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Ta có: \(\left|x-2\right|-\left|2x+3\right|-x=-2\)

=>|x-2|-|2x+3|=x-2(1)

Trường hợp 1: x<-3/2

(1) trở thành 2-x-(-2x-3)=x-2

=>2-x+2x+3=x-2

=>x+5=x-2(loại)

Trường hợp 2: -3/2<=x<2

(1) trở thành 2-x-2x-3=x-2

=>-3x-1=x-2

=>-4x=-1

hay x=1/4(nhận)

Trường hợp 3: x>=2

(1) trở thành x-2-2x-3=x-2

=>-2x-3=0

=>-2x=3

hay x=-3/2(loại)

10 tháng 3 2022

a)TH1: \(2x-3>0;3x+2>0\)

\(=>2x-3-3x-2=0\\ =>-x-5=0\\ =>-x=5=>x=-5\)

TH2: \(2x-3< 0;3x+2< 0\)

\(=>-2x+3+3x+2=0\\ =>x+5=0\\ =>x=-5\)

Cả 2 TH ra \(x=-5=>x=-5\)

b)TH1 \(\dfrac{1}{2}x>0\)

\(=>\dfrac{1}{2}x=3-2x\\ =>3-2x-\dfrac{1}{2}x=0\\ =>\dfrac{4}{2}x-\dfrac{1}{2}x=3\\ =>\dfrac{3}{2}x=3\\ =>x=2\)

TH2 \(\dfrac{1}{2}x< 0\)

\(=>-\dfrac{1}{2}x=3-2x\\ =>3-2x+\dfrac{1}{2}x=0\\ =>\dfrac{4}{2}x+\dfrac{1}{2}x=3\\ =>\dfrac{5}{2}x=3\\ =>x=\dfrac{6}{5}\)

\(=>x=2;\dfrac{6}{5}\)

:V lập 2 ý là làm đc á em 

12 tháng 10 2023

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\(\left[{}\begin{matrix}2x-\dfrac{2}{3}=\dfrac{1}{3}\\2x-\dfrac{2}{3}=\dfrac{-1}{3}\end{matrix}\right.\left[{}\begin{matrix}2x=1\\2x=\dfrac{1}{3}\end{matrix}\right.\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{1}{6}\end{matrix}\right.\)

a)\(\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{5}{2}\\x+\dfrac{1}{2}=-\dfrac{5}{2}\end{matrix}\right.\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

 

29 tháng 1 2022

Chia nhỏ ra

a: =>1/2x=7/2-2/3=21/6-4/6=17/6

=>x=17/3

b: =>2/3:x=-7-1/3=-22/3

=>x=2/3:(-22/3)=-1/11

c: =>1/3x+2/5x-2/5=0

=>11/15x=2/5

hay x=6/11

d: =>2x-3=0 hoặc 6-2x=0

=>x=3/2 hoặc x=3

a: =>2x-1=-2

=>2x=-1

hay x=-1/2

b: \(\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\-\dfrac{2}{5}x-7=0\end{matrix}\right.\Leftrightarrow x\in\left\{-\dfrac{2}{3};-\dfrac{35}{2}\right\}\)

c: x/8=9/4

nên x/8=18/8

hay x=18

d: \(\Leftrightarrow\left(x-3\right)^2=36\)

=>x-3=6 hoặc x-3=-6

=>x=9 hoặc x=-3

e: =>-1,7x=6,12

hay x=-3,6

h: =>x-3,4=27,6

hay x=31

22 tháng 2 2022

a) \(\dfrac{1}{3}\div\left(2x-1\right)=\dfrac{-1}{6}\)

\(\left(2x-1\right).\dfrac{1}{3}\div\left(2x-1\right)=\left(2x-1\right)\left(-\dfrac{1}{6}\right)\)

\(\dfrac{1}{3}=\left(2x-1\right)\left(-\dfrac{1}{6}\right)\)

\(\dfrac{1}{3}=-1\left(2x-1\right)\div6\)

\(\dfrac{1}{3}=-2x+1\div6\)

\(x=-\dfrac{1}{2}\)

b) \(\left(3x+2\right)\left(\dfrac{-2}{5}x-7\right)=0\)

\(TH1:3x+2=0\)

\(3x=0-2\)

\(3x=-2\)

\(x=\dfrac{-2}{3}\)

\(TH2:\left(-\dfrac{2}{5}x-7\right)=0\)

\(\left(\dfrac{-2}{5}x-7\right)=0\)

\(\left(\dfrac{-2x}{5}+\dfrac{5\left(-7\right)}{5}\right)=0\)

\(\left(\dfrac{-2x-35}{5}\right)=0\)

\(-2x-35=0\)

\(-2x=0+35\)

\(x=-\dfrac{35}{2}\)

c) \(\dfrac{x}{8}=\dfrac{9}{4}\)

\(\Leftrightarrow x=\dfrac{9.8}{4}=\dfrac{72}{4}=18\)

\(x=18\)

d) \(\dfrac{x-3}{2}=\dfrac{18}{x-3}\)

\(x-3=18+2\)

\(x=20-3\)

\(x=17\)

e) \(4,5x-6,2x=6,12\)

\(\dfrac{9x}{2}-6,2.x=6,12\)

\(\dfrac{9x}{2}+\dfrac{-31x}{5}=6,12\)

\(\dfrac{5.9x}{10}+\dfrac{2\left(-31\right)x}{10}=6.12\)

\(\dfrac{45x-62x}{10}=6.12\)

\(=-17x\div10=6.12\)

\(-17x=10.6.12\)

\(x=-3,6\)

h) \(11,4-\left(x-3,4\right)=-16,2\)

\(x-3,4=-16,2+11,4\)

\(x-3,4=-4,8\)

\(x=-1,4\)

 

a: =>x-3/4=1/6-1/2=1/6-3/6=-2/6=-1/3

=>x=-1/3+3/4=-4/12+9/12=5/12

b: =>x(1/2-5/6)=7/2

=>-1/3x=7/2

hay x=-21/2

c: (4-x)(3x+5)=0

=>4-x=0 hoặc 3x+5=0

=>x=4 hoặc x=-5/3

d: x/16=50/32

=>x/16=25/16

hay x=25

e: =>2x-3=-1/4-3/2=-1/4-6/4=-7/4

=>2x=-7/4+3=5/4

hay x=5/8

a: =>11(x-3)=6(x-5)

=>11x-33=6x-30

=>5x=3

=>x=3/5

b: =>(4/3-1/4x-5/12)-2x=8/5*5/3=8/3

=>-9/4x+11/12=8/3

=>-9/4x=32/12-11/12=21/12=7/4

=>x=-7/9

c: =>1/2x-1/3-2/3x-1=x

=>-1/6x-4/3=x

=>-7/6x=4/3

=>x=-4/3:7/6=-4/3*6/7=-24/21=-8/7

d: =>1-2x-3x+1=7/2

=>-5x=3/2

=>x=-3/10

27 tháng 8 2023

làm ơn giúp 🙏🙏🙏

a: =>1/3x+2/5x-2/5=0

=>11/15x-2/5=0

=>11/15x=2/5

=>x=2/5:11/15=2/5*15/11=30/55=6/11

b: =>-5x-1-1/2x+1/3=x

=>-11/2x-2/3-x=0

=>-13/2x=2/3

=>x=-2/3:13/2=-2/3*2/13=-4/39

c: (x+1/2)(2/3-2x)=0

=>x+1/2=0 hoặc 2/3-2x=0

=>x=1/3 hoặc x=-1/2

d: 9(3x+1)^2=16

=>(3x+1)^2=16/9

=>3x+1=4/3 hoặc 3x+1=-4/3

=>3x=1/3 hoặc 3x=-7/3

=>x=1/9 hoặc x=-7/9

17 tháng 8 2020

Bài làm:

Ta có: \(x\left(x-3\right)-2x\left(x-1\right)=-x^2+5x\)

\(\Leftrightarrow x^2-3x-2x^2+2x+x^2-5x=0\)

\(\Leftrightarrow-6x=0\)

\(\Rightarrow x=0\)

17 tháng 8 2020

Ta có: \(x\left(x-3\right)-2x\left(x-1\right)=x^2-3x-2x^2+2x=-x^2-x\)

=> \(-x^2-x=-x^2+5x\)

=> \(-x^2+x^2=5x+x\)

=> 6x=0

=> x=0

Đúng hong tar? '-'