a) \(x^2-12x+11=0\)
\(\Leftrightarrow x^2-2.6.x+36-25=0\)
\(\Leftrightarrow\left(x-6\right)^2-25=0\)
\(\Leftrightarrow\left(x-6\right)^2=25=5^2=\left(-5\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=5\\x-6=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=11\\x=1\end{matrix}\right.\)
Vậy : \(x\in\left\{11,1\right\}\)
c) \(4x^2-12x-7=0\)
\(\Leftrightarrow\left(2x\right)^2-2.2x.3+9-16=0\)
\(\Leftrightarrow\left(2x-3\right)^2-16=0\)
\(\Leftrightarrow\left(2x-3\right)^2=16=4^2=\left(-4\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=4\\2x-3=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=7\\2x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{2}\\x=-\frac{1}{2}\end{matrix}\right.\)
Vậy : \(x\in\left\{\frac{7}{2},-\frac{1}{2}\right\}\)
Câu b) và d) xíu em làm sau, em hơi bận chút !!
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b) \(4x^2-4x-3=0\)
\(\Leftrightarrow\left(2x\right)^2-2.2x.1+1-4=0\)
\(\Leftrightarrow\left(2x-1\right)^2-4=0\)
\(\Leftrightarrow\left(2x-1\right)^2=4=2^2=\left(-2\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=2\\2x-1=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=3\\2x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=-\frac{1}{2}\end{matrix}\right.\)
Vậy : \(x\in\left\{\frac{3}{2},-\frac{1}{2}\right\}\)
d) \(x^3-6x^2=8-12x\)
\(\Leftrightarrow x^3-6x^2-\left(8-12x\right)=0\)
\(\Leftrightarrow x^3-6x^2-8+12x=0\)
\(\Leftrightarrow x^3-3.x^2.2+3.x.2^2-2^3=0\)
\(\Leftrightarrow\left(x-2\right)^3=0\)
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)
Vậy : \(x=2\)
P/s : Hằng đẳng thức với lập phương khó thật, rối câu d) mãi mới nghĩ ra >>
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