Giải phương trình a. (x+1)(x+2)=(2-x)(x+2) b. $\frac{2-x}{2016}$ -1= $\frac{1-x}{2017}$ – $\frac{x}{2018}$ c. $\frac{x-19}{1999}$ + $\frac{x-23}{1995

Giải thích các bước giải:

a.$(x+1)(x+2)=(2-x)(x+2)$ 

$\to (x+1)(x+2)-(2-x)(x+2)=0$

$\to (x+2)(x+1-2+x)=0$

$\to (x+2)(2x-1)=0\to x\in\{-2,\dfrac 12\}$

b.Ta có $\dfrac{2-x}{2016}-1=\dfrac{1-x}{2017}-\dfrac{x}{2018}$

$\to \dfrac{2}{2016}-\dfrac{x}{2016}-1=\dfrac{1}{2017}-\dfrac{x}{2017}-\dfrac{x}{2018}$

$\to \dfrac{x}{2017}+\dfrac{x}{2018}-\dfrac{x}{2016}=\dfrac{1}{2017}-\dfrac{2}{2016}+1$

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

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

c.$\dfrac{x-19}{1999}+\dfrac{x-23}{1995}+\dfrac{x+82}{700}=5$

$\to \dfrac{x-19}{1999}-1+\dfrac{x-23}{1995}-1+\dfrac{x+82}{700}-3=0$

$\to \dfrac{x-2018}{1999}+\dfrac{x-2018}{1995}+\dfrac{x-2018}{700}=0$

$\to (x-2018)( \dfrac{1}{1999}+\dfrac{1}{1995}+\dfrac{1}{700}=0)$

$\to x=2018$

d.$x^3-3x^2+4=0$

$\to (x+1)(x-2)^2=0\to x\in\{2,-1\}$