Ta có :\(x+6\sqrt{x+8}+4\sqrt{6-2x}=27\Leftrightarrow6\sqrt{x+8}+4\sqrt{6-2x}=27-x\) ( Điều kiện :-8 <= x <= 3 ) 

Mà VT = \(6\sqrt{x+8}+4\sqrt{6-2x}=2\sqrt{9.\left(x+8\right)}+2\sqrt{4.\left(6-2x\right)}\)
Áp dụng cauchy :
\(2\sqrt{9.\left(x+8\right)}\le9+\left(x+8\right)\) 
\(2\sqrt{4.\left(6-2x\right)}\le4+\left(6-2x\right)\)

=> \(VT=2\sqrt{9.\left(x+8\right)}+2\sqrt{4.\left(6-2x\right)}\le9+x+8+4+6-2x\le27-2x\)

=> \(VT\le VP\)

Dấu " = " xảy ra <=> \(\hept{\begin{cases}9=x+8\\4=6-2x\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\x=1\end{cases}\Leftrightarrow}x=1\left(tm\right)}\)

Vậy phương trình đã cho có 1 nghiệm x = 1 

ĐKXĐ: \(\left\{{}\begin{matrix}x+8>=0\\6-2x>=0\end{matrix}\right.\Leftrightarrow-8< =x< =3\)

\(PT\Leftrightarrow1=6\sqrt{x+8}-18+x+4\sqrt{6-2x}-8\)

\(\Leftrightarrow6\cdot\dfrac{x+8-9}{\sqrt{x+8}+3}+x-1+4\cdot\dfrac{6-2x-4}{\sqrt{6-2x}+2}=0\)

=>\(\left(x-1\right)\left(\dfrac{6}{\sqrt{x+8}+3}+1-\dfrac{4}{\sqrt{6-2x}+2}\right)=0\)

=>x-1=0

=>x=1

Lời giải:

a. ĐKXĐ: $x\geq -9$

PT $\Leftrightarrow x+9=7^2=49$

$\Leftrightarrow x=40$ (tm)

b. ĐKXĐ: $x\geq \frac{-3}{2}$

PT $\Leftrightarrow 4\sqrt{2x+3}-\sqrt{4(2x+3)}+\frac{1}{3}\sqrt{9(2x+3)}=15$

$\Leftrightarrow 4\sqrt{2x+3}-2\sqrt{2x+3}+\sqrt{2x+3}=15$

$\Leftrgihtarrow 3\sqrt{2x+3}=15$

$\Leftrightarrow \sqrt{2x+3}=5$

$\Leftrightarrow 2x+3=25$

$\Leftrightarrow x=11$ (tm)

c.

PT \(\Leftrightarrow \left\{\begin{matrix} 2x+1\geq 0\\ x^2-6x+9=(2x+1)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ 3x^2+10x-8=0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ (3x-2)(x+4)=0\end{matrix}\right.\)

\(\Leftrightarrow x=\frac{2}{3}\)

d. ĐKXĐ: $x\geq 1$

PT \(\Leftrightarrow \sqrt{(x-1)+4\sqrt{x-1}+4}-\sqrt{(x-1)+6\sqrt{x-1}+9}=9\)

\(\Leftrightarrow \sqrt{(\sqrt{x-1}+2)^2}-\sqrt{(\sqrt{x-1}+3)^2}=9\)

\(\Leftrightarrow \sqrt{x-1}+2-(\sqrt{x-1}+3)=9\)

\(\Leftrightarrow -1=9\) (vô lý)

Vậy pt vô nghiệm.

a: ĐKXĐ: \(x^2-5x-6>=0\)

=>(x-6)(x+1)>=0

=>\(\left[{}\begin{matrix}x>=6\\x< =-1\end{matrix}\right.\)

\(\sqrt{x^2-5x-6}=x-2\)

=>\(\left\{{}\begin{matrix}x-2>=0\\x^2-5x-6=\left(x-2\right)^2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=2\\x^2-5x-6=x^2-4x+4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=6\\-5x-6=-4x+4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=6\\-x=10\end{matrix}\right.\)

=>\(x\in\varnothing\)

b: ĐKXĐ: \(x\in R\)

\(\sqrt{x^2-8x+16}=4-x\)

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

=>|x-4|=4-x

=>x-4<=0

=>x<=4

c: ĐKXĐ: \(x^2-2x>=0\)

=>x(x-2)>=0

=>\(\left[{}\begin{matrix}x>=2\\x< =0\end{matrix}\right.\)

\(\sqrt{x^2-2x}=2-x\)

=>\(\left\{{}\begin{matrix}x^2-2x=\left(2-x\right)^2\\x< =2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x^2-2x=x^2-4x+4\\x< =2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x=4\\x< =2\end{matrix}\right.\Leftrightarrow x=2\left(nhận\right)\)

d: ĐKXĐ: x>=-27/2

\(\sqrt{2x+27}-6=x\)

=>\(\sqrt{2x+27}=x+6\)

=>\(\left\{{}\begin{matrix}x>=-6\\\left(x+6\right)^2=2x+27\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=-6\\x^2+12x+36-2x-27=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=-6\\x^2+10x+9=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=-6\\\left(x+9\right)\left(x+1\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-6\\x\in\left\{-9;-1\right\}\end{matrix}\right.\)

=>x=-1

Kết hợp ĐKXĐ, ta được: x=-1

a.

\(\sqrt{x^2-5x-6}=x-2\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2\ge0\\x^2-5x-6=\left(x-2\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x^2-5x-6=x^2-4x+4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x=-10\left(ktm\right)\end{matrix}\right.\)

Vậy pt đã cho vô nghiệm

b.

\(\sqrt{x^2-8x+16}=4-x\)

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

\(\Leftrightarrow\left|x-4\right|=-\left(x-4\right)\)

\(\Leftrightarrow x-4\le0\)

\(\Rightarrow x\le4\)

Cách nhóm thích hợp sẽ giúp chúng ta có

PT\(\Leftrightarrow-\left(x-1\right)^2\left(\frac{1}{\left(\sqrt{x+8}+3\right)^2}+\frac{4}{\left(\sqrt{6-2x}+2\right)^2}\right)=0\)

Bây giờ thì cái ngoặc to hiển nhiên > 0

Vậy x = 1

#Ý tưởng là từ bài chụy Ther em thấy cái ngoặc to cũng có nghiệm x = 1 nên em tiếp tục nhân liên hợp cho nó đẹp:D

_{P/s: Đây là cmt thứ 4000 của em! Mọi người có chúc gì cho em hok:))}

\(PT\Leftrightarrow x-1+6\left(\sqrt{x+8}-3\right)+4\left(\sqrt{6-2x}-2\right)=0\)

\(\Leftrightarrow x-1+\frac{6\left(x-1\right)}{\sqrt{x+8}+3}+\frac{4\left(2-2x\right)}{\sqrt{6-2x}+2}=0\)

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

\(\Leftrightarrow x=1\)(tm)

a: ĐKXĐ: \(x\in R\)

\(\sqrt{x^2-4x+4}=7\)

=>\(\sqrt{\left(x-2\right)^2}=7\)

=>|x-2|=7

=>\(\left[{}\begin{matrix}x-2=7\\x-2=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-5\end{matrix}\right.\)

b: ĐKXĐ: x>=-3

\(\sqrt{4x+12}-3\sqrt{x+3}+\dfrac{4}{3}\cdot\sqrt{9x+27}=6\)

=>\(2\sqrt{x+3}-3\sqrt{x+3}+\dfrac{4}{3}\cdot3\sqrt{x+3}=6\)

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

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

=>x+3=4

=>x=1(nhận)

c, \(\sqrt{9x-9}-2\sqrt{x-1}=8\left(đk:x\ge1\right)\)

\(< =>\sqrt{9\left(x-1\right)}-2\sqrt{x-1}=8\)

\(< =>\sqrt{9}.\sqrt{x-1}-2\sqrt{x-1}=8\)

\(< =>3\sqrt{x-1}-2\sqrt{x-1}=8\)

\(< =>\sqrt{x-1}=8< =>\sqrt{x-1}=\sqrt{8}^2=\left(-\sqrt{8}\right)^2\)

\(< =>\orbr{\begin{cases}x-1=8\\x-1=-8\end{cases}< =>\orbr{\begin{cases}x=9\left(tm\right)\\x=-7\left(ktm\right)\end{cases}}}\)

d, \(\sqrt{x-1}+\sqrt{9x-9}-\sqrt{4x-4}=4\left(đk:x\ge1\right)\)

\(< =>\sqrt{x-1}+\sqrt{9\left(x-1\right)}-\sqrt{4\left(x-1\right)}=4\)

\(< =>\sqrt{x-1}+\sqrt{9}.\sqrt{x-1}-\sqrt{4}.\sqrt{x-1}=4\)

\(< =>\sqrt{x-1}+3\sqrt{x-1}-2\sqrt{x-1}=4\)

\(< =>\sqrt{x-1}\left(1+3-2\right)=4< =>2\sqrt{x-1}=4\)

\(< =>\sqrt{x-1}=\frac{4}{2}=2=\sqrt{2}^2=\left(-\sqrt{2}\right)^2\)

\(< =>\orbr{\begin{cases}x-1=2\\x-1=-2\end{cases}< =>\orbr{\begin{cases}x=3\left(tm\right)\\x=-1\left(ktm\right)\end{cases}}}\)