\(ĐKXĐ:\orbr{\begin{cases}x\ge3+\sqrt{3}\\x\le3-\sqrt{3}\end{cases}}\)

\(\Leftrightarrow\sqrt{\left(3x-7\right)^2}-1=3\sqrt{x^2-6x+6}\)

\(\Leftrightarrow\left|3x-7\right|-1=3\sqrt{x^2-6x+6}\)

- Với \(x\ge3+\sqrt{3}\)

\(\Leftrightarrow3x-8=3\sqrt{x^2-6x+6}\)

\(\Leftrightarrow9x^2-48x+64=9\left(x^2-6x+6\right)\)

\(\Rightarrow x=-\frac{10}{3}\left(l\right)\)

- Với \(x\le3-\sqrt{3}\)

\(\Leftrightarrow2-x=\sqrt{x^2-6x+6}\)

\(\Leftrightarrow x^2-4x+4=x^2-6x+6\)

\(\Rightarrow x=1\) ( t/m)

Chúc bạn học tốt !!!

Mn giúp e với ak

a) \(\sqrt{x^2-6x+9}\)

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

\(=\sqrt{\left(x-3\right)^2}\) ≥0,∀x

⇒x∈\(R\)

b) \(\sqrt{x^2-2x+1}\)

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

\(=\sqrt{\left(x-1\right)^2}\) ≥0,∀x

⇒x∈\(R\)

a: Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)

\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)

\(\Leftrightarrow3\sqrt{x+5}=6\)

\(\Leftrightarrow x+5=4\)

hay x=-1

b: Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)

\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)

\(\Leftrightarrow\sqrt{x-1}=17\)

\(\Leftrightarrow x-1=289\)

hay x=290

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

\(\Leftrightarrow3x+2=\left(2-\sqrt{3}\right)^2\)

\(\Leftrightarrow3x+2=7-4\sqrt{3}\)

\(\Leftrightarrow3x=7-2-4\sqrt{3}\)

\(\Leftrightarrow3x=5-4\sqrt{3}\)

\(\Leftrightarrow x=\dfrac{5}{3}-\dfrac{4\sqrt{3}}{3}\)

\(\Leftrightarrow x=\dfrac{5-4\sqrt{3}}{3}\)

b) \(\sqrt{x^2-4x+4}=49\)

\(\Leftrightarrow\sqrt{\left(x-2\right)^2}=49\)

\(\Leftrightarrow\left|x-2\right|=49\)\

\(\Leftrightarrow\left[{}\begin{matrix}x-2=49\\-x+2=49\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=51\\x=-47\end{matrix}\right.\)

c) \(\sqrt{x+1}=x-1\)

ĐKXĐ: \(x-1\ge0\Rightarrow x\ge1\)

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

\(\Leftrightarrow x+1=x^2-2x+1\)

\(\Leftrightarrow-x^2+2x+x=-1+1\)

\(\Leftrightarrow3x-x^2=0\)

\(\Leftrightarrow x\left(3-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\3-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(lo\text{ại}\right)\\x=3\left(nh\text{ậ}n\right)\end{matrix}\right.\)

d)e) lát mình làm sau

b: Ta có: \(\sqrt{x^2-6x+9}-\dfrac{\sqrt{6}+\sqrt{3}}{\sqrt{2}+1}=0\)

\(\Leftrightarrow x^2-6x+9=3\)

\(\Leftrightarrow x^2-6x+6=0\)

\(\text{Δ}=\left(-6\right)^2-4\cdot1\cdot6=36-24=12\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{6-2\sqrt{3}}{2}=3-\sqrt{3}\\x_2=3+\sqrt{3}\end{matrix}\right.\)

\(\sqrt{9x^2-6x+1}+\sqrt{25-30x+9x^2}\)

\(=\sqrt{\left(3x-1\right)^2}+\sqrt{\left(5-3x\right)^2}\)

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

\(\ge\left|3x-1+5-3x\right|=4\)