a) \(x^2-12x+11\)\(=0\)
\(\Leftrightarrow\left(x-6\right)^2-25=0\)
\(\Leftrightarrow\left(x-6+5\right)\left(x-6-5\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-11=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=11\end{matrix}\right.\)
a)\(x^2-12x+11=0\)
\(x^2-x-11x+11=0\)
\(\left(x^2-x\right)-\left(11x-11\right)=0\)
\(x\left(x-1\right)-11\left(x-1\right)=0\)
\(\left(x-1\right)\left(x-11\right)=0\)
\(=>\left[{}\begin{matrix}x-1=0\\x-11=0\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=1\\x=11\end{matrix}\right.\)
b)\(4x^2-4x-3=0\)
\(4x^2-2x+6x-3=0\)
\(2x\left(2x-1\right)+3\left(3x-1\right)=0\)
\(\left(2x-1\right)\left(2x+3\right)=0\)
\(=>\left[{}\begin{matrix}2x-1=0\\2x+3=0\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=0,5\\x=-1,5\end{matrix}\right.\)\
c)\(4x^2-12x-7=0\)
\(4x^2-14x+2x-7=0\)
\(2x\left(2x-7\right)+\left(2x-7\right)=0\)
\(\left(2x-7\right)\left(2x+1\right)=0\)
\(=>\left[{}\begin{matrix}2x-7=0\\2x+1=0\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=3,5\\x=-0,5\end{matrix}\right.\)
d)\(x^3-6x^2=8-12x\)
\(=>\left(x^3-6x^2\right)-\left(8-12x\right)=0\)
\(=>x^3-6x^2-8+12x=0\)
\(x^3-3x^2.2+3x.2^2-2^3=0\)
\(\left(x-2\right)^3=0\)
\(=>x-2=0\)
\(=>x=2\)
a ) x\(^2\) - 12x + 11 =0
⇔ x\(^2\) - 11x - x + 11 = 0
⇔ x ( x - 11 ) - ( x - 11 ) = 0
⇔ ( x - 1 ) ( x - 11 ) =0
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=11\end{matrix}\right.\)
b ) 4x\(^2\) - 4x - 3 = 0
⇔ 4x\(^2\) - 6x + 2x - 3 = 0
⇔ 2x ( 2x - 3 ) + ( 2x - 3 ) = 0
⇔ ( 2x + 1 ) ( 2x - 3 ) =0
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=\frac{3}{2}\end{matrix}\right.\)
c ) 4x\(^2\) - 12x - 7 = 0
⇔ 4x\(^2\) - 14x + 2x - 7 = 0
⇔ 2x ( 2x + 1 ) - 7 ( 2x + 1 ) = 0
⇔ ( 2x + 1 ) ( 2x - 7 ) = 0
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\2x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=\frac{7}{2}\end{matrix}\right.\)
d ) x\(^3\) - 6x\(^2\) = 8 - 12x
⇔ x\(^3\) - 6x\(^2\) - 8 + 12x = 0
⇔ ( x\(^3\) - 8 ) - ( 6x\(^2\) - 12x ) =0
⇔ (x - 2 ) ( x\(^2\) + 2x + 4 ) - 6x ( x - 2 ) = 0
⇔ ( x- 2 ) ( x\(^2\) + 2x + 4 - 6x ) = 0
⇔ ( x - 2 ) ( x\(^2\) - 4x + 4 ) =0
⇔ ( x - 2 ) ( x- 2 )\(^2\) = 0
⇔ x = 2
