+2)}-\dfrac{2(x+2)}{(x-2)(x+2)}+\dfrac{x-2}{(x+2(x-2)}).\dfrac{x+2}{2}$
$A=(\dfrac{x}{(x-2)(x+2)}-\dfrac{2x+4}{(x-2)(x+2)}+\dfrac{x-2}{(x+2(x-2)}).\dfrac{x+2}{2}$
$A=(\dfrac{x-2x-4+x-2}{(x-2)(x+2)}).\dfrac{x+2}{2}$
$A=\dfrac{-6}{(x-2)(x+2)}.\dfrac{x+2}{2}$
$A=\dfrac{-6}{x-2}.\dfrac{1}{2}$
$A=\dfrac{-3}{x-2}$

Trả lời
2. ngoclandiep
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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