K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

\(2x-\frac{1}{3}+2=0\)

\(2x-\frac{1}{3}=0-2\)

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

\(x=\frac{-5}{3}:2\)

\(x=\frac{-5}{3}.\frac{1}{2}\)

\(x=\frac{-5}{6}\)

~HT~

2x-1/3x+2=0

=>5/3x+2=0

<=>5/3x=-2

<=>x=-6/5

18 tháng 10 2023

\(\dfrac{1}{2}-3x+\left|x-1\right|=0\\ \Rightarrow3x+\left|x-1\right|=\dfrac{1}{2}-0\\ \Rightarrow3x+\left|x-1\right|=\dfrac{1}{2}\\ \Rightarrow\left|x-1\right|=\dfrac{1}{2}-3x\\ \Rightarrow\left[{}\begin{matrix}x-1=\dfrac{1}{2}-3x\\x-1=-\dfrac{1}{2}+3x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x+3x=\dfrac{1}{2}+1\\x-3x=-\dfrac{1}{2}+1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}4x=\dfrac{3}{2}\\2x=\dfrac{1}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{8}\\x=\dfrac{1}{4}\end{matrix}\right.\)

__

\(\dfrac{1}{2}\left|2x-1\right|+\left|2x-1\right|=x+1\\ \Rightarrow\left|2x-1\right|\cdot\left(\dfrac{1}{2}+1\right)=x+1\\ \Rightarrow\left|2x-1\right|\cdot\dfrac{3}{2}=x+1\\ \Rightarrow\left|2x-1\right|=x+1:\dfrac{3}{2}\\ \Rightarrow\left|2x-1\right|=x+\dfrac{2}{3}\\ \Rightarrow\left[{}\begin{matrix}2x-1=x+\dfrac{2}{3}\\2x-1=-x-\dfrac{2}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x-x=\dfrac{2}{3}+1\\2x+x=-\dfrac{2}{3}+1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\3x=\dfrac{1}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=\dfrac{1}{9}\end{matrix}\right.\)

27 tháng 11 2019

1)Tìm x

a) (x+1)(x-2)<0

=>Có 2TH:

TH1:

x+1<0=>x< -1

x-2>0=>x>2

=>Vô lí 

TH2:

x+1>0=>x> -1

x-2<0=>x<2

=> -1<x<2

Vậy x thuộc {0;1}

b) Tương tự a thôi ạ. 

c) (x-2)(3x+2)

=> Có hai TH:

TH1:

x-2<0=>x<2

3x+2<0=>3x< -2=>x< -2/3

=>x< -2/3

TH2:

x-2>0=>x>2

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

=>x>2

Vậy x< -2/3 hoặc x>2

2)Tìm x

x.x=x

<=>x²-x=0

<=>x(x-1)=0

<=>x=0 hoặc x=1

28 tháng 11 2019

Cảm ơn nha Linh

1 tháng 12 2016

1,X=-1 hoặc 3

2,Tìm x sao cho (x+3) và (3x-2) ko bằng 0

7 tháng 7 2016

1) x (x-2016) + 2015 (2016-x) = 0

 x (x-2016) - 2015 (x- 2016) = 0

(x-2015)(x-2016) =0

\(\Rightarrow\orbr{\begin{cases}x-2015=0\\x-2016=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2015\\x=2016\end{cases}}}\)

Vậy x= 2015; 2016

2) -5x (x-15) + (15-x) = 0

-5x (x-15) - (x-15) =0

(-5x -1) (x-15) =0

\(\Rightarrow\orbr{\begin{cases}-5x-1=0\\x-15=0\end{cases}\Rightarrow\orbr{\begin{cases}-5x=1\\x=15\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{5}\\x=15\end{cases}}}\)

Vậy x= -1/5; 15

3) 3x (3x-7) - (7-3x) =0

3x(3x-7) + (3x -7) =0

(3x+1) (3x-7) =0

\(\Rightarrow\orbr{\begin{cases}3x+1=0\\3x-7=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=-1\\3x=7\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{3}\\x=\frac{7}{3}\end{cases}}}\)

Vậy x= -1/3 ; 7/3

DT
2 tháng 10 2023

loading...  Bạn tham khảo.

`#040911`

`a)`

`2x^2 - 3x = 0`

`\Rightarrow x(2x - 3) = 0`

`\Rightarrow`\(\left[{}\begin{matrix}x=0\\2x-3=0\end{matrix}\right.\)

`\Rightarrow`\(\left[{}\begin{matrix}x=0\\2x=3\end{matrix}\right.\)

`\Rightarrow`\(\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)

Vậy, \(x\in\left\{0;\dfrac{3}{2}\right\}\)

`b)`

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

Bạn xem lại đề

`c)`

\(x^3-x^2=0\\ \Rightarrow x^2\cdot\left(x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x^2=0\\x-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

Vậy, \(x\in\left\{0;1\right\}.\)

5 tháng 9 2023

\(a,2x^2-3x=0\\ \Leftrightarrow x\left(2x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\\ b,Xem.lại,đề\\ c,x^3-x^2=0\\ \Leftrightarrow x^2.\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

18 tháng 8 2020

+) \(2x\left(x-4\right)-x\left(2x+3\right)+22=0\)

\(\Leftrightarrow2x^2-8x-2x^2-3x+22=0\)

\(\Leftrightarrow-11x+22=0\)

\(\Leftrightarrow-11\left(x-2\right)=0\)

\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)

+) \(\left(2x+3\right)\left(3x+2\right)+2\left(1-3x\right)\left(x+\frac{1}{2}\right)=1\)

\(\Leftrightarrow6x^2+4x+9x+6+\left(2-6x\right)\left(x+\frac{1}{2}\right)=1\)

\(\Leftrightarrow6x^2+13x+6+2x+1-6x^2-3x=1\)

\(\Leftrightarrow12x+7=1\)

\(\Leftrightarrow x=\frac{-1}{2}\)

18 tháng 8 2020

2x( x - 4 ) - x( 2x + 3 ) + 22 = 0

<=> 2x2 - 8x - 2x2 - 3x + 22 = 0

<=> -11x + 22 = 0

<=> -11x = -22

<=> x = 2

( 2x + 3 )( 3x + 2 ) + 2( 1 - 3x )( x + 1/2 ) = 1

<=> 6x2 + 13x + 6 + 2( -3x2 - 1/2x + 1/2 ) = 1

<=> 6x2 + 13x + 6 - 6x2 - x + 1 = 1

<=> 12x + 7 = 1 

<=> 12x = -6

<=> x = -6/12 = -1/2