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

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

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

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

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

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)

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

cách khác đơn giản hơn nhiều 

Đk:\(x\ge1\)

\(pt\Leftrightarrow\sqrt{2\left(x-1\right)\left(x+4\right)}+\sqrt{2\left(x-1\right)\left(x+3\right)}-3\sqrt{x+4}-3\sqrt{x+3}-1=0\)

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

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

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

Xét Ư(1)={1;-1}={....}

Dễ nhé, tự làm nốt

Đk: \(x\ge1\)

\(pt\Leftrightarrow\sqrt{2x^2+6x-8}+\sqrt{2x^2+4x-6}-3\sqrt{x+4}-3\sqrt{x+3}-1=0\)

\(\Leftrightarrow\sqrt{2x^2+6x-8}-\frac{10}{3}\sqrt{x+3}+\frac{1}{3}\sqrt{x+3}-1\sqrt{2x^2+4x-6}-3\sqrt{x+4}=0\)

\(\Leftrightarrow\frac{2x^2+6x-8-\frac{100}{9}\left(x+3\right)}{\sqrt{2x^2+6x-8}+\frac{10}{3}\sqrt{x+3}}+\frac{x-6}{3\left(\sqrt{x+3}+3\right)}+\frac{2x^2+4x-6-9\left(x+4\right)}{\sqrt{2x^2+4x-6}+3\sqrt{x+4}}=0\)

Để đỡ rối ta đặt mấy cái mẫu \(\hept{\begin{cases}N=\sqrt{2x^2+6x-8}+\frac{10}{3}\sqrt{x+3}>0\\H=\sqrt{x+3}+3>0\\T=\sqrt{2x^2+4x-6}+3\sqrt{x+4}>0\end{cases}}\)

\(\Leftrightarrow\frac{18x^2-46x-372}{9N}+\frac{x-6}{3H}+\frac{2x^2-5x-42}{T}=0\)

\(\Leftrightarrow\left(x-6\right)\left(\frac{18x+62}{9N}+\frac{1}{3H}+\frac{2x+7}{T}\right)=0\)

Dễ  thấy: \(\forall x\ge1\) thì \(\frac{18x+62}{9N}+\frac{1}{3H}+\frac{2x+7}{T}>0\)

\(\Rightarrow x-6=0\Rightarrow x=6\) (thỏa mãn)

Lời giải:

a. ĐKXĐ: $x\geq 4$

PT $\Leftrightarrow \sqrt{(x-4)+4\sqrt{x-4}+4}=2$

$\Leftrightarrow \sqrt{(\sqrt{x-4}+2)^2}=2$

$\Leftrightarrow |\sqrt{x-4}+2|=2$

$\Leftrightarrow  \sqrt{x-4}+2=2$

$\Leftrightarrow \sqrt{x-4}=0$

$\Leftrightarrow x=4$ (tm)

b. ĐKXĐ: $x\in\mathbb{R}$

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

$\Leftrightarrow |2x-1|=|x-3|$

\(\Rightarrow \left[\begin{matrix} 2x-1=x-3\\ 2x-1=3-x\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=-2\\ x=\frac{4}{3}\end{matrix}\right.\)

c.

PT \(\Rightarrow \left\{\begin{matrix} 2x-1\geq 0\\ 2x^2-2x+1=(2x-1)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{1}{2}\\ 2x^2-2x=0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{1}{2}\\ 2x(x-1)=0\end{matrix}\right.\Rightarrow x=1\)

ta có:

pt trên \(< =>x^2+6x+1=\left(2x+1\right)\sqrt{x^2+2x+3}\)

\(< =>\left[\left(x^2+6x\right)+1\right]^2=\left(2x+1\right)^2.\left(x^2+2x+3\right)\)

\(< =>x^4+12x^3+36x^2+2.\left(x^2+6x\right)+1=\left(4x^2+4x+1\right)\left(x^2+2x+3\right)\)

\(< =>x^4+12x^3+38x^2+12x+1=\)

\(4x^4+8x^3+12x^2+4x^3+8x^2+12x+x^2+2x+3\)

\(=4x^4+12x^3+21x^2+14x+3\)

\(< =>-3x^4+17x^2-2x-2=0\)

\(< =>-\left(x^2+2x-1\right)\left(3x^2-6x+2\right)=0\)

đến đây dễ rùi bạn tự giải nhé 

\(\text{Đ}K:x^2+2x+3\ge0\\ x^2+6x+1=\left(2x+1\right)\cdot\sqrt{x^2+2x+3}\\ \Leftrightarrow x^2+2x+3+4x+2=\left(2x+1\right)\cdot\sqrt{x^2+2x+3+4}\)

\(\text{ Đặt }\)\(m=\sqrt{x^2+2x+3};n=2x+1\) \(\text{ phương trình trở thành :}\)

\(m^2+2n=mn+4\\ \Leftrightarrow m^2-4-mn+2n=0\\ \Leftrightarrow\left(m-2\right)\left(m+2\right)-n\left(m-2\right)=0\\ \Leftrightarrow\left(m-2\right)\left(m-n-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}m=2\\m-n=-2\end{matrix}\right.\)

`\text{ Với}` \(m=2\\ \Leftrightarrow\sqrt{x^2+2x+3}=2\Leftrightarrow x^2+2x-1=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}-1\left(N\right)\\x=-\sqrt{2}-1\left(N\right)\end{matrix}\right.\)

`\text{Với}`\(m-n=-2\Leftrightarrow\sqrt{x^2+2x+3}-\left(2x+1\right)=-2\\ \Leftrightarrow\sqrt{x^2+2x+3}=-2+2x+1=2x-1\\ \Leftrightarrow x^2+2x+3=4x^2-4x+1\\ \Leftrightarrow3x^2-6x-2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3+\sqrt{15}}{3}\left(N\right)\\x=\dfrac{3-\sqrt{15}}{3}\left(L\right)\end{matrix}\right.\)

weo hay thế:33