c) \(\sqrt{\left(x-2\right)^2}=10\)

\(x-2=10\)

\(x=12\)

d) \(\sqrt{9x^2-6x+1}=15\)

\(\sqrt{\left(3x\right)^2-2.3x.1+1^2}=15\)

\(\sqrt{\left(3x-1\right)^2}=15\)

\(3x-1=15\)

\(3x=16\)

\(x=\dfrac{16}{3}\)

a) \(đk:x\ge0\)

\(pt\Leftrightarrow3\sqrt{2x}+4\sqrt{2x}-3\sqrt{2x}=12\)

\(\Leftrightarrow4\sqrt{2x}=12\Leftrightarrow\sqrt{2x}=3\Leftrightarrow2x=9\Leftrightarrow x=\dfrac{9}{2}\left(tm\right)\)

b) \(đk:x\ge-2\)

\(pt\Leftrightarrow3\sqrt{x+2}+12\sqrt{x+2}-2\sqrt{x+2}=26\)

\(\Leftrightarrow13\sqrt{x+2}=26\)

\(\Leftrightarrow\sqrt{x+2}=2\Leftrightarrow x+2=4\Leftrightarrow x=2\left(tm\right)\)

c) \(pt\Leftrightarrow\left|x-2\right|=10\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=10\\x-2=-10\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-8\end{matrix}\right.\)

d) \(pt\Leftrightarrow\sqrt{\left(3x-1\right)^2}=15\)

\(\Leftrightarrow\left|3x-1\right|=15\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=15\\3x-1=-15\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{16}{3}\\x=-\dfrac{14}{3}\end{matrix}\right.\)

e) \(đk:x\ge\dfrac{8}{3}\)

\(pt\Leftrightarrow3x+4=9x^2-48x+64\)

\(\Leftrightarrow9x^2-51x+60=0\)

\(\Leftrightarrow3\left(x-4\right)\left(5x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=\dfrac{5}{3}\left(ktm\right)\end{matrix}\right.\)

1: Đặt \(a=9x^2-6x\)

=>\(45x^2-30x=5\left(9x^2-6x\right)=5a\)

\(\sqrt{9x^2-6x+2}+\sqrt{45x^2-30x+9}=\sqrt{6x-9x^2+8}\)

=>\(\sqrt{a+2}+\sqrt{5a+9}=\sqrt{-a+8}\)

=>\(\sqrt{a+2}-1+\sqrt{5a+9}-2=\sqrt{-a+8}-3\)

=>\(\frac{a+2-1}{\sqrt{a+2}+1}+\frac{5a+9-4}{\sqrt{5a+9}+2}=\frac{-a+8-9}{\sqrt{-a+8}+3}\)

=>\(\left(a+1\right)\left(\frac{1}{\sqrt{a+2}+1}+\frac{5}{\sqrt{5a+9}+2}+\frac{1}{\sqrt{a+8}+3}\right)=0\)

=>a+1=0

=>a=-1

=>\(9x^2-6x=-1\)

=>\(9x^2-6x+1=0\)

=>\(\left(3x-1\right)^2=0\)

=>3x-1=0

=>3x=1

=>x=1/3

2: Đặt \(x^2-2x=a\)

\(\sqrt{2x^2-4x+3}+\sqrt{3x^2-6x+7}=2-x^2+2x\)

=>\(\sqrt{2\left(x^2-2x\right)+3}+\sqrt{3\left(x^2-2x\right)+7}=-\left(x^2-2x\right)+2\)

=>\(\sqrt{2a+3}+\sqrt{3a+7}=-a+2\)

=>\(\sqrt{2a+3}-1+\sqrt{3a+7}-2=-a+2-3\)

=>\(\frac{2a+2}{\sqrt{2a+3}+1}+\frac{3a+7-4}{\sqrt{3a+7}+2}=-a-1\)

=>\(\left(a+1\right)\left(\frac{2}{\sqrt{2a+3}+1}+\frac{3}{\sqrt{3a+7}+2}+1\right)=0\)

=>a+1=0

=>\(x^2-2x+1=0\)

=>\(\left(x-1\right)^2=0\)

=>x-1=0

=>x=1

\(3T=\left(\sqrt{x^2-6x+19}-\sqrt{x^2-6x+10}\right)\left(\sqrt{x^2-6x+19}+\sqrt{x^2-6x+10}\right)\)

\(=x^2-6x+19-\left(x^2-6x+10\right)=9\)

\(\Rightarrow T=3\)

Bài 1: 

a: Ta có: \(\sqrt{3x^2}=\sqrt{12}\)

\(\Leftrightarrow3x^2=12\)

\(\Leftrightarrow x^2=4\)

hay \(x\in\left\{2;-2\right\}\)

b: Ta có: \(\sqrt{\left(x-2\right)^2}=3\)

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

b) Đặt \(\sqrt{x^2-6x+6}=a\left(a\ge0\right)\)

\(\Rightarrow a^2+3-4a=0\)

=> (a - 3).(a - 1) = 0

=> \(\left[{}\begin{matrix}a=3\\a=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2-6x+6}=3\\\sqrt{x^2-6x+6}=1\end{matrix}\right.\)

Bình phương lên giải tiếp nhé!

c) Tương tư câu b nhé

a) giải pt ra ta được  : x=-1

b) giải pt ra ta được  : x=2

c)giải pt ra ta được  : x vô ngiệm

d)giải pt ra ta được  : x=vô ngiệm

~~~~~~~~~~~ai đi ngang qua nhớ để lại k ~~~~~~~~~~~~~

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