suy ra :

1/2x - 3 = 0 => 1/2x = 3 => x = 6 

hoặc x + (-7/4) = 0 => x = 0 - (-7/4) = 0+7/4 =7/4

vậy x =6 hoặc x=7/4

\(\left(2x+3\right)\left(x-1\right)+\left(2x-3\right)\left(1-x\right)=0\)

\(\left(2x+3\right)\left(x-1\right)-\left(2x-3\right)\left(x-1\right)=0\)

\(\left(x-1\right)\left(2x+3-2x+3\right)=0\)

\(\left(x-1\right)\cdot6=0\)

\(\Rightarrow x-1=0\)

\(\Rightarrow x=1\)

(2x+3).(x-1) + (2x-3).(1-x) = 0

(2x+3).(x-1) - (2x+3).(1-x) = 0

(2x+3).[(x-1) - (1-x)] = 0

(2x+3).( x - 1 -1 + x) = 0

(2x+1). ( 2x - 2) = 0

(2x+1).2.(x-1) = 0

=> 2x+1 = 0 => 2x = -1 => x = -1/2

x-1=0 => x = 1

CÔNG CHÚA ÔRI làm buồn cười quá

Dòng 6 từ ( 2x + 3 ) chuyển thành ( 2x + 1 )

Với lại sai từ đầu rồi

+ ( 2x - 3 ) = - ( 2x + 3 ) á ??

\(a,3x-2\left(x-3\right)=0\\ \Leftrightarrow3x-2x+6=0\\ \Leftrightarrow x=-6\\ b,\left(x+1\right)\left(2x-3\right)=\left(2x-1\right)\left(x+5\right)\\ \Leftrightarrow2x^2+2x-3x-3=2x^2-x+10x-5\\ \Leftrightarrow2x^2-x-3=2x^2+9x-5\\ \Leftrightarrow10x-2=0\\ \Leftrightarrow x=\dfrac{1}{5}\\ c,ĐKXĐ:x\ne\pm1\\ \dfrac{2x}{x-1}-\dfrac{x}{x+1}=1\\ \Leftrightarrow\dfrac{2x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\\ \Leftrightarrow\dfrac{2x^2+2x-x^2+x-x^2+1}{\left(x+1\right)\left(x-1\right)}=0\)

\(\Rightarrow3x+1=0\\ \Leftrightarrow x=-\dfrac{1}{3}\left(tm\right)\)

\(d,\left(2x+3\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x+3=0\\3x-5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{5}{3}\end{matrix}\right.\\ e,ĐKXĐ:x\ne\pm2\\ \dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\\ \Leftrightarrow\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-22}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\dfrac{x^2-4x+4-3x-6-2x+22}{\left(x-2\right)\left(x+2\right)}=0\\ \Rightarrow x^2-9x+20=0\\ \Leftrightarrow\left(x^2-5x\right)-\left(4x-20\right)=0\\ \Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x-5\right)\\ \Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=5\left(tm\right)\end{matrix}\right.\)

\(\left(x-\dfrac{3}{2}\right)\times\left(2x+1\right)>0\)

Th1:

\(x-\dfrac{3}{2}>0\Leftrightarrow x>\dfrac{3}{2}\)

\(2x+1>0\Leftrightarrow2x>1\Leftrightarrow x>\dfrac{1}{2}\)

( 1 )

Th2: 

\(x-\dfrac{3}{2}< 0\Leftrightarrow x< \dfrac{3}{2}\)

\(2x+1< 0\Leftrightarrow2x< -1\Leftrightarrow x< -\dfrac{1}{2}\)

( 2 )

Từ ( 1 ) và ( 2 ), ta có:

\(\Rightarrow x< -\dfrac{1}{2};x>\dfrac{3}{2}\)

\(\left(2-x\right)\times\left(\dfrac{4}{5}-x\right)< 0\)

Th1:

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

\(\dfrac{4}{5}-x< 0\Leftrightarrow x< \dfrac{4}{5}\)

( Loại )

Th2:

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

\(\dfrac{4}{5}-x>0\Leftrightarrow x>\dfrac{4}{5}\)

=> \(\dfrac{4}{5}< x< 2\)

`@` `\text {Ans}`

`\downarrow`

\(\left(\dfrac{x}{3}+\dfrac{1}{2}\right)\left(75\%-1\dfrac{1}{2}x\right)=0\)

`=>`\(\left[{}\begin{matrix}\dfrac{x}{3}+\dfrac{1}{2}=0\\\dfrac{75}{100}-\dfrac{3}{2}x=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}\dfrac{x}{3}=-\dfrac{1}{2}\\\dfrac{3}{2}x=\dfrac{75}{100}\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}2x=-1\cdot3\\x=\dfrac{75}{100}\div\dfrac{3}{2}\end{matrix}\right.\)

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

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

Vậy, `x={-3/2; 1/2}.`

=>2x+1/2=0 hoặc 2x-3=0

=>x=-1/4 hoặc x=3/2

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

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

( 1/2 + 2x ) = 0

( 2x - 3 ) = 0

( 1/2 + 2x ) = 0

\(\Leftrightarrow\) ( 2x + 1/2 ) = 0

2x + 1/2 - 1/2 = 0 - 1/2

2x = -1/2

2x/2 = -1/2 : 2

x = -1/4

2x - 3 = 0

2x - 3 + 3 = 0 + 3

2x = 3

2x/2 = 3/2

x = 3/2

\(\Rightarrow\) x₁ = -1/4 ; x₂ = 3/2