Ta có: \(\left(\dfrac{2}{\sqrt{x}-2}+\dfrac{3}{2\sqrt{x}+1}-\dfrac{5\sqrt{x}-7}{2x-3\sqrt{x}-2}\right):\dfrac{2\sqrt{x}+3}{5x-10\sqrt{x}}\)

\(=\dfrac{4\sqrt{x}+2+3\sqrt{x}-6-5\sqrt{x}+7}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)

\(=\dfrac{2\sqrt{x}+3}{2\sqrt{x}+1}\cdot\dfrac{5\sqrt{x}}{2\sqrt{x}+3}\)

\(=\dfrac{5\sqrt{x}}{2\sqrt{x}+1}\)

1) đk: \(x\ge1\)

Ta có: \(\sqrt{x-1}-\sqrt{2x\left(x-1\right)}=0\)

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

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

\(\Leftrightarrow2x^2-3x+1=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\left(ktm\right)\\x=1\left(tm\right)\end{cases}}\)

Vậy x = 1

2) đk: \(x\ge\frac{1}{2}\)

Ta có: \(\sqrt{5x^2}=2x-1\)

\(\Leftrightarrow5x^2=\left(2x-1\right)^2\)

\(\Leftrightarrow5x^2=4x^2-4x+1\)

\(\Leftrightarrow x^2+4x-1=0\)

\(\Leftrightarrow\left(x+2\right)^2-5=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=-2+\sqrt{5}\left(ktm\right)\\x=-2-\sqrt{5}\left(ktm\right)\end{cases}}\)

=> PT vô nghiệm

3) đk: \(x\ge-1\)

Ta có: \(\sqrt{x+1}+\sqrt{9x+9}=4\)

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

\(\Leftrightarrow4\sqrt{x+1}=4\)

\(\Leftrightarrow x+1=1\)

\(\Rightarrow x=0\)

4) đk: \(x\ge2\)

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

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

\(\Leftrightarrow x-2=x\left(x-2\right)\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=1\left(ktm\right)\\x=2\left(tm\right)\end{cases}}\)

Vậy x = 2

6) đk: \(x\ge-\frac{7}{5}\)

Ta có: \(\frac{\sqrt{2x-3}}{\sqrt{x-1}}=2\)

\(\Leftrightarrow\frac{2x-3}{x-1}=2\)

\(\Leftrightarrow2x-3=2x-2\)

\(\Leftrightarrow0x=1\) vô lý

=> PT vô nghiệm

Xin lỗi mk ghi nhầm phần

Phần 6 ban nãy là phần 5 và cho mk sửa lại

5) đk: \(x\ge\frac{3}{2}\)

Ta có: \(\frac{\sqrt{2x-3}}{\sqrt{x-1}}=2\)

\(\Leftrightarrow\frac{2x-3}{x-1}=4\)

\(\Leftrightarrow2x-3=4x-4\)

\(\Leftrightarrow2x=1\)

\(\Rightarrow x=\frac{1}{2}\left(ktm\right)\)

Vậy PT vô nghiệm

6) đk: \(x\ge-\frac{7}{5}\)

Ta có: \(\frac{\sqrt{5x+7}}{\sqrt{x+3}}=4\)

\(\Leftrightarrow\frac{5x+7}{x+3}=16\)

\(\Leftrightarrow16x+48=5x+7\)

\(\Leftrightarrow11x=-41\)

\(\Rightarrow x=-\frac{41}{11}\)(ktm)

Vậy PT vô nghiệm

7) đk: \(x\ge1\)

Ta có: \(\sqrt{5x-5}-\sqrt{35}=0\)

\(\Leftrightarrow\sqrt{5x-5}=\sqrt{35}\)

\(\Leftrightarrow5x-5=35\)

\(\Leftrightarrow5x=40\)

\(\Rightarrow x=8\left(tm\right)\)

Vậy x = 8

a: ĐKXĐ: x>=5

\(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\cdot\sqrt{9x-45}=4\)

=>\(2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4\)

=>\(2\sqrt{x-5}=4\)

=>\(\sqrt{x-5}=2\)

=>x-5=4

=>x=9(nhận)

b: ĐKXĐ: x>=1/2

\(\sqrt{2x-1}-\sqrt{8x-4}+5=0\)

=>\(\sqrt{2x-1}-2\sqrt{2x-1}+5=0\)

=>\(5-\sqrt{2x-1}=0\)

=>\(\sqrt{2x-1}=5\)

=>2x-1=25

=>2x=26

=>x=13(nhận)

c: \(\sqrt{x^2-10x+25}=2\)

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

=>\(\left|x-5\right|=2\)

=>\(\left[{}\begin{matrix}x-5=2\\x-5=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\)

d: \(\sqrt{x^2-14x+49}-5=0\)

=>\(\sqrt{x^2-2\cdot x\cdot7+7^2}=5\)

=>\(\sqrt{\left(x-7\right)^2}=5\)

=>|x-7|=5

=>\(\left[{}\begin{matrix}x-7=5\\x-7=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=2\end{matrix}\right.\)

\(a,\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\left(đkxđ:x\ge5\right)\\ \Leftrightarrow\sqrt{4\left(x-5\right)}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\\ \Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\\ \Leftrightarrow2\sqrt{x-5}=4\\ \Leftrightarrow\sqrt{x-5}=2\\ \Leftrightarrow x-5=4\\ \Leftrightarrow x=9\left(tm\right)\)

\(b,\sqrt{2x-1}-\sqrt{8x-4}+5=0\left(đkxđ:x\ge\dfrac{1}{2}\right)\\ \Leftrightarrow\sqrt{2x-1}-\sqrt{4\left(2x-1\right)}=-5\\ \Leftrightarrow\sqrt{2x-1}-2\sqrt{2x-1}=-5\\ \Leftrightarrow-\sqrt{2x-1}=-5\\ \Leftrightarrow\sqrt{2x-1}=5\\ \Leftrightarrow2x-1=25\\ \Leftrightarrow2x=26\\ \Leftrightarrow x=13\left(tm\right)\)

\(c,\sqrt{x^2-10x+25}=2\\ \Leftrightarrow\sqrt{\left(x-5\right)^2}=2\\ \Leftrightarrow\left|x-5\right|=2\\ \Leftrightarrow\left[{}\begin{matrix}x-5=2\\x-5=-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\)

\(d,\sqrt{x^2-14x+49}-5=0\\ \Leftrightarrow\sqrt{\left(x-7\right)^2}=5\\ \Leftrightarrow\left|x-7\right|=5\\ \Leftrightarrow\left[{}\begin{matrix}x-7=5\\x-7=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=12\\x=2\end{matrix}\right.\)

\(a)ĐKXĐ:x\ge5\\ \sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\\ \Leftrightarrow\sqrt{4\left(x-5\right)}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\\ \Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\\ \Leftrightarrow2\sqrt{x-5}=4\\ \Leftrightarrow\sqrt{x-5}=\dfrac{4}{2}\\ \Leftrightarrow\sqrt{x-5}=2\\ \Leftrightarrow\left(\sqrt{x-5}\right)^2=2^2\\ \Leftrightarrow x-5=4\\ \Leftrightarrow x=4+5\\ \Leftrightarrow x=9\left(tmđk\right)\)

Vậy \(S=\left\{9\right\}\)

\(b)ĐKXĐ:x\ge2\\ \sqrt{2x-1}-\sqrt{8x-4}+5=0\\ \Leftrightarrow\sqrt{2x-1}-\sqrt{8x-4}=0-5\\ \Leftrightarrow\sqrt{2x-1}-\sqrt{4\left(2x-1\right)}=-5\\ \Leftrightarrow\sqrt{2x-1}-2\sqrt{2x-1}=-5\\ \Leftrightarrow-\sqrt{2x-1}=-5\\ \Leftrightarrow-\left(\sqrt{2x-1}\right)=\left(-5\right)^2\\ \Leftrightarrow-2x+1=-25\\ \Leftrightarrow-2x=\left(-25\right)-1\\ \Leftrightarrow-2x=-26\\ \Leftrightarrow x=\dfrac{-26}{-2}\\ \Leftrightarrow x=13\left(tmđk\right)\)

Vậy \(S=\left\{13\right\}\)

\(c)\sqrt{x^2-10x+25}=2\\ \Leftrightarrow\sqrt{\left(x-5\right)^2}=2\\ \Leftrightarrow\left|x-5\right|=2\\ \Leftrightarrow\left[{}\begin{matrix}x-5=2\\x-5=-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2+5\\x=\left(-2\right)+5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\)

Vậy: \(S=\left\{7;3\right\}\)

\(d)\sqrt{x^2-14x+49}-5=0\\ \Leftrightarrow\sqrt{x^2-14x+49}=0+5\\ \Leftrightarrow\sqrt{x^2-14x+49}=5\\ \Leftrightarrow\sqrt{\left(x-7\right)^2}=5\\ \Leftrightarrow\left|x-7\right|=5\\ \Leftrightarrow\left[{}\begin{matrix}x-7=5\\x-7=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5+7\\x=\left(-5\right)+7\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=12\\x=2\end{matrix}\right.\)

Vậy \(S=\left\{12;2\right\}.\)

@Akai Haruma @Nguyễn Huy Tú