Tham khảo:

1) Giải phương trình : \(11\sqrt{5-x}+8\sqrt{2x-1}=24+3\sqrt{\left(5-x\right)\left(2x-1\right)}\) - Hoc24

\(\sqrt{x+3}+2\sqrt{x}=2+\sqrt{x\left(x+3\right)}\left(đk:x\ge0\right)\)

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

\(\Leftrightarrow5x+3=4+x^2+3x\)

\(\Leftrightarrow x^2-2x+1=0\)

\(\Leftrightarrow\left(x-1\right)^2=0\)

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\left(tm\right)\)

ĐKXĐ: \(\left[{}\begin{matrix}x=0\\x\ge3\end{matrix}\right.\)

Với \(x=0\) là nghiệm

Với \(x\ge3\), chia 2 vế cho \(\sqrt{x}\) ta được:

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

\(\Leftrightarrow\sqrt{x+1}+\sqrt{x+2}-\sqrt{x-3}=0\)

\(\Leftrightarrow\sqrt{x+1}+\dfrac{5}{\sqrt{x+2}+\sqrt{x-3}}=0\) (vô nghiệm do vế trái luôn dương)

Vậy pt có nghiệm duy nhất \(x=0\)

ĐK: \(x\ge1\)

\(pt\Leftrightarrow2\sqrt{\left(x-1\right)\left(x+2\right)}-\sqrt{x-1}-6\sqrt{x+2}+3=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2\sqrt{x+2}=1\\\sqrt{x-1}=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4\left(x+2\right)=1\\x-1=9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{4}\left(l\right)\\x=10\left(tm\right)\end{matrix}\right.\)

Vậy ...

\(\left(\sqrt{x+3}-\sqrt{x+1}\right)\left(x^2+\sqrt{x^2+4x+3}\right)=2x\left(đk:x\ge0\right)\)

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

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

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

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{5}}{2}\left(tm\right)\\x=\dfrac{1-\sqrt{5}}{2}\left(ktm\right)\\x=\dfrac{1+\sqrt{13}}{2}\left(tm\right)\\x=\dfrac{1-\sqrt{13}}{2}\left(ktm\right)\end{matrix}\right.\)

\(ĐK:x\ge\dfrac{1}{2}\\ PT\Leftrightarrow2x-2\sqrt{2x^2+5x-3}=1+x\sqrt{2x-1}-2x\sqrt{x+3}\\ \Leftrightarrow\left(2x-2\right)-\left(2\sqrt{2x^2+5x-3}-4\right)=\left(x\sqrt{2x-1}-x\right)-\left(2x\sqrt{x+3}-4x\right)-3x+3\\ \Leftrightarrow2\left(x-1\right)-\dfrac{2\left(2x^2+5x-7\right)}{\sqrt{2x^2+5x-3}+4}=\dfrac{x\left(2x-2\right)}{\sqrt{2x-1}+1}-\dfrac{2x\left(x-1\right)}{\sqrt{x+3}+4x}-3\left(x-1\right)\\ \Leftrightarrow2\left(x-1\right)-\dfrac{2\left(x-1\right)\left(2x+7\right)}{\sqrt{2x^2+5x-3}+4}-\dfrac{2x\left(x-1\right)}{\sqrt{2x-1}+1}+\dfrac{2x\left(x-1\right)}{\sqrt{x+3}+4x}+3\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left[2-\dfrac{2\left(2x+7\right)}{\sqrt{2x^2+5x-3}+4}-\dfrac{2x}{\sqrt{2x-1}+2}+\dfrac{2x}{\sqrt{x+3}+4x}+3\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\2-\dfrac{2\left(2x+7\right)}{\sqrt{2x^2+5x-3}+4}-\dfrac{2x}{\sqrt{2x-1}+2}+\dfrac{2x}{\sqrt{x+3}+4x}+3=0\left(1\right)\end{matrix}\right.\)

Với \(x\ge\dfrac{1}{2}\Leftrightarrow-\dfrac{2\left(2x+7\right)}{\sqrt{2x^2+5x-3}+4}>-\dfrac{2\cdot8}{4}=-4\)

\(-\dfrac{2x}{\sqrt{2x-1}+2}>-\dfrac{1}{2};\dfrac{2x}{\sqrt{x+3}+4x}>0\)

Do đó \(\left(1\right)>2-4-\dfrac{1}{2}+3=\dfrac{1}{2}>0\) nên (1) vô nghiệm

Vậy PT có nghiệm duy nhất \(x=1\)

ĐKXĐ: \(-1\le x\le1\)

Xét \(\sqrt{\left(1+x\right)^3}-\sqrt{\left(1-x\right)^3}=\left(\sqrt{1+x}-\sqrt{1-x}\right)\left[\left(1+x\right)+\left(1-x\right)+\sqrt{\left(1+x\right)\left(1-x\right)}\right]\)

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

Khi đó phương trình đề trở thành:

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

Vì \(2+\sqrt{1-x^2}>0\)nên ta có thể chia 2 vế cho \(2+\sqrt{1-x^2}\):

\(\Rightarrow\sqrt{1+\sqrt{1-x^2}}\left(\sqrt{1+x}-\sqrt{1-x}\right)=\frac{1}{\sqrt{3}}\),Bình phương 2 vế:

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

\(\Leftrightarrow\left(1+\sqrt{1-x^2}\right)\left(2-2\sqrt{1-x^2}\right)=\frac{1}{3}\Leftrightarrow2\left(1+\sqrt{1-x^2}\right)\left(1-\sqrt{1-x^2}\right)=\frac{1}{3}\)\(\Leftrightarrow1-\left(1-x^2\right)=\frac{1}{3}\Leftrightarrow x^2=\frac{1}{6}\Leftrightarrow x=\pm\frac{1}{\sqrt{6}}\)

Ta xét phương trình đề: vế phải luôn không âm vì vậy vế trái phải không âm 

Khi đó \(\sqrt{\left(1+x\right)^3}-\sqrt{\left(1-x\right)^3}\ge0\Leftrightarrow1+x\ge1-x\Leftrightarrow x\ge0\)

Vậy ta chỉ nhận nghiệm duy nhất là \(x=\frac{1}{\sqrt{6}}\)

ĐKXĐ : \(x\ge-2\)

\(\sqrt{1+\left(x+2\right).\sqrt{1+\left(x+3\right).\left(x+5\right)}}=2023x+1\)

\(\Leftrightarrow\sqrt{1+\left(x+2\right).\sqrt{x^2+8x+16}}=2023x+1\)

\(\Leftrightarrow\sqrt{1+\left(x+2\right).\left(x+4\right)}=2023x+1\) (Do \(x\ge-2\Rightarrow x+4>0\))

\(\Leftrightarrow\sqrt{x^2+6x+9}=2023x+1\)

\(\Leftrightarrow x+3=2023x+1\) (Do \(x\ge-2\Rightarrow x+3>0\)

\(\Leftrightarrow x=\dfrac{1}{1011}\)(tm) 

Vậy tập nghiệm \(S=\left\{\dfrac{1}{1011}\right\}\)