a) \(\sqrt{1-6x+9x^2}=9\)

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

\(\Leftrightarrow\left|1-3x\right|=9\)

\(\Leftrightarrow\left[{}\begin{matrix}1-3x=9\\1-3x=-9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=1-9\\3x=1+9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=-8\\3x=10\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{8}{3}\\x=\dfrac{10}{3}\end{matrix}\right.\)

b) \(\sqrt{2x-3}-\sqrt{x+1}=0\) (\(x\ge\dfrac{3}{2}\))

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

\(\Leftrightarrow2x-3=x+1\)

\(\Leftrightarrow2x-x=1+3\)

\(\Leftrightarrow x=4\left(tm\right)\)

c) \(\sqrt{9x^2+12+4}-2=3x\)

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

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

\(\Leftrightarrow3x+2\ge0\)

\(\Leftrightarrow3x\ge-2\)

\(\Leftrightarrow x\ge-\dfrac{2}{3}\)

a: =>|3x-1|=9

=>3x-1=9 hoặc 3x-1=-9

=>x=-8/3 hoặc x=10/3

b: =>căn 2x-3=căn x+1

=>2x-3=x+1

=>x=4

c: =>|3x+2|=3x+2

=>3x+2>=0

=>x>=-2/3

\(A=\sqrt{1-6x+9x^2}+\sqrt{9x^2-12x+4}\)

\(A=\sqrt{1^2-2\cdot3x\cdot1+\left(3x\right)^2}+\sqrt{\left(3x\right)^2-2\cdot2\cdot3x+2^2}\)

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

\(A=\left|1-3x\right|+\left|3x-2\right|\)

\(A=\left|1-3x+3x-2\right|\)

\(A=\left|-1\right|=1\)

Dấu "=" xảy ra \(\left(1-3x\right)\left(3x-2\right)\ge0\)

\(\Rightarrow\dfrac{1}{3}\le x\le\dfrac{2}{3}\)

Vậy: \(A_{min}=1\) khi \(\dfrac{1}{3}\le x\le\dfrac{2}{3}\)

\(A=\sqrt{1-6x+9x^2}+\sqrt{9x^2-12x+4}\)

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

\(=\left|1-3x\right|+\left|3x-2\right|\)

\(\ge\left|1-3x+3x-2\right|=\left|-1\right|=1\)

Dấu "=" xảy ra \(\Leftrightarrow\left(1-3x\right)\left(3x-2\right)\ge0\Leftrightarrow\frac{1}{3}\le x\le\frac{2}{3}\)

Vậy \(A_{min}=1\) tại \(\frac{1}{3}\le x\le\frac{2}{3}\)

Xin lỗi cậu tớ mới học lớp 7 thôi

6) ĐKXĐ: \(x\le-6\)

\(\sqrt{\left(x+6\right)^2}=-x-6\Leftrightarrow\left|x+6\right|=-x-6\)

\(\Leftrightarrow x+6=x+6\left(đúng\forall x\right)\)

Vậy \(x\le-6\)

7) ĐKXĐ: \(x\ge\dfrac{2}{3}\)

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

\(\Leftrightarrow3x-2=3x-2\left(đúng\forall x\right)\)

Vậy \(x\ge\dfrac{2}{3}\)

8) ĐKXĐ: \(x\ge5\)

\(pt\Leftrightarrow\sqrt{\left(4-3x\right)^2}=2x-10\)\(\Leftrightarrow\left|4-3x\right|=2x-10\)

\(\Leftrightarrow4-3x=10-2x\Leftrightarrow x=-6\left(ktm\right)\Leftrightarrow S=\varnothing\)

9) ĐKXĐ: \(x\ge\dfrac{3}{2}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-3=2x-3\left(x\ge3\right)\\x-3=3-2x\left(\dfrac{3}{2}\le x< 3\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=2\left(tm\right)\end{matrix}\right.\)