mình sửa đề câu 1 

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

\(ĐK:x\le12\)

Đặt \(\hept{\begin{cases}\sqrt[3]{24+x}=a\\\sqrt{12-x}=b\end{cases}\left(b\ge0\right)\Rightarrow}a^3+b^2=36\)

PT trở thành a+b=6

Ta có hệ phương trình \(\hept{\begin{cases}a+b=6\\a^3+b^2=36\end{cases}\Leftrightarrow}\hept{\begin{cases}b=6-a\\a^3+a^2-12a=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}b=6-a\\a\left(a-3\right)\left(a+4\right)=0\end{cases}}\)

Đến đây đơn giản rồi nhé

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

\(ĐK:\orbr{\begin{cases}x\le0\\x\ge3\end{cases}}\)

Đặt \(\sqrt{x^2-3x}=a\left(a\ge0\right)\)

\(PT\Leftrightarrow a^2-6+a=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=2\left(tm\right)\\a=-3\left(loai\right)\end{cases}}\)

\(\Rightarrow x^2-3x-4=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=-1\end{cases}}\)

ĐKXĐ :  -1 <= x <= 3 

XH : \(\left(-x^2+4x+12\right)-\left(x^2+2x+3\right)=2x+9>0\)

=> VT > 0 

VÌ -1 <=x <=3  => VT = \(\sqrt{x+2}\sqrt{6-x}-\sqrt{x+1}.\sqrt{3-x}\)

Áp dụng BĐT \(\left(ab-cd\right)^2\le\left(a^2-c^2\right)\left(b^2-d^2\right)\) ta có :

\(VT^2=\left(\sqrt{x+2}\sqrt{6-x}-\sqrt{x+1}\sqrt{3-x}\right)^2\ge\left(x+2-x-1\right)\left(6-x-3+x\right)=1.3=3\)

=> VT \(\ge\sqrt{3}\) dấu bằng xảy ra khi \(\left(x+2\right)\left(6-x\right)=\left(x+1\right)\left(3-x\right)\) <=> x = 0 

VP = \(\sqrt{3}-x^2\le\sqrt{3}\)

Dấu bằng xảy ra khi x = 0 

Để VT bằng VP => x = 0 

ĐK: \(-3\le x\le2\)

Đặt: \(\left\{{}\begin{matrix}\sqrt{x+3}=a\\\sqrt{2-x}=b\end{matrix}\right.\left(a,b\ge0\right)\)

\(PT\Leftrightarrow a+b-ab=1\)

\(\Leftrightarrow\left(a-1\right)\left(1-b\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}a=1\left(tm\right)\\b=1\left(tm\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x+3}=1\\\sqrt{2-x}=1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x+3=1\\2-x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\) (tm)

Vậy....

ĐKXĐ: \(x\ge2\)

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

\(\Leftrightarrow2\left(x-3\right)+\frac{9\left(x-2\right)-\left(x+6\right)}{3\sqrt{x-2}+\sqrt{x+6}}=0\)

\(\Leftrightarrow2\left(x-3\right)+\frac{8\left(x-3\right)}{3\sqrt{x-2}+\sqrt{x+6}}=0\)

\(\Leftrightarrow\left(x-3\right)\left(2+\frac{8}{3\sqrt{x-2}+\sqrt{x+6}}\right)=0\) (1)

Với \(x\ge2\Rightarrow2+\frac{8}{3\sqrt{x-2}+\sqrt{x+6}}>0\)

(1) <=> x-3=0 <=> x=3 (tm ĐKXĐ)

Vậy x=3