a, Tìm số nguyên $\frac{3x-5}{x+1}$ ≤ 1 b, Giải phương trình: $\frac{1}{2016x+1}$ – $\frac{1}{2017x+2}$ = $\frac{1}{2018x+4}$ – $\frac{1}{2

Đáp án: $a.x\in\{0,1,2,3\} $                 b.$x\in\{-1,\dfrac{-12117\pm\sqrt{146240649}}{16140}\}$

Giải thích các bước giải:

a.Ta có :
$\dfrac{3x-5}{x+1}\le 1$

$\to\dfrac{3x-5}{x+1}-1\le 0$

$\to \dfrac{3x-5-(x+1)}{x+1}\le 0$

$\to \dfrac{2x-6}{x+1}\le 0$

$\to \dfrac{x-3}{x+1}\le 0$
$\to -1<x\le 3$

$\to x\in\{0,1,2,3\}$ vì $x\in Z$ 

b.Ta có :

$\dfrac{1}{2016x+1}-\dfrac{1}{2017x+2}=\dfrac{1}{2018x+4}-\dfrac{1}{2019x+5}$

$\to \dfrac{2017x+2-(2016x+1)}{(2016x+1)(2017x+2)}=\dfrac{2019x+5-(2018x+4)}{(2018x+4)(2019x+5)}$

$\to \dfrac{x+1}{(2016x+1)(2017x+2)}=\dfrac{x+1}{(2018x+4)(2019x+5)}$

$\to x=-1$

Hoặc $(2016x+1)(2017x+2)=(2018x+4)(2019x+5)$

$\to 4066272x^2+6049x+2=4074342x^2+18166x+20$

$\to 8070x^2+12117x+18=0$

$\to x=\dfrac{-12117\pm\sqrt{146240649}}{16140}$