Toán X-1 phần x-3 trừ 1-x phần x+3 cộng 2(1-x) phần x2-9 02/08/2021 By Delilah X-1 phần x-3 trừ 1-x phần x+3 cộng 2(1-x) phần x2-9
x-1/x-3 – 1-x/x+3 + 2(1-x)/x^2-9 =(x-1)(x+3)/(x-3)(x+3) – (1-x)(x-3)/(x-3)(x+3) + 2-2x/(x-3)(x+3) =x^2+3x-x-3-x+3+x^2-3x+2-2x/(x-3)(x+3) =2x^2-4x+2/(x-3)(x+3) =2(x^2-2x+1)/(x-3)(x+3) =2(x-1)^2/(x-3)(x+3) Trả lời
ĐKXĐ: $x\neq±3$ $\frac{x-1}{x-3}-\frac{1-x}{x+3}+\frac{2 (1-x)}{x^2-9}$ = $\frac{(x-1)(x+3)}{(x-3)(x+3)}-\frac{(1-x)(x-3)}{(x-3)(x+3)}+\frac{2-2x}{(x-3)(x+3)}$ = $\frac{x^2+3x-x-3-(x-3-x^2+3x)+2-2x}{(x-3)(x+3)}$ = $\frac{x^2+3x-x-3-x+3+x^2-3x+2-2x}{(x-3)(x+3)}$ = $\frac{2x^2-4x+2}{(x-3)(x+3)}$ = $\frac{2(x^2-2x+1)}{(x-3)(x+3)}$ = $\frac{2(x-1)^2}{(x-3)(x+3)}$ Trả lời
x-1/x-3 – 1-x/x+3 + 2(1-x)/x^2-9
=(x-1)(x+3)/(x-3)(x+3) – (1-x)(x-3)/(x-3)(x+3) + 2-2x/(x-3)(x+3)
=x^2+3x-x-3-x+3+x^2-3x+2-2x/(x-3)(x+3)
=2x^2-4x+2/(x-3)(x+3)
=2(x^2-2x+1)/(x-3)(x+3)
=2(x-1)^2/(x-3)(x+3)
ĐKXĐ: $x\neq±3$
$\frac{x-1}{x-3}-\frac{1-x}{x+3}+\frac{2 (1-x)}{x^2-9}$
= $\frac{(x-1)(x+3)}{(x-3)(x+3)}-\frac{(1-x)(x-3)}{(x-3)(x+3)}+\frac{2-2x}{(x-3)(x+3)}$
= $\frac{x^2+3x-x-3-(x-3-x^2+3x)+2-2x}{(x-3)(x+3)}$
= $\frac{x^2+3x-x-3-x+3+x^2-3x+2-2x}{(x-3)(x+3)}$
= $\frac{2x^2-4x+2}{(x-3)(x+3)}$
= $\frac{2(x^2-2x+1)}{(x-3)(x+3)}$
= $\frac{2(x-1)^2}{(x-3)(x+3)}$