Thiếu điều kiện.

a) \(\sqrt{x}=2\Rightarrow x=4\)

b) \(\sqrt{x-3}=5\) \(\left(x\ge3\right)\)

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

\(\Leftrightarrow x-3=25\)

\(\Leftrightarrow x=28\)

c) \(\dfrac{\sqrt{x}+2}{\sqrt{x}+5}=\dfrac{2}{3}\) \(\left(x>0\right)\)

\(\Leftrightarrow3\sqrt{x}+6=2\sqrt{x}+10\)

\(\Leftrightarrow\sqrt{x}=4\)

\(\Leftrightarrow x=16\)

d) \(x-2\sqrt{x}+1=0\)

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

\(\Leftrightarrow x=1\)

\(\left(2x-1\right)^5=x^5\)

\(2x-1=x\)

\(2x-x=1\)

\(x=1\)

_________________________

\(2x^2+2=20\)

\(2x^2=20-2\)

\(2x^2=18\)

\(x^2=\dfrac{18}{2}\)

\(x^2=9\)

\(x=\pm3\)

Ta có: \(n⋮4\Rightarrow n=4k\) \((k\ge0 ; k\in \mathbb N)\)

\(n< 2017\Rightarrow4k< 2017\Rightarrow k=504,25\)

\(\xrightarrow[]{k\ge0}k=0;1;2;...504\)

KL: tập hợp có 505 phần tử.

\(\left\{{}\begin{matrix}2\sqrt{x}+\dfrac{1}{y-3}=5\\3\sqrt{x}=5+\dfrac{1}{y-3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x}+\dfrac{1}{y-3}=5\\3\sqrt{x}-\dfrac{1}{y-3}=5\end{matrix}\right.\) 

ĐK: \(x\ge0;y\ge3\).

Đặt \(\sqrt{x}=a;\dfrac{1}{y-3}=b\)

\(\Rightarrow\left\{{}\begin{matrix}2a+b=5\\3a-b=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\)

Trả ẩn: \(\left\{{}\begin{matrix}\sqrt{x}=2\\\dfrac{1}{y-3}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=4\end{matrix}\right.\)

Vậy hệ pt có nghiệm: \(\left(x;y\right)=\left(4;4\right)\).

B. (any dùng trong câu pđ)

Check lại đi, bạn chưa rút gọn hết.

\(\dfrac{1}{\sqrt{5}+\sqrt{3}}-\dfrac{1}{\sqrt{5}-\sqrt{3}}\)

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

\(=\dfrac{\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}}{2}\)

\(=\dfrac{-2\sqrt{3}}{2}=-\sqrt{3}\)

\(\left(4x+1\right)\left(2x-5\right)=\left(2x-5\right)\left(2x+3\right)\)

\(\Leftrightarrow8x^2-20x+2x-5=4x^2+6x-10x-15\)

\(\Leftrightarrow8x^2-4x^2-20x+2x-6x+10x=-15+5\)

\(\Leftrightarrow4x^2-14x=-10\)

\(\Leftrightarrow4x^2-14x+10=0\)

\(\left(2x-5\right)\left(2x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\2x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=1\end{matrix}\right.\)

\(x^2-2+x+4=-0\)

\(\Leftrightarrow x^2+x+4=0\)

\(\Rightarrow x\in\varnothing\)

Vậy phương trình vô nghiệm.