=> 3 - 2x = 4x - 5 

hoặc 3- 2x = - (4x - 5)

+) Trường hợp: 3 -2x = 4x - 5 => 3 + 5 = 4x + 2x => 8 = 6x => x = 4/3
+) Trường hợp: 3 - 2x = - (4x - 5) => 3 - 2x = - 4x + 5 => 4x - 2x = 5 - 3 => 2x = 2 => x = 1

Vậy x = 4/3 hoặc x = 1

nếu ai có cách khác thì giúp mình với

\(\left(x+3\right)^3-3\cdot\left(3x+1\right)^2+\left(2x+1\right)\cdot\left(4x^2-2x+1\right)=54\)

\(\Leftrightarrow x^3+9x^2+27x+27-3\cdot\left(9x^2+6x+1\right)+8x^3-4x^2+2x+4x^2-2x+1=54\)

\(\Leftrightarrow x^3+9x^2+27x+27-27x^2-18x-3+8x^3-4x^2+2x+4x^2-2x+1=54\)

\(\Leftrightarrow9x^3-18x^2+9x-29=0\)

\(\Leftrightarrow x=2,208024627\)

2x+|x+3|=4x+27

=> \(\left|x+3\right|\)=(4x+27)-2x

=>\(\left|x+3\right|\)=4x-2x+27-2x

=>\(\left|x+3\right|\)=2x+27-2x

=>\(\left|x+3\right|\)=27

TH1 : x+3=27=> x=24

TH2 : x+3=-27 => x =-30

Hic hic mình rất muốn làm nhưng mình bận.

x=-24

a: \(P=\left(\dfrac{-\left(x+3\right)}{x-3}+\dfrac{x-3}{x+3}+\dfrac{4x^2}{x^2-9}\right):\dfrac{2x+1-x-3}{x+3}\)

\(=\dfrac{-x^2-6x-9+x^2-6x+9+4x^2}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x-2}\)

\(=\dfrac{4x^2-12x}{x-3}\cdot\dfrac{1}{x-2}=\dfrac{4x}{x-2}\)

b: \(2x^2-5x+2=0\)

=>(x-2)(2x-1)=0

=>x=1/2

Thay x=1/2 vào P, ta được:

\(P=\left(4\cdot\dfrac{1}{2}\right):\left(\dfrac{1}{2}-2\right)=2:\dfrac{-3}{2}=\dfrac{-4}{3}\)

a, 12 - (2\(x^2\) - 3) = 7

            2\(x^2\)  - 3  =  12  - 7

           2\(x^2\) - 3  = 5

           2\(x^2\)  = 8

             \(x^2\)   = 4

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

a) \(12-\left(2x^2-3\right)=7\\ 12-2x^2+3=7\\ 15-2x^2=7\\ 2x^2=15-7=8\\ x^2=8:2=4\\ x=\pm2\)

b) \(3x^2-12=2x^2+4\\ 3x^2-2x^2=12+4\\ x^2=16\\ x=\pm4\)

b, 3\(x^2\) - 12  = 2\(x^2\) + 4

    3\(x^2\) - 2\(x^2\) = 12 + 4

     \(x^2\) = 16

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

\(4x^2-9-x\left(2x-3\right)=0\)

\(\Leftrightarrow4x^2-9-2x^2+3x=0\)

\(\Leftrightarrow2x^2+3x-9=0\)     \(\Leftrightarrow\left(x+3\right)\left(2x-3\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x=-3\\x=\frac{3}{2}\end{cases}}\)

a. \(\frac{3}{4}x-\frac{4}{5}.x=\frac{-2}{3}\)

\(\left(\frac{3}{4}-\frac{4}{5}\right)\) \(.x\) = \(\frac{-2}{3}\)

\(\frac{-1}{20}.x=\frac{-2}{3}\)

\(x=\frac{-2}{3}:\frac{-1}{20}\)

x =\(\frac{-40}{3}\)

a) \(\frac{3}{4}.x\) - \(\frac{4}{5}.x\) = \(\frac{-2}{3}\)

=> \(x.\left(\frac{3}{4}-\frac{4}{5}\right)\) = \(\frac{-2}{3}\)

=> \(x.\frac{-1}{20}\) = \(\frac{-2}{3}\)

=> \(x=\frac{-2}{3}:\frac{-1}{20}\)

=> \(x=\frac{40}{3}\)