1.

\(\Leftrightarrow6x^2-12x+7-6\sqrt{6x^2-12x+7}-7=0\)

Đặt \(\sqrt{6x^2-12x+7}=t>0\)

\(\Rightarrow t^2-6t-7=0\Rightarrow\left[{}\begin{matrix}t=-1\left(loại\right)\\t=7\end{matrix}\right.\)

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

\(\Leftrightarrow6x^2-12x+7=49\Rightarrow x=1\pm2\sqrt{2}\)

2.

\(\Delta'=\left(m+1\right)^2-m^2-3=2m-2>0\Rightarrow m>1\)

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1x_2=m^2+3\end{matrix}\right.\)

\(\left(x_1+x_2\right)^2-2x_1x_2=2x_1x_2+8\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-4x_1x_2-8=0\)

\(\Leftrightarrow4\left(m+1\right)^2-4\left(m^2+3\right)-8=0\)

\(\Leftrightarrow2m-4=0\Rightarrow m=2\)

\(ĐK:x\in R\)

Đặt \(x^2-2x=a\), PTTT:

\(-a+\sqrt{6a+7}=0\\ \Leftrightarrow\sqrt{6a+7}=a\\ \Leftrightarrow a^2-6a-7=0\\ \Leftrightarrow\left[{}\begin{matrix}a=7\\a=-1\left(loại.do.a=\sqrt{6a+7}\ge0\right)\end{matrix}\right.\\ \Leftrightarrow a=7\\ \Leftrightarrow x^2-2x-7=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1+2\sqrt{2}\\x=1-2\sqrt{2}\end{matrix}\right.\)

a: ĐKXĐ: \(x^2-6x+6\ge0\)

=>\(x^2-6x+9-3\ge0\)

=>\(\left(x-3\right)^2-3\ge0\)

=>\(\left(x-3\right)^2\ge3\)

=>\(\left[\begin{array}{l}x-3\ge\sqrt3\\ x-3\le-\sqrt3\end{array}\right.\Rightarrow\left[\begin{array}{l}x\ge\sqrt3+3\\ x\le-\sqrt3+3\end{array}\right.\)

Ta có: \(x^2-6x+9=4\sqrt{x^2-6x+6}\)

=>\(x^2-6x+6-4\cdot\sqrt{x^2-6x+6}+3=0\)

=>\(\left(\sqrt{x^2-6x+6}-3\right)\left(\sqrt{x^2-6x+6}-1\right)=0\)

TH1: \(\sqrt{x^2-6x+6}-3=0\)

=>\(\sqrt{x^2-6x+6}=3\)

=>\(x^2-6x+6=9\)

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

=>\(x^2-6x+9-12=0\)

=>\(\left(x-3\right)^2=12\)

=>\(\left[\begin{array}{l}x-3=2\sqrt3\\ x-3=-2\sqrt3\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\sqrt3+3\left(nhận\right)\\ x=3-2\sqrt3\left(nhận\right)\end{array}\right.\)

TH2: \(\sqrt{x^2-6x+6}-1=0\)

=>\(x^2-6x+6=1\)

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

=>(x-1)(x-5)=0

=>\(\left[\begin{array}{l}x=1\left(nhận\right)\\ x=5\left(nhận\right)\end{array}\right.\)

b: ĐKXĐ: x∈R

\(x^2-x+8-4\sqrt{x^2-x+4}=0\)

=>\(x^2-x+4-4\cdot\sqrt{x^2-x+4}+4=0\)

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

=>\(\sqrt{x^2-x+4}-2=0\)

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

=>\(x^2-x+4=4\)

=>\(x^2-x=0\)

=>x(x-1)=0

=>x=0 hoặc x=1

c: \(x^2+\sqrt{4x^2-12x+44}=3x+4\)

=>\(x^2-3x-4+2\sqrt{x^2-3x+11}=0\)

=>\(x^2-3x+11+2\sqrt{x^2-3x+11}-15=0\)

=>\(\left(\sqrt{x^2-3x+11}+5\right)\left(\sqrt{x^2-3x+11}-3\right)=0\)

=>\(\sqrt{x^2-3x+11}-3=0\)

=>\(\sqrt{x^2-3x+11}=3\)

=>\(x^2-3x+11=9\)

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

=>(x-1)(x-2)=0

=>x=1(nhận) hoặc x=2(nhận)

pt <=>\(\sqrt{6x^2-12x+7}-\left(x^2-2x\right)=0\)

<=>\(\sqrt{6\left(x^2-2x+1\right)+1}-\left(x^2-2x+1\right)+1=0\)

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

Đặt \(\left(x-1\right)^2=a\left(a\ge0\right)\)

Có \(\sqrt{6a+1}-a=-1\)

<=> \(\sqrt{6a+1}=a-1\)

=> \(6a+1=a^2-2a+1\)

<=> \(a^2-2a-6a+1-1=0\)

<=>\(a^2-8a=0\) <=>a(a-8)=0

=> \(\left[{}\begin{matrix}a=0\\a=8\end{matrix}\right.\) <=>\(\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(x-1\right)^2=8\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=1\left(ktm\right)\\x=2\sqrt{2}+1\left(tm\right)\\x=1-2\sqrt{2}\left(tm\right)\end{matrix}\right.\)

阮芳邵族 bạn có thể thấy trong căn luôn > hoặc = 1 => bt trong căn >0

=>luôn t/m với mọi x.

Điều kiện xác định của pt : \(6x^2-12x+7\ge0\) => Với mọi số thực thì pt xác định

Ta có : \(2x-x^2+\sqrt{6x^2-12x+7}=0\)

\(\Leftrightarrow-\left(6x^2-12x+7\right)+6\sqrt{6x^2-12x+7}+7=0\)

Đặt \(t=\sqrt{6x^2-12x+7},t\ge0\) . pt trở thành : \(-t^2+6t+7=0\) \(\Leftrightarrow\left[\begin{array}{nghiempt}t=7\left(\text{nhận}\right)\\t=-1\left(\text{loại}\right)\end{array}\right.\)

Với \(t=7\) ta có pt : \(6x^2-12x+7=49\)

\(\Leftrightarrow6x^2-12x-42=0\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=1-2\sqrt{2}\\x=1+2\sqrt{2}\end{array}\right.\)

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

<=>\(t^2-7=6x^2-12x\)

\(\Leftrightarrow\dfrac{t^2-7}{6}=x^2-2x\)

Ta có pt mới:

\(\dfrac{7-t^2}{6}+t=0\)

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

\(\Leftrightarrow t^2-2\cdot t\cdot3+9-9-7=0\)

\(\Leftrightarrow\left(t-3\right)^2=16\)

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

Với t=7

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

<=>6x²-12x+7=49

<=>6x²-12x-42=0

<=>x²-2x-7=0

<=>(x-1)²=8

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