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 >>

\(\)

a, 3x² +12x=0

    3x(x+4)=0

=> 3x=0 hoặc x+4=0

=> x=0 hoặc x= -4

Vậy x=0; x= -4

b, 4x³ = 4x

    4x³- 4x=0

    4x(x²- 1) =0

     4x(x-1)(x+1)=0

     => 4x=0 hoặc x-1=0 hoặc x+1=0

     => x=0 hoặc x=1 hoặc x=-1

    Vậy x=0; x=1;x=-1

c, ( x-1)(x+1)+2=0

     x²- 1+2=0

     x²+1=0

      x² = -1

  => x vô nghiệm

a: \(A=\left(\dfrac{2\left(2x+1\right)}{2\left(2x+4\right)}-\dfrac{x}{3x-6}-\dfrac{2x^3}{3x^3-12x}\right):\dfrac{6x+13x^2}{24x-12x^2}\)

\(=\left(\dfrac{2x+1}{2\left(x+2\right)}-\dfrac{x}{3\left(x-2\right)}-\dfrac{2x^3}{3x\left(x^2-4\right)}\right):\dfrac{x\left(13x+6\right)}{x\left(24-12x\right)}\)

\(=\left(\dfrac{2x+1}{2\left(x+2\right)}-\dfrac{x}{3\left(x-2\right)}-\dfrac{2x^2}{3\left(x-2\right)\left(x+2\right)}\right):\dfrac{13x+6}{-12\left(x-2\right)}\)

\(=\dfrac{3\left(2x+1\right)\left(x-2\right)-2x\left(x+2\right)-4x^2}{6\left(x+2\right)\left(x-2\right)}\cdot\dfrac{-12\left(x-2\right)}{13x+6}\)

\(=\dfrac{3\left(2x^2-3x-2\right)-2x^2-4x-4x^2}{x-2}\cdot\dfrac{-2}{13x+6}\)

\(=\dfrac{6x^2-9x-6-6x^2-4x}{x-2}\cdot\dfrac{-2}{13x+6}\)

\(=\dfrac{-\left(13x+6\right)\cdot\left(-2\right)}{\left(13x+6\right)\left(x-2\right)}=\dfrac{2}{x-2}\)

b: Để A>0 thì x-2>0

hay x>2

Để A>-1 thì A+1>0

\(\Leftrightarrow\dfrac{2+x-2}{x-2}>0\)

=>x/x-2>0

=>x>2 hoặc x<0

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

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

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

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

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

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

\(4x^5-12x^3=0\Leftrightarrow4x^3\left(x^2-3\right)=0\Leftrightarrow4x^3\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)=0\)

\(\Leftrightarrow x=0\text{ hoặc }x=\sqrt{3}\text{ hoặc }x=-\sqrt{3}\)

a) 3x² + 12x =0

<=> 3x( x+ 4)=0

<=> \(\orbr{\begin{cases}3x=0\\x+4=0\end{cases}}\) <=>\(\orbr{\begin{cases}x=0\\x=-4\end{cases}}\) 

d) \(4x^3=4x\)

<=> \(4x^3-4x=0\) 

<=> 4x( x² -1) =0

<=> 4x ( x - 1) ( x+ 1) =0

<=> 4x=0 hoac x-1=0 hoac x+1=0

<=> x=0 hoac x=1 hoac x=-1

a,  4x^2 - 4x = -1

\(\Leftrightarrow\)4x^2 - 4x + 1 = 0

\(\Leftrightarrow\)(2x-1)²              =0 

\(\Leftrightarrow\)2x - 1          = 0 

\(\Leftrightarrow\)x                = 1/2

b, \(\Leftrightarrow\)( 2x + 1)^3 = 0

\(\Leftrightarrow\)2x + 1 = 0 

\(\Leftrightarrow\)x       = -1/2

đúng thì

a) \(4x^2-4x=-1\)

\(\Leftrightarrow4x^2-4x+1=0\)

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

\(\Leftrightarrow2x-1=0\)

\(\Leftrightarrow x=\frac{1}{2}\)

b) \(8x^3+12x^2+6x+1=0\)

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

\(\Leftrightarrow2x+1=0\)

\(\Leftrightarrow x=-\frac{1}{2}\)

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

\(d,x\left(x-3\right)-7x+21=0\)

\(\Leftrightarrow x\left(x-3\right)-7\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-7\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x-7=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=7\end{cases}}}\)

\(a,2x\left(x-7\right)+5x-35=0\)

 \(\Leftrightarrow2x\left(x-7\right)+5\left(x-7\right)=0\)

 \(\Leftrightarrow\left(x-7\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\2x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-\frac{5}{2}\end{cases}}}\)

\(c,4x^2+12x+9=0\)

\(\Leftrightarrow4x^2+6x+6x+9=0\)

\(\Leftrightarrow2x\left(2x+3\right)+3\left(2x+3\right)=0\)

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

\(\Leftrightarrow2x+3=0\)

\(\Leftrightarrow x=-\frac{3}{2}\)