x^2-5x+6/4-x^2 nhỏ hơn hoặc bằng x-3/2-x Kẻ bảng giùm em 02/11/2021 Bởi aikhanh x^2-5x+6/4-x^2 nhỏ hơn hoặc bằng x-3/2-x Kẻ bảng giùm em
Ta có $\dfrac{x^2 – 5x + 6}{4-x^2} \leq \dfrac{x-3}{2-x}$ $<-> \dfrac{(x-2)(x-3)}{(2-x)(2+x)} \leq \dfrac{x-3}{2-x}$ $<-> \dfrac{3-x}{2+x} \leq \dfrac{x-3}{2-x}$ $<-> \dfrac{x-3}{2-x} – \dfrac{3-x}{x+2} \geq 0$ $<-> \dfrac{(x-3)(x+2) – (3-x)(2-x)}{(2-x)(2+x)} \geq 0$ $<-> \dfrac{x^2 -x – 6 – (x^2 -5x + 6)}{(2-x)(2+x)} \geq 0$ $<-> \dfrac{-6x – 12}{(2-x)(2+x)} \geq 0$ $<-> \dfrac{-6}{2-x} \geq 0$ $<-> 2-x \leq 0$ $<-> x \leq 2$ Vậy $x \leq 2$. Bình luận
Ta có
$\dfrac{x^2 – 5x + 6}{4-x^2} \leq \dfrac{x-3}{2-x}$
$<-> \dfrac{(x-2)(x-3)}{(2-x)(2+x)} \leq \dfrac{x-3}{2-x}$
$<-> \dfrac{3-x}{2+x} \leq \dfrac{x-3}{2-x}$
$<-> \dfrac{x-3}{2-x} – \dfrac{3-x}{x+2} \geq 0$
$<-> \dfrac{(x-3)(x+2) – (3-x)(2-x)}{(2-x)(2+x)} \geq 0$
$<-> \dfrac{x^2 -x – 6 – (x^2 -5x + 6)}{(2-x)(2+x)} \geq 0$
$<-> \dfrac{-6x – 12}{(2-x)(2+x)} \geq 0$
$<-> \dfrac{-6}{2-x} \geq 0$
$<-> 2-x \leq 0$
$<-> x \leq 2$
Vậy $x \leq 2$.