Toán Lớp 8: ($\frac{2}{x-1}$-$\frac{2}{x+1}$):$\frac{4}{x^2-2x+1}$
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Toán Lớp 8: ($\frac{2}{x-1}$-$\frac{2}{x+1}$):$\frac{4}{x^2-2x+1}$
Comments ( 2 )
$(\frac{2}{x-1}-\frac{2}{x+1}):\frac{4}{x^2-2x+1}$
=$(\frac{2(x+1)}{(x-1)(x+1)}-\frac{2(x-1)}{(x-1)(x+1)}):\frac{4}{(x-1)^2}$
=$\frac{2x+2-2x+2}{(x-1)(x+1)}.\frac{(x-1)^2}{4}$
=$\frac{4}{(x-1)(x+1)}.\frac{(x-1)^2}{4}$
=$\frac{x-1}{x+1}$
Giải đáp:
(x – 1)/(x + 1)
Lời giải và giải thích chi tiết:
(2/(x – 1) – 2/(x + 1)) : 4/(x^2 – 2x + 1)
= ((2(x + 1))/((x – 1)(x + 1)) – (2(x – 1))/((x – 1)(x + 1))) : 4/(x – 1)^2
= (2(x + 1) – 2(x – 1))/((x – 1)(x + 1)) . (x – 1)^2/4
= (2x + 2 – 2x + 2)/((x – 1)(x + 1)) . (x – 1)^2/4
= 4/((x – 1)(x + 1)) . (x – 1)^2/4
= (x – 1)/(x + 1)