\(x\left(x-3\right)-12+4x=0\)

\(\Leftrightarrow x^2-3x-12+4x=0\)

\(\Leftrightarrow x^2+x-12=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+4=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-4\end{cases}}\)

\(x^3-4x^2-4x+1\)

\(=x^3-5x^2+x+x^2-5x+1\)

\(=x\left(x^2-5x+1\right)+\left(x^2-5x+1\right)\)

\(=\left(x+1\right)\left(x^2-5x+1\right)\)

Mà \(x^3-4x^2-4x+1=A\left(x^2-5x+1\right)\)

\(\Rightarrow A=x+1\)

 a) 2 (x-1) -3 (x+2) = 5x-9

2x -2 - 3x -6 = 5x -9

2x - 3x - 5x = -9 +2 +6

-6x = -1

x= 1/6

b) 4x (x+1) -2x (x+2) =0

4x^2 + 4x - 2x^2 - 4x =0

2x^2 =0

x^2 =0

x=0

(x²-4x+4) + (y²+6y+9) = 0

bạn làm tiếp nhé, dáp số x=2, y=-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

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

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

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

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

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

\(\Leftrightarrow x=2,208024627\)