1: \(9x^2-4-\left(2-3x\right)\left(x+3\right)=0\)
=>\(\left(3x-2\right)\left(3x+2\right)+\left(3x-2\right)\left(x+3\right)=0\)
=>\(\left(3x-2\right)\left(3x+2+x+3\right)=0\)
=>(3x-2)(4x+5)=0
=>\(\left[{}\begin{matrix}3x-2=0\\4x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{4}\end{matrix}\right.\)
2: \(9x^2-6x+1-x\left(3x-1\right)=0\)
=>\(\left(3x-1\right)^2-x\left(3x-1\right)=0\)
=>(3x-1)(3x-1-x)=0
=>(3x-1)(2x-1)=0
=>\(\left[{}\begin{matrix}3x-1=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=\dfrac{1}{2}\end{matrix}\right.\)
`1) 9x^2 - 4 - (2 - 3x)(x+3) = 0`
`=> (3x)^2 - 2^2 - (2 - 3x)(x+3) = 0`
`=> (3x-2)(3x+2) - (2 - 3x)(x+3) = 0
`=> (3x-2)(3x+2) +(3x - 2)(x+3) = 0`
`=>(3x -2 )(3x + 2 + x + 3) = 0`
`=> (3x - 2)(4x + 5) =0`
`TH1`:
`3x - 2 = 0`
`=> 3x = 2`
`=> x = 2/3`
`TH2`:
`4x + 5 = 0`
`=> 4x = -5`
`=> x = -5/4`
Vậy `x`∈`{2/3 ; -5/4}`
`2) 9x^2 - 6x + 1 - x(3x-1) = 0`
`=> (3x)^2 - 2.3x.1 + 1^2 - x(3x-1) = 0`
`=> (3x-1)^2 - x(3x-1) = 0`
`=> (3x-1)(3x-1 - x) = 0`
`=>(3x-1)(2x-1) = 0`
`TH1:`
`3x-1= 0`
`=>3x = 1 `
`=> x = 1/3`
`TH2:`
`2x - 1 = 0`
`=> 2x = 1`
`=> x = 1/2`
Vậy `x`∈`{1/3 ; 1/2}`
