ĐKXĐ: \(\begin{cases}x+3\ge0\\ 3-2x\ge0\end{cases}\Rightarrow\begin{cases}x\ge-3\\ 2x\le3\end{cases}\Rightarrow-3\le x\le\frac32\)

\(x+4\sqrt{x+3}+2\sqrt{3-2x}=11\)

=>\(x-1+4\sqrt{x+3}-8+2\sqrt{3-2x}-2=0\)

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

=>\(x-1+4\cdot\frac{x+3-4}{\sqrt{x+3}+2}+2\cdot\frac{3-2x-1}{\sqrt{3-2x}+1}=0\)

=>\(x-1+4\cdot\frac{x-1}{\sqrt{x+3}+2}+2\cdot\frac{-2x+2}{\sqrt{3-2x}+1}=0\)

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

=>x-1=0

=>x=1(nhận)

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

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

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

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

\(\Rightarrow\orbr{\begin{cases}\sqrt{x+1}=0\\\sqrt{x+3}+\sqrt{x}=\sqrt{3x+1}\end{cases}}\)

Suy ra x=-1 pt còn lại bình lên là thấy vô nghiệm

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

\(ĐK:\left\{{}\begin{matrix}1-x\ge0\\x-2\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le1\\x\ge2\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

Vậy PT vô nghiệm

Sau gõ latex.

\(\sqrt{x-5}-\left(x-\dfrac{14}{3}+\sqrt{x-5}\right)=3\\ \Leftrightarrow\sqrt{x-5}-x+\dfrac{14}{3}-\sqrt{x-5}=3\\ \Leftrightarrow-x=3-\dfrac{14}{3}=-\dfrac{5}{3}\\ \Rightarrow x=-\dfrac{5}{3}:\left(-1\right)=\dfrac{5}{3}\)

\(\sqrt{9.\left(x-1\right)^2}-12=0\)

=> 3.(x - 1) - 12 = 0

=> 3x - 15 = 0

=> 3x = 15

=> x = 5

b) \(\sqrt{4.\left(3-x\right)}=16\) (ĐKXĐ: x ≤ 3)

\(\Rightarrow\sqrt{3-x}=8\)

=> 3 - x = 64

=> x = -61