/x-3/=9-2x

= (1) x-3 = 9-2x

= 3x=12

x=4

(2) x+3= 9-2x

= 3x= 6

x=2

/x-3/=9-2x

\(\sqrt{x^{ }2-6x+9}=4-x\)
\(\sqrt{\left(x-3\right)^{ }2}=4-x\)
x-3=4-x
x+x=4+3
2x=7
x=\(\dfrac{7}{2}\)

Lời giải:
a.

PT \(\Leftrightarrow \left\{\begin{matrix} 4-x\geq 0\\ x^2-6x+9=(4-x)^2=x^2-8x+16\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\leq 4\\ 2x=7\end{matrix}\right.\Leftrightarrow x=\frac{7}{2}\)

b.

ĐKXĐ: $x\geq \frac{3}{2}$

PT \(\Leftrightarrow \sqrt{(2x-3)+2\sqrt{2x-3}+1}+\sqrt{(2x-3)+8\sqrt{2x-3}+16}=5\)

\(\Leftrightarrow \sqrt{(\sqrt{2x-3}+1)^2}+\sqrt{(\sqrt{2x-3}+4)^2}=5\)

\(\Leftrightarrow |\sqrt{2x-3}+1|+|\sqrt{2x-3}+4|=5\)

\(\Leftrightarrow \sqrt{2x-3}+1+\sqrt{2x-3}+4=2\sqrt{2x-3}+5=5\)

\(\Leftrightarrow \sqrt{2x-3}=0\Leftrightarrow x=\frac{3}{2}\)

a: Ta có: \(\sqrt{x^2-6x+9}=4-x\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-3=4-x\left(x\ge3\right)\\x-3=x-4\left(x< 3\right)\left(loại\right)\end{matrix}\right.\)

\(\Leftrightarrow2x=7\)

hay \(x=\dfrac{7}{2}\left(nhận\right)\)

x-1 + x-3 =1 <=> 2x -4=1 tu giai not

ĐKXĐ: \(x\ge0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-3=2x\left(x\ge3\right)\\x-3=-2x\left(0\le x< 3\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-3\left(ktm\right)\\x=1\left(tm\right)\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-3=2x\left(x\ge3\right)\\x-3=-2x\left(x< 3\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)

Lời giải:

a. Đề thiếu

b. PT $\Leftrightarrow \sqrt{(x-1)^2}+\sqrt{(x-2)^2}=3$

$\Leftrightarrow |x-1|+|x-2|=3$
Nếu $x\geq 2$ thì pt trở thành:
$x-1+x-2=3$

$\Leftrightarrow 2x-3=3$

$\Leftrightarrow x=3$ (tm)

Nếu $1\leq x< 2$ thì:

$x-1+2-x=3\Leftrightarrow 1=3$ (vô lý)

Nếu $x< 1$ thì:

$1-x+2-x=3$

$\Leftrightarrow x=0$ (tm)

a: \(\Leftrightarrow x^2+6x+9=0\)

\(\Leftrightarrow\left(x+3\right)^2=0\)

=>x+3=0

hay x=-3

b: \(\Leftrightarrow x^2+x-12-6x+4=x^2-8x+16\)

=>-7x+8=-8x+16

=>x=8

ĐK: \(\forall x\in R\)

PT\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{5}{2}\\x^2-6x+9=4x^2-20x+25\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{5}{2}\\3x^2-14x+16=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(l\right)\\x=\dfrac{8}{3}\left(tm\right)\end{matrix}\right.\)

Điều kiện :  

\(\left\{{}\begin{matrix}x^2-6x+9\ge0\\2x-5\ge0\end{matrix}\right.\)⇔ \(x\ge\dfrac{5}{2}\)

Ta có : 

\(\left(\sqrt{x^2-6x+9}\right)^2=\left(2x-5\right)^2\)

⇔ \(x^2-6x+9=4x^2-20x+25\)

⇔ \(3x^2-14x+16=0\)

⇔\(\left\{{}\begin{matrix}x=2\left(loại\right)\\x=\dfrac{8}{3}\left(tm\right)\end{matrix}\right.\)

Vì \(VT\ge0\Rightarrow VP\ge0\Rightarrow x\ge\dfrac{5}{2}\)

\(\sqrt{x^2-6x+9}=2x-5\Rightarrow\sqrt{\left(x-3\right)^2}=2x-5\)

\(\Rightarrow\left|x-3\right|=2x-5\)

Xét \(x\ge3\Rightarrow x-3=2x-5\Rightarrow x-2=0\Rightarrow x=2\) (loại)

Xét \(x< 3\Rightarrow\dfrac{5}{2}\le x< 3\Rightarrow3-x=2x-5\Rightarrow3x-8=0\Rightarrow x=\dfrac{8}{3}\) 

Vậy \(x=\dfrac{8}{3}\) là nghiệm của pt...

\(PT\Leftrightarrow\left|x-3\right|=2x+1\Leftrightarrow\left[{}\begin{matrix}x-3=2x+1\left(x\ge3\right)\\3-x=2x+1\left(x< 3\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-4\left(ktm\right)\\x=\dfrac{2}{3}\left(tm\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{2}{3}\)