$ a/ 12x(x – 5) – 3x(4x - 10) = 120$

`<=>12x^2-60x-12x^2+30x=120`

`<=>-30x=120`

`<=>x=-4`

Vậy `x=-4`

$b/ 9x(x + 4) – 5x(3x + 2) = 112 - 2x(3x + 1)$

`<=>9x^2+36x-15x^2-10x=112-6x^2-2x`

`<=>-6x^2+26x=112-6x^2-2x`

`<=>28x=112`

`<=>x=4`

Vậy `x=4`

$c/ 3x(1 – x) - 5x(3x + 7) = 154 + 9x(5 – 2x)$

`<=>3x-3x^2-15x^2-35x=154+45x-18x^2`

`<=>-32x-18x^2=154+45x-18x^2`

`<=>77x=-154`

`<=>x=-2`

Vậy `x=-2`

(1-3x²)-(x-2)(9x+1)=(3x-4)(3x+4)-9(x+3)²

⇒1-3x²-(9x²+x-18x-2)=9x²-16-9(x²+6x+9)

⇒1-3x²-(9x²-17x-2)= -56x-97

⇒1-3x²-9x²+17x+2=-56x-97

⇒3-12x²+17x=-56x-97

⇒3-12x²+17x+56x+97=0

⇒-12x²+73x+100=0

⇒-(12x²-73x-100)=0

a)
(3x+1)^2 -9x (x-1) =46
9x²+6x+1-9x²+9x = 46
15x+1=46
15x=45
x=3
b)
5x (x-3) = 7 (3-x)
5x ( x - 3 ) - 7 ( 3 - x ) = 0
5x ( x - 3 ) + 7 ( x - 3 ) = 0
(5x + 7 ) ( x - 3 ) = 0
5x+7=0 hoặc x-3 = 0 
5x=-7    hoặc x=3
  x= -7 / 5 hoặc x=3

trình bày đầy đủ các bước

a: Ta có: \(\left(x+1\right)^2-3\left(x+1\right)=0\)

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

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

b: Ta có: \(2\left(3x-2\right)^2=9x^2-4\)

\(\Leftrightarrow2\left(3x-2\right)^2-\left(3x-2\right)\left(3x+2\right)=0\)

\(\Leftrightarrow\left(3x-2\right)\left(6x-4-3x-2\right)=0\)

\(\Leftrightarrow\left(3x-2\right)\left(3x-6\right)=0\)

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