=>3x-2x^2-3+2x+2x^2+6x-x-3=14

=>10x-6=14

=>x=2

\(\left(x-1\right)\left(3-2x\right)+\left(2x-1\right)\left(x+3\right)=14\)

\(\Leftrightarrow3x-2x^2-3+2x+2x^2+6x-x-3=14\)

\(\Leftrightarrow\left(-2x^2+2x^2\right)+\left(3x+2x+6x-x\right)-\left(3+3\right)=14\)

\(\Leftrightarrow10x-6=14\)

\(\Leftrightarrow10x=14+6\)

\(\Leftrightarrow10x=20\)

\(\Leftrightarrow x=20:10\)

\(\Leftrightarrow x=2\)

Vậy: \(x=2\)

Hình hiển thị bị lỗi rồi. Bạn nên gõ hẳn đề ra để được hỗ trợ tốt hơn nhé.

d) \(\left|2x-3\right|=x-3\)

TH1: \(\left|2x-3\right|=2x-3\) với \(2x-3\ge0\Leftrightarrow x\ge\dfrac{3}{2}\)

Pt trở thành:

\(2x-3=x-3\) (ĐK: \(x\ge\dfrac{3}{2}\) )

\(\Leftrightarrow2x-x=-3+3\)

\(\Leftrightarrow x=0\left(ktm\right)\)

TH2: \(\left|2x-3\right|=-\left(2x-3\right)\) với \(2x-3< 0\Leftrightarrow x< \dfrac{3}{2}\)

Pt trở thành:

\(-\left(2x-3\right)=x-3\)

\(\Leftrightarrow-2x+3=x-3\)

\(\Leftrightarrow-2x-x=-3-3\)

\(\Leftrightarrow-3x=-6\)

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

Vậy Pt vô nghiệm

\(\left(x+1\right)\left(2x-2\right)-3>-5x-\left(2x+1\right)\left(3x-x\right)\)

\(\Leftrightarrow2x^2-2x+2x-2-3>-5x-\left(6x^2-2x^2+3x-x\right)\)

\(\Leftrightarrow2x^2-2x+2x-5>-5x-6x^2+2x^2-3x+x\)

\(\Leftrightarrow2x^2-5+5x+6x^2-2x^2+3x-x>0\)

\(\Leftrightarrow6x^2-2x>5\)

\(\Leftrightarrow2x\left(3x-1\right)>5\)

\(\Leftrightarrow\left[{}\begin{matrix}2x>5\\3x-1>5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{5}{2}\\x>2\end{matrix}\right.\)

\(2x^3+2x+x^2+1=0\\ \Rightarrow\left(2x^3+2x\right)+\left(x^2+1\right)=0\\ \Rightarrow2x\left(x^2+1\right)+\left(x^2+1\right)=0\\ \Rightarrow\left(2x+1\right)\left(x^2+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x^2=-1\left(vô.lí\right)\end{matrix}\right.\)

Vậy \(x=-\dfrac{1}{2}\)

\(\dfrac{-2}{3}\left(x-\dfrac{1}{4}\right)=\dfrac{1}{3}\left(2x-1\right)\)

\(\Leftrightarrow\dfrac{-2}{3}x+\dfrac{1}{6}=\dfrac{2}{3}x-\dfrac{1}{3}\)

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

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

\(\Leftrightarrow x=\dfrac{3}{8}\)

Vậy \(x=\dfrac{3}{8}\)

gio con noc ha ?!

<=> 2x^2 +x-4x-2-5x-15=2x^2-6x+4+8x-2-2x

      2x^2-8x-17-2x^2-2=0

     -8x-19=0

x=-19/8

(x-2).(2x+1)-5(x+3)=2x(x-3)+4(1+2x)-2(1+x)

3x-4x+x-2-5x-15=3x-6x+4+8x-2-2x

-5x-17=3x+2

-19=8x

-19/8=x

Vậy x=-19/8

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

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

\(\dfrac{3x-2}{6}\) - \(\dfrac{4x-6}{6}\) = \(\dfrac{6x}{6}\)

3\(x-2-4x\) + 6 = 6\(x\) 

-\(x\) + 4 - 6\(x\) = 0

7\(x\)  = 4

    \(x\) =  \(\dfrac{4}{7}\) 

Sửa đề: \(\dfrac{2x-1}{3}=\dfrac{27}{2x-1}\)

ĐKXĐ: x<>1/2

\(\dfrac{2x-1}{3}=\dfrac{27}{2x-1}\)

=>\(\left(2x-1\right)^2=3\cdot27=81\)

=>\(\left[{}\begin{matrix}2x-1=9\\2x-1=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=10\\2x=-8\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=5\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)