hình như bạn ghi sai đề phải k ạ 

Sai đề :D

a) \(\sqrt{9-12x+4x^2}=4+x\Leftrightarrow\sqrt{\left(3-2x\right)^2}=4+x\)

\(\Leftrightarrow\left|3-2x\right|=4+x\)

th1: \(3-2x\ge0\Leftrightarrow2x\le3\Leftrightarrow\Leftrightarrow x\le\dfrac{3}{2}\)

\(\Rightarrow\left|3-2x\right|=4+x\Leftrightarrow3-2x=4+x\Leftrightarrow3x=-1\Leftrightarrow x=\dfrac{-1}{3}\left(tmđk\right)\)

th2: \(3-2x< 0\Leftrightarrow2x>3\Leftrightarrow x>\dfrac{3}{2}\)

\(\Rightarrow\left|3-2x\right|=4+x\Leftrightarrow2x-3=4+x\Leftrightarrow x=7\left(tmđk\right)\)

vậy \(x=\dfrac{-1}{3};x=7\)

b) \(\sqrt{4-4x+x^2}=\left(x-1\right)^2+x-6\)

\(\Leftrightarrow\sqrt{\left(2-x\right)^2}=x^2-2x+1+x-6\)

\(\Leftrightarrow\left|2-x\right|=x^2-x-5\)

th1: \(2-x\ge0\Leftrightarrow x\le2\)

\(\Rightarrow\left|2-x\right|=x^2-x-5\Leftrightarrow2-x=x^2-x-5\)

\(\Leftrightarrow x^2=7\Leftrightarrow\left\{{}\begin{matrix}x=\sqrt{7}\left(loại\right)\\x=-\sqrt{7}\left(tmđk\right)\end{matrix}\right.\)

th2: \(2-x< 0\Leftrightarrow x>2\)

\(\Rightarrow\left|2-x\right|=x^2-x-5\Leftrightarrow x-2=x^2-x-5\)

\(\Leftrightarrow x^2-2x-3=0\Leftrightarrow x^2+x-3x-3=0\)

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

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

vậy \(x=-\sqrt{7};x=3\)

a) \(\sqrt{9-12x+4x^2}=4+x\)

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

\(\Leftrightarrow\left|3-2x\right|=4+x\)

\(\Leftrightarrow\left[{}\begin{matrix}3-2x=4+x\\3-2x=-4-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\x=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=7\end{matrix}\right.\)

Vậy \(x_1=-\dfrac{1}{3};x_2=7\).

b) \(\sqrt{4-4x+x^2}=\left(x-1\right)^2+x-6\)

\(\Leftrightarrow\sqrt{\left(2-x\right)^2}=x^2-2x+1+x-6\)

\(\Leftrightarrow\left|2-x\right|=x^2-x-5\)

\(\Leftrightarrow\left[{}\begin{matrix}2-x=x^2-x-5\\2-x=-x^2+x+5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2=7\\x^2=2x+3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\left(l\right)\\x=-\sqrt{7}\\x=3\\x=-1\left(l\right)\end{matrix}\right.\)

Vậy \(x_1=-\sqrt{7};x_2=3\).

ĐKXĐ:...

\(VT\le\dfrac{\left(x^2+x-1\right)+1}{2}+\dfrac{x-x^2+1+1}{2}=x+1\)

\(=2x-x+1\le x^2+1-x+1=VP\)

Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}x^2+x-1=1\\x-x^2+1=1\end{matrix}\right.\Leftrightarrow x=1\)

\(\left(x+3\right)^2=x^2+6x+9\le x^2+\left(9x^2+1\right)+9=10\left(x^2+1\right)\)

Suy ra: \(P=\dfrac{x+3}{\sqrt{x^2+1}}\le\sqrt{10}\)

Vậy \(MaxP=\sqrt{10}\) (khi \(x=\dfrac{1}{3}\))

a,  \(\Leftrightarrow\sqrt{\left(3-2x\right)^2=4+x}\)

\(\Leftrightarrow\left|3-2x\right|=4+x\)

\(\Leftrightarrow\orbr{\begin{cases}3-2x=4+x\\3-2x=-4-x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3x=-1\\x=7\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{3}\\x=7\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\orbr{\begin{cases}x=\sqrt{7}\\x=-\sqrt{7}\end{cases}}\\\left(x-3\right)\left(x-1\right)=0\end{cases}}\)