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.

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

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

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

\(\Leftrightarrow x-4\le0\)

\(\Rightarrow x\le4\)

c.

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}x\le2\\x^2-2x=x^2-4x+4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le2\\2x=4\end{matrix}\right.\)

\(\Rightarrow x=2\)

d.

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

\(\Leftrightarrow\left\{{}\begin{matrix}x+6\ge0\\x+27=\left(x+6\right)^2\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-6\\x^2+11x+9=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-11+\sqrt{85}}{2}\\x=\dfrac{-11-\sqrt{85}}{2}\left(loại\right)\end{matrix}\right.\)

\(a,ĐK:x\ge\dfrac{5}{2}\\ PT\Leftrightarrow2x-5=4\Leftrightarrow x=\dfrac{9}{2}\left(tm\right)\\ b,PT\Leftrightarrow\left|x-3\right|=7\Leftrightarrow\left[{}\begin{matrix}x-3=7\\3-x=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-4\end{matrix}\right.\\ c,ĐK:x\le4\\ PT\Leftrightarrow\left|x-8\right|=4-x\\ \Leftrightarrow\left[{}\begin{matrix}x-8=4-x\left(x\ge8\right)\\8-x=4-x\left(x\le8\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x\in\varnothing\left(trái.vs.ĐK\right)\\0x=4\left(ktm\right)\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

a) \(\sqrt{2x-5}=2\)

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

\(\Leftrightarrow\) \(2x-5=4\)

\(\Leftrightarrow\) 2x = 9

\(\Leftrightarrow\) x = \(\dfrac{9}{2}\)

 Chúc bạn học tốt

\(1.\sqrt{16-8x+x^2}=4-x\)

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

\(4-x-4+x=0\)

= 0 phương trình vô nghiệm.

\(2.\sqrt{4x^2-12x+9}=2x-3\)

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

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

= 0 phương trình vô nghiệm.

a: Ta có: \(\sqrt{16-8x+x^2}=4-x\)

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

hay \(x\le4\)

b: Ta có: \(\sqrt{4x^2-12x+9}=2x-3\)

\(\Leftrightarrow\left|2x-3\right|=2x-3\)

hay \(x\ge\dfrac{3}{2}\)

a/ \(\sqrt{16-8x+x^2}=4-x\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le4\\\sqrt{\left(4-x\right)^2}=4-x\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le4\\\left|4-x\right|=4-x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\le4\\\left[{}\begin{matrix}4-x=4-x\left(loại\right)\\4-x=x-4\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow x=4\)

Vậy...

b/ \(\sqrt{4x^2-12x+9}=2x-3\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{2}{3}\\\sqrt{\left(2x-3\right)^2}=2x-3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{2}{3}\\\left[{}\begin{matrix}2x-3=2x-3\left(loại\right)\\2x-3=3-2x\end{matrix}\right.\end{matrix}\right.\)

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

Vậy...

a, \(\sqrt{4x^2+20x+25}\) + \(\sqrt{x^2-8x+16}\) = \(\sqrt{x^2+18x+81}\)

⇔ 4x² + 20x + 25 + \(2\sqrt{\left(4x^2+20x+25\right)\left(x^2-8x+16\right)}\) = x² + 18x + 81

⇔ 4x² + 20x + 25 - x² - 18x - 81 + \(2\sqrt{\left(2x+5\right)^2.\left(x-4\right)^2}\) = 0

⇔ 3x² + 2x - 56 + 2.(2x + 5) . (x - 4) = 0

⇔ 3x² + 2x - 56 + (4x + 10) . (x - 4) = 0

⇔ 3x² + 2x - 56 + 4x² - 16x + 10x - 40 = 0

⇔ 7x² - 4x - 96 = 0

x₁ = 4 ( nhận )

x₂ = \(\frac{-24}{7}\) ( nhận )

Vậy: S = {4; \(\frac{-24}{7}\)}

a) \(\sqrt{3x+10}=4\left(đk:x\ge-\dfrac{10}{3}\right)\Leftrightarrow3x+10=16\Leftrightarrow x=2\)

b) \(\sqrt{9x^2-6x+1}=\sqrt{x^2+8x+16}\Leftrightarrow\sqrt{\left(3x-1\right)^2}=\sqrt{\left(x+4\right)^2}\Leftrightarrow3x-1=x+4\Leftrightarrow2x=5\Leftrightarrow x=\dfrac{5}{2}\)

c) \(\sqrt{2x+1}=3\left(đk:x\ge-\dfrac{1}{2}\right)\Leftrightarrow2x+1=9\Leftrightarrow x=4\)

d) \(\sqrt{2x+1}+1=x\left(đk:x\ge1\right)\Leftrightarrow\sqrt{2x+1}=x-1\Leftrightarrow2x+1=x^2-2x+1\Leftrightarrow x^2-4x=0\Leftrightarrow x\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)\(\Leftrightarrow x=4\)(do \(x\ge1\))

a: Ta có: \(\sqrt{3x+10}=4\)

\(\Leftrightarrow3x+10=16\)

\(\Leftrightarrow3x=6\)

hay x=2

b: Ta có: \(\sqrt{9x^2-6x+1}=\sqrt{x^2+8x+16}\)

\(\Leftrightarrow\left|3x-1\right|=\left|x+4\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=x+4\\3x-1=-x-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\4x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{3}{4}\end{matrix}\right.\)

c: Ta có: \(\sqrt{2x+1}=3\)

\(\Leftrightarrow2x+1=9\)

\(\Leftrightarrow x=4\)

\(C=\sqrt{9x^2}-2x=\left|3x\right|-2x=-3x-2x=-5x\)

\(D=x-4+\sqrt{16-8x+x^2}=x-4+\left|4-x\right|=x-4+x-4=2x-8\)

\(C=\sqrt{9x^2}-2x=-3x-2x=-5x\)

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

a) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\) (ĐK: \(x\ge1\)) 

\(\Leftrightarrow\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}+2=0\)

\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)

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

\(\Leftrightarrow\sqrt{x-1}=\dfrac{2}{2}\)

\(\Leftrightarrow\sqrt{x-1}=1\)

\(\Leftrightarrow x-1=1\)

\(\Leftrightarrow x=2\left(tm\right)\)

b) \(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\) (ĐK: \(x\ge-1\))

\(\Leftrightarrow\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)

\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)

\(\Leftrightarrow4\sqrt{x+1}=16\)

\(\Leftrightarrow\sqrt{x+1}=4\)

\(\Leftrightarrow x+1=16\)

\(\Leftrightarrow x=15\left(tm\right)\)

2: ĐKXĐ: x>=0

\(\sqrt{3x}-2\sqrt{12x}+\dfrac{1}{3}\cdot\sqrt{27x}=-4\)

=>\(\sqrt{3x}-2\cdot2\sqrt{3x}+\dfrac{1}{3}\cdot3\sqrt{3x}=-4\)

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

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

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

=>3x=4

=>\(x=\dfrac{4}{3}\left(nhận\right)\)

3: 

ĐKXĐ: x>=0

\(3\sqrt{2x}+5\sqrt{8x}-20-\sqrt{18}=0\)

=>\(3\sqrt{2x}+5\cdot2\sqrt{2x}-20-3\sqrt{2}=0\)

=>\(13\sqrt{2x}=20+3\sqrt{2}\)

=>\(\sqrt{2x}=\dfrac{20+3\sqrt{2}}{13}\)

=>\(2x=\dfrac{418+120\sqrt{2}}{169}\)

=>\(x=\dfrac{209+60\sqrt{2}}{169}\left(nhận\right)\)

4: ĐKXĐ: x>=-1

\(\sqrt{16x+16}-\sqrt{9x+9}=1\)

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

=>\(\sqrt{x+1}=1\)

=>x+1=1

=>x=0(nhận)

5: ĐKXĐ: x<=1/3

\(\sqrt{4\left(1-3x\right)}+\sqrt{9\left(1-3x\right)}=10\)

=>\(2\sqrt{1-3x}+3\sqrt{1-3x}=10\)

=>\(5\sqrt{1-3x}=10\)

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

=>1-3x=4

=>3x=1-4=-3

=>x=-3/3=-1(nhận)

6: ĐKXĐ: x>=3

\(\dfrac{2}{3}\sqrt{x-3}+\dfrac{1}{6}\sqrt{x-3}-\sqrt{x-3}=-\dfrac{2}{3}\)

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

=>\(\sqrt{x-3}\cdot\dfrac{-1}{6}=-\dfrac{2}{3}\)

=>\(\sqrt{x-3}=\dfrac{2}{3}:\dfrac{1}{6}=\dfrac{2}{3}\cdot6=\dfrac{12}{3}=4\)

=>x-3=16

=>x=19(nhận)