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)\(4x^2-12x-7=0\)

\(4x^2-14x+2x-7=0\)

\(2x\left(2x-7\right)+\left(2x-7\right)=0\)

\(\left(2x-7\right)\left(2x+1\right)=0\)

\(=>\left[{}\begin{matrix}2x-7=0\\2x+1=0\end{matrix}\right.\)

\(=>\left[{}\begin{matrix}x=3,5\\x=-0,5\end{matrix}\right.\)

1. \(4x^2-49=0\)

\(\Leftrightarrow\left(2x+7\right)\left(2x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+7=0\Leftrightarrow x=-\dfrac{7}{2}\\2x-7=0\Leftrightarrow x=\dfrac{7}{2}\end{matrix}\right.\)

Vậy: \(x=-\dfrac{7}{2}\) hoặc \(x=\dfrac{7}{2}\)

===========

2. \(x^2+36=12x\)

\(\Leftrightarrow x^2-12x+36=0\)

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

\(\Leftrightarrow x=6\)

Vậy: \(x=6\)

===========

3. \(10\left(x-5\right)-8x\left(5-x\right)=0\)

\(\Leftrightarrow10\left(x-5\right)+8x\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(10+8x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\Leftrightarrow x=5\\10+8x=0\Leftrightarrow x=-\dfrac{5}{4}\end{matrix}\right.\)

Vậy: \(x=5\) hoặc \(x=-\dfrac{5}{4}\)

1: Ta có: \(4x^2-49=0\)

\(\Leftrightarrow\left(2x-7\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

2: Ta có: \(x^2+36=12x\)

\(\Leftrightarrow x^2-12x+36=0\)

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

\(\Leftrightarrow x-6=0\)

hay x=6

1,=\(x^2-3x-2x^2+6x=-x^2+3x\)

2,=\(3x^2-x-5+15x=3x^2+14x-5\)

3,=\(5x+15-6x^2-6x=-6x^2-x+15\)

4,=\(4x^2+12x-x-3=4x^2+11x-3\)

5: =>(x+5)^3=0

=>x+5=0

=>x=-5

6: =>(2x-3)^2=0

=>2x-3=0

=>x=3/2

7: =>(x-6)(x-10)=0

=>x=10 hoặc x=6

8: \(\Leftrightarrow x^3-12x^2+48x-64=0\)

=>(x-4)^3=0

=>x-4=0

=>x=4

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 !!

Làm tiếp nha >>>

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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-_- bài này hôm qua lm rùi

a,4x^2-4x+1=0

  4x^2-2x-2x+1=0

  2x (2x-1)-(2x-1)=0

  (2x-1)(2x-1)=0

  (2x-1)^2=0

=>2x-1=0 <=> x=1/2

\(\left(x+1\right)\left(x+2\right)-\left(x+2\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x+1-x-3\right)=0\)

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

\(\Leftrightarrow x=-2\)

\(4x^2-12x+5=4x^2-10x-2x+5=2x\left(2x-5\right)-\left(2x-5\right)=\left(2x-1\right)\left(2x-5\right)\)

\(\left(x+1\right)\left(x+2\right)-\left(x+2\right)\left(x+3\right)=0\)

\(\left(x+2\right)\left(-2\right)=0\)\(\Rightarrow x+2=0\) hay \(x=-2\)

b: \(\Leftrightarrow48x^2-12x-20x+5-48x^2+36x=30\)

\(\Leftrightarrow4x=25\)

hay \(x=\dfrac{25}{4}\)