\(a,Đk:x\ge0\\ PT\Leftrightarrow4x-8\sqrt{x}-7\sqrt{x}+14=0\\ \Leftrightarrow\left(\sqrt{x}-2\right)\left(4\sqrt{x}-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{49}{4}\end{matrix}\right.\left(tm\right)\\ b,ĐK:x\ge0\\ PT\Leftrightarrow\sqrt{x+1}-\sqrt{3x}+1-4x^2=0\\ \Leftrightarrow\dfrac{1-2x}{\sqrt{x+1}+\sqrt{3x}}+\left(1-2x\right)\left(2x+1\right)=0\\ \Leftrightarrow\left(1-2x\right)\left(\dfrac{1}{\sqrt{x+1}+\sqrt{3x}}+2x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\\dfrac{1}{\sqrt{x+1}+\sqrt{3x}}+2x+1=0\left(1\right)\end{matrix}\right.\)

Với \(x\ge0\Leftrightarrow\left(1\right)>0\)

Vậy PT có nghiệm \(x=\dfrac{1}{2}\)

ĐKXĐ: \(\left[{}\begin{matrix}x< -1\\x>1\end{matrix}\right.\)

- Với \(x< -1\Rightarrow VT< 0< 2\sqrt{2}\Rightarrow\) ptvn

- Với \(x>1\), bình phương 2 vế:

\(x^2+\dfrac{x^2}{x^2-1}+\dfrac{2x^2}{\sqrt{x^2-1}}=8\)

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

Đặt \(\sqrt{\dfrac{x^4}{x^2-1}}=t>0\)

\(\Rightarrow t^2+2t-8=0\Rightarrow\left[{}\begin{matrix}t=2\\t=-4\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow\dfrac{x^4}{x^2-1}=4\Rightarrow x^4-4x^2+4=0\)

\(\Rightarrow x^2=2\Rightarrow x=\sqrt{2}\)

ĐKXĐ: \(x\ge2\)

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

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

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

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

\(\Leftrightarrow x=3\)

a) Ta có: \(\left(x-\sqrt{2}\right)+3\left(x^2-2\right)=0\)

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

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

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

Vậy: \(S=\left\{\sqrt{2};\dfrac{-3\sqrt{2}-1}{3}\right\}\)

b) Ta có: \(x^2-5=\left(2x-\sqrt{5}\right)\left(x+\sqrt{5}\right)\)

\(\Leftrightarrow\left(x+\sqrt{5}\right)\left(x-\sqrt{5}\right)-\left(2x-\sqrt{5}\right)\left(x+\sqrt{5}\right)=0\)

\(\Leftrightarrow\left(x+\sqrt{5}\right)\left(x-\sqrt{5}-2x+\sqrt{5}\right)=0\)

\(\Leftrightarrow-x\left(x+\sqrt{5}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\x+\sqrt{5}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\sqrt{5}\end{matrix}\right.\)

Vậy: \(S=\left\{0;-\sqrt{5}\right\}\)

ĐK:  \(x\ge0\) hoặc  \(x\le-1\)

Đặt:  \(\sqrt{x^2+1}=a;\) \(\sqrt{x^2+x}=b\)   \(\left(a,b\ge0\right)\)

Khi đó pt đcho trở thành: 

 \(a-b=b^2-a^2\)

<=>  \(\left(a-b\right)\left(a+b+1\right)=0\)

đến đây tự lm
p/s: bài này có nhiều cách, bn tham khảo