b) X=5

a/ \(x^2-6x+9=4\sqrt{x^2-6x+6}\)

\(\Leftrightarrow\left(x^2-6x+6\right)-4\sqrt{x^2-6x+6}+3=0\)

Đặt \(a=\sqrt{x^2-6x+6},a\ge0\)

PT trên trở thành \(a^2-4a+3=0\Leftrightarrow\left(a-3\right)\left(a-1\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}a=1\\a=3\end{array}\right.\)

Tới đây bạn tự giải nhé ^^

b/ \(x^2+\sqrt{x^2+11}=31\Leftrightarrow\left(x^2+11\right)+\sqrt{x^2+11}-42=0\)

Đặt \(b=\sqrt{x^2+11},b>0\) thì pt trở thành

\(b^2+b-42=0\Leftrightarrow\left(b-6\right)\left(b+7\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}b=6\\b=-7\end{array}\right.\) (chú ý loại b = -7)

Tới đây bạn tự giải nhé ^^

X=1

\(\sqrt{6x^2-12x+7}=x^2-2x\)

\(\Leftrightarrow\sqrt{6x^2-12x+7}=\dfrac{6x^2-12x+7-7}{6}\left(1\right)\)

Đặt \(\sqrt{6x^2-12x+7}=t\left(t\ge0\right)\)

\(\left(1\right)\Leftrightarrow t=\dfrac{t^2}{6}-\dfrac{7}{6}\)

\(\Leftrightarrow t^2-6t-7=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=7\left(TM\right)\\t=-1\left(loại\right)\end{matrix}\right.\)

t=7\(\Rightarrow\sqrt{6x^2-12x+7}=7\)

\(\Leftrightarrow6x^2-12x+7=49\)

\(\Leftrightarrow x^2-2x-7=0\)

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

\(\sqrt{x^2-4x+5}=2x^2-8x\)

\(\Leftrightarrow\sqrt{x^2-4x+5}=2\left(x^2-4x+5\right)-10\)(1)

đặt \(t=\sqrt{x^2-4x+5}\) (t\(\ge\)0)

\(\left(1\right)\Leftrightarrow t=2t^2-10\)

\(\Leftrightarrow\left[{}\begin{matrix}t=-2\left(loại\right)\\t=\dfrac{5}{2}\left(TM\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x^2-4x+5}=\dfrac{5}{2}\)

\(\Leftrightarrow x-4-\dfrac{5}{4}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4+\sqrt{21}}{2}\left(TM\right)\\x=\dfrac{4-\sqrt{21}}{2}\left(TM\right)\end{matrix}\right.\)