Điều kiện:`x>=2`

Ta có:

`sqrt{x+6}-sqrt{x-2}=(x+6-x+2)/(sqrt{x+6}+sqrt{x-2})`

`=8/(\sqrt{x+6}+sqrt{x-2})`

`pt<=>8/(sqrt{x+6}+sqrt{x-2})(1+sqrt{(x-2)(x+6)})=8`

`<=>(1+sqrt{(x-2)(x+6)})/(sqrt{x+6}+sqrt{x-2})=1`

`<=>1+sqrt{(x-2)(x+6)}=sqrt{x+6}+sqrt{x-2}`

`<=>sqrt{(x-2)(x+6)}-sqrt{x+6}=sqrt{x-2}-1`

`<=>sqrt{x+6}(sqrt{x-2}-1)=sqrt{x-2}-1`

`<=>(sqrt{x-2}-1)(sqrt{x+6}-1)=0`

Vì `x>=2=>x+6>=8=>sqrt{x+6}>=2sqrt2`

`=>sqrt{x+6}-1>=2sqrt2-1>0`

`<=>sqrt{x-2}=1`

`<=>x=3(tm)`

Vậy `S={3}`

Tham khảo:

Giải phương trình: \(\sqrt{12-\dfrac{3}{x^2}}+\sqrt{4x^2-\dfrac{3}{x^2}}=4x^2\) - Hoc24

\(ĐK:x\ge2\\ PT\Leftrightarrow2\sqrt{x-2}+\sqrt{x-2}=6\\ \Leftrightarrow3\sqrt{x-2}=6\\ \Leftrightarrow\sqrt{x-2}=2\\ \Leftrightarrow x-2=4\\ \Leftrightarrow x=6\left(tm\right)\)

a, ĐKXĐ:...

\(\sqrt{5x+10}=8-x\\ \Leftrightarrow5x+10=64-16x+x^2\\ \Leftrightarrow x^2-21x+54=0\)

.....

b, ĐKXĐ:...

\(\sqrt{4x^2+x-12}=3x-5\\ \Leftrightarrow4x^2+x-12=9x^2-30x+25\\ \Leftrightarrow5x^2-31x+37=0\)

.....

\(\sqrt{4x^2-4x+1}=\sqrt{x^2+10x+25}\)

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

\(\Leftrightarrow\left|2x-1\right|=\left|x+5\right|\)

\(\Leftrightarrow\orbr{\begin{cases}2x-1=x+5\\2x-1=-\left(x+5\right)\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x-1=x+5\\2x-1=-x-5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=-\frac{4}{3}\end{cases}}\)

a) 

\(\sqrt{x+3}+2\sqrt{4\left(x+3\right)}-\frac{1}{3}\sqrt{9\left(x+3\right)}=8\)  

\(\sqrt{x+3}+2\cdot2\sqrt{x+3}-\frac{1}{3}\cdot3\sqrt{x+3}=8\)    

\(\sqrt{x+3}+4\sqrt{x+3}-\sqrt{x+3}=8\)    

\(4\sqrt{x+3}=8\)          

\(\sqrt{x+3}=2\) 

\(\orbr{\begin{cases}2\ge0\left(llđ\right)\\x+3=2^2\end{cases}}\) 

\(x+3=4\) 

\(x=1\) 

b) 

\(\orbr{\begin{cases}x^2+10x+25\ge0\\4x^2-4x+1=x^2+10x+25\end{cases}}\) 

\(\orbr{\begin{cases}\left(x+5\right)^2\ge0\left(lld\right)\\3x^2-6x-24=0\end{cases}}\) 

\(\orbr{\begin{cases}x=6\\x=-\frac{4}{3}\end{cases}}\)        

a) \(3\left(x-3\right)-5\left(-x+1\right)=x+6\)

\(\Leftrightarrow3x-9+5x-5-x-6=0\)

\(\Leftrightarrow7x=20\)

\(\Rightarrow x=\frac{20}{7}\)

b) \(\left|4x-2\right|=8\Leftrightarrow\orbr{\begin{cases}4x-2=8\\4x-2=-8\end{cases}}\Leftrightarrow\orbr{\begin{cases}4x=10\\4x=-6\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{3}{2}\end{cases}}\)

c) \(-3\left|6x+1\right|=-12\)

\(\Leftrightarrow\left|6x+1\right|=4\Leftrightarrow\orbr{\begin{cases}6x+1=4\\6x+1=-4\end{cases}}\Leftrightarrow\orbr{\begin{cases}6x=3\\6x=-5\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{5}{6}\end{cases}}\)

                                                   Bài giải

a, \(3\left(x-3\right)-5\left(-x+1\right)=x+6\)

\(3x-9+5x-5-x-6=0\)

\(7x-20=0\)

\(7x=20\)

\(x=\frac{20}{7}\)

b, \(\left|4x-2\right|=8\)

\(4x-2=\pm8\)

\(\Rightarrow\orbr{\begin{cases}4x-2=-8\\4x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}4x=-6\\4x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)

Vậy \(x\in\left\{-3\text{ ; }2\right\}\)

c, \(-3\left|6x+1\right|=-12\)

\(\left|6x+1\right|=4\)

\(6x+1=\pm4\)

\(\Rightarrow\orbr{\begin{cases}6x+1=-4\\6x+1=4\end{cases}}\Rightarrow\orbr{\begin{cases}6x=-5\\6x=3\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{5}{6}\\x=\frac{1}{2}\end{cases}}\)

\(\Rightarrow\text{ }x\in\left\{-\frac{5}{6}\text{ ; }\frac{1}{2}\right\}\)