\(x^2-6x+9=4.\sqrt{x^2-6x+6}\)\(ĐK:x^2-6x+6\ge0\)

Đặt \(\sqrt{x^2-6x+6}=t\)\(\left(ĐK:t\ge0\right)\)

\(\Leftrightarrow t^2=x^2-6x+6\)

\(\Leftrightarrow x^2-6x=t-6\)thay vào pt ta được : 

\(\Leftrightarrow t^2-6+9=4t\)

\(\Leftrightarrow t^2-4t+3=0\)\(\Leftrightarrow\orbr{\begin{cases}t=1\\t=3\end{cases}}\)

Với \(t=1\Rightarrow\sqrt{x^2-6x+6}=1\)

                  \(\Leftrightarrow x^2-6x+5=0\)

                   \(\Leftrightarrow\orbr{\begin{cases}x=1\left(TM\right)\\x=5\left(TM\right)\end{cases}}\)

Với \(t=3\Rightarrow\sqrt{x^2-6x+6}=3\)

                   \(\Leftrightarrow x^2-6x+6=0\)

                    \(\Leftrightarrow\orbr{\begin{cases}x=3+\sqrt{6}\left(TM\right)\\x=3-\sqrt{6}\left(TM\right)\end{cases}}\)

\(\Leftrightarrow x^2-6x+8=6\sqrt{2x+1}-18\left(Đk:x\ge-\dfrac{1}{2}\right)\)

\(\Leftrightarrow\left(x-2\right)\left(x-4\right)=\dfrac{12\left(x-4\right)}{\sqrt{2x+1}+3}\left(\sqrt{2x+1}+3>0\right)\)

+) \(x=4\left(TM\right)\)

+) \(x\ne4\Rightarrow x-2=\dfrac{12}{\sqrt{2x+1}+3}\)

            \(\Leftrightarrow x-4=\dfrac{12-2\left(\sqrt{2x+1}+3\right)}{\sqrt{2x+1}+3}\)

             \(\Leftrightarrow x-4+\dfrac{2\left(x-4\right)}{\left(\sqrt{2x+1}+3\right)^2}=0\)

             \(\Leftrightarrow1+\dfrac{2}{\left(\sqrt{2x+1}+3\right)^2}=0\left(x\ne4\right)\)

    Vì \(\dfrac{2}{\left(\sqrt{2x+1}+3\right)^2}>0\forall x\) => VT>0

=> phương trình vô nghiệm

Vậy \(S=\left\{4\right\}\)

a) \(=x\left(x-5\right)+\left(x-5\right)^2=\left(x-5\right)\left(x+x-5\right)=\left(x-5\right)\left(2x-5\right)\)

b) \(=x^2-2.x.10+10^2=\left(x-10\right)^2\)

c) \(=x\left(x+3\right)+2\left(x+3\right)=\left(x+3\right)\left(x+2\right)\)

\(|x-6|=-5x+9\)

Xét \(x\ge6\)thì \(pt< =>x-6=-5x+9\)

\(< =>x-6+5x-9=0\)

\(< =>6x-15=0\)

\(< =>x=\frac{15}{6}\)(ktm)

Xét \(x< 6\)thì \(pt< =>x-6=5x-9\)

\(< =>4x-9+6=0\)

\(< =>4x-3=0< =>x=\frac{3}{4}\)(tm)

Vậy ...

\(|x+1|=x^2+x\)

Xét \(x\ge-1\)thì \(pt< =>x+1=x^2+x\)

\(< =>x^2+x-x-1=0\)

\(< =>\left(x-1\right)\left(x+1\right)=0\)

\(< =>\orbr{\begin{cases}x=1\\x=-1\end{cases}\left(tm\right)}\)

Xét \(x< -1\)thì \(pt< =>-x-1=x^2+x\)

\(< =>x^2+2x+1=0\)

\(< =>\left(x+1\right)^2=0\)

\(< =>x=-1\left(ktm\right)\)

Vậy ...

a: Sửa đề: PT x^2-2x-m-1=0

Khi m=2 thì Phương trình sẽ là:

x^2-2x-2-1=0

=>x^2-2x-3=0

=>(x-3)(x+1)=0

=>\(\left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

b:

\(\text{Δ}=\left(-2\right)^2-4\left(-m-1\right)\)

\(=4+4m+4=4m+8\)

Để phương trình có hai nghiệm dương thì

\(\left\{{}\begin{matrix}4m+8>0\\2>0\\-m-1>0\end{matrix}\right.\Leftrightarrow-2< m< -1\)

\(\sqrt{x_1}+\sqrt{x_2}=2\)

=>\(x_1+x_2+2\sqrt{x_1x_2}=4\)

=>\(2+2\sqrt{-m-1}=4\)

=>\(2\sqrt{-m-1}=2\)

=>-m-1=1

=>-m=2

=>m=-2(loại)