,`#3107.101107`
`4x^2 - 12x = -9`
`<=> 4x^2 - 12x + 9 = 0`
`<=> (2x)^2 - 2*2x*3 + 3^2 = 0`
`<=> (2x - 3)^2 = 0`
`<=> 2x - 3 = 0`
`<=> 2x = 3`
`<=> x = 3/2`
Vậy, `x = 3/2`
____
`x(x - 1) + 2x - 2 = 0`
`<=> x(x - 1) + (2x - 2) = 0`
`<=> x(x - 1) + 2(x - 1) = 0`
`<=> (x + 2)(x - 1) = 0`
`<=>`\(\left[{}\begin{matrix}x+2=0\\x-1=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
Vậy, `x \in {-2; 1}`
_____
`2x^3 + 3x^2 + 2x + 3 = 0`
`<=> (2x^3 + 2x) + (3x^2 + 3) = 0`
`<=> 2x(x^2 + 1) + 3(x^2 + 1) = 0`
`<=> (2x + 3)(x^2 + 1) = 0`
`<=>`\(\left[{}\begin{matrix}2x+3=0\\x^2+1=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}2x=-3\\x^2=-1\left(\text{vô lý}\right)\end{matrix}\right.\)
`<=> x = -3/2`
Vậy, `x = -3/2.`
`(x+8)^2 =121`
`<=> (x+8)^2=11^2`
\(\Leftrightarrow\left[{}\begin{matrix}x+8=11\\x+8=-11\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-19\end{matrix}\right.\)
__
`4x^2 -12x=-9`
`<=> 4x^2 -12x+9=0`
`<=>(2x-3)^2=0`
`<=>2x-3=0`
`<=>2x=3`
`<=>x=3/2`
__
`x(x-1)+2x-2=0`
`<=> x(x-1)+2(x-1)=0`
`<=>(x-1)(x+2)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
__
`2x^3 +3x^2 +2x+3=0`
`<=>(2x^3+3x^2)+(2x+3)=0`
`<=>x^2(2x+3)+(2x+3)=0`
`<=>(2x+3)(x^2+1)=0`
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\x^2+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=-3\\x^2=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x\in\varnothing\end{matrix}\right.\)
\(4x^2-12x=-9\)
\(\Leftrightarrow4x^2-12x+9=0\)
\(\Leftrightarrow\left(2x\right)^2-2\cdot2x\cdot3+3^2=0\)
\(\Leftrightarrow (2x-3)^2=0\)
\(\Leftrightarrow 2x-3=0\)
\(\Leftrightarrow 2x=3\)
\(\Leftrightarrow x=\dfrac{3}{2}\)
\(---\)
\(x(x-1)+2x-2=0\)
\(\Leftrightarrow x(x-1)+2(x-1)=0\)
\(\Leftrightarrow (x-1)(x+2)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
\(---\)
\(2x^3+3x^2+2x+3=0\)
\(\Leftrightarrow (2x^3+3x^2)+(2x+3)=0\)
\(\Leftrightarrow x^2(2x+3)+(2x+3)=0\)
\(\Leftrightarrow (2x+3)(x^2+1)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\x^2+1=0\end{matrix}\right.\)
\(\Leftrightarrow2x+3=0\) (vì \(x^2+1>0\forall x\))
\(\Leftrightarrow2x=-3\)
\(\Leftrightarrow x=-\dfrac{3}{2}\)
#\(Toru\)
