1: Ta có: \(\sqrt{3x-5}=2\)

\(\Leftrightarrow3x-5=4\)

hay x=3

2: Ta có: \(\sqrt{25\left(x-1\right)}=20\)

\(\Leftrightarrow x-1=16\)

hay x=17

quy đồng r khử mẫu là ok

\(ĐK:x\ne0;2\)

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

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

\(\Leftrightarrow2\left(x^2-4\right)-x\left(2x+3\right)=0\)

\(\Leftrightarrow2x^2-8-2x^2-3x=0\)

\(\Leftrightarrow-3x=8\Leftrightarrow x=-\dfrac{8}{3}\left(tm\right)\)

\(\dfrac{2\left(x-2\right)\left(x+2\right)}{2x\left(x-2\right)}=\dfrac{x\left(2x+3\right)}{2x\left(x-2\right)}\)

\(\Leftrightarrow\) \(2x^2-8=2x^2+3x\)

\(\Leftrightarrow-3x=8\\ \Leftrightarrow x=-\dfrac{8}{3}\)

Vậy S = \(\left\{-\dfrac{8}{3}\right\}\)

ĐKXĐ:\(x\ne\pm1\)

\(\dfrac{x-1}{x+1}-\dfrac{x+1}{x-1}=\dfrac{14}{x^2-1}\\ \Leftrightarrow\dfrac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}-\dfrac{\left(x+1\right)^2}{\left(x+1\right)\left(x-1\right)}-\dfrac{14}{\left(x+1\right)\left(x-1\right)}=0\\ \Rightarrow x^2-2x+1-x^2-2x-1-14=0\\ \Leftrightarrow-4x-14=0\\ \Leftrightarrow x=-\dfrac{7}{2}\left(tm\right)\)

\(\Leftrightarrow\dfrac{3\left(2x-1\right)-36}{12}=\dfrac{2\left(1+x\right)}{12}\)

\(\Leftrightarrow\dfrac{6x-3-36}{12}=\dfrac{2+2x}{12}\)

\(\Leftrightarrow6x-39=2+2x\)

\(\Leftrightarrow4x=41\Leftrightarrow x=\dfrac{41}{4}\)

\(\left(2-x\right)^2=\left(2x-1\right)\left(2x+1\right)\\ \Leftrightarrow4-4x+x^2=4x^2-1\\ \Leftrightarrow4x^2-1-x^2+4x-4=0\\ \Leftrightarrow3x^2+4x-5=0\)

Đến đây mik thấy nghiệm rất xấu bạn xem đề đúng chx nhé

\(3.\left(2-x\right)^2=\left(2x-1\right)\left(2x+1\right)\)

\(\Leftrightarrow3.\left(2-x\right)=2x+1\)

\(\Leftrightarrow6-3x=2x+1\)

\(\Leftrightarrow-3x-2x=1-6\)

\(\Leftrightarrow-5x=-5\)

\(\Leftrightarrow x=1\)

Vậy: S = {1}

\(3\left(2-x\right)^2=\left(2x-1\right)\left(2x+1\right)\\ \Leftrightarrow3\left(4-4x+x^2\right)=4x^2-1\\ \Leftrightarrow12-12x+3x^2=4x^2-1\\ \Leftrightarrow4x^2-1-3x^2+12x-12=0\\ \Leftrightarrow x^2+12x-13=0\\ \Leftrightarrow x^2+13x-x-13=0\\ \Leftrightarrow x\left(x+13\right)-\left(x+13\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+13\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-13\end{matrix}\right.\)

\(\left(2x-5\right)\left(x-7\right)-\left(x-7\right)\left(5+x\right)=0\\ \left(x-7\right)\left(2x-5-5-x\right)=0\\ \left(x-7\right)\left(x-10\right)=0\\ \left\{{}\begin{matrix}x-7=0\\x-10=0\end{matrix}\right.\left\{{}\begin{matrix}x=7\\x=10\end{matrix}\right.\)

\(\left(2x-5\right)\left(x-7\right)=\left(x-7\right)\left(5+x\right)\\ \Leftrightarrow\left(2x-5\right)\left(x-7\right)-\left(x-7\right)\left(5+x\right)=0\\ \Leftrightarrow\left(x-7\right)\left(2x-5-5-x\right)=0\\ \Leftrightarrow\left(x-7\right)\left(x-10\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-7=0\\x-10=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=7\\x=10\end{matrix}\right.\)

\(Đk:x\ne0;3\)

\(\Leftrightarrow\dfrac{x+3}{x-3}=\dfrac{18}{x\left(x-3\right)}+\dfrac{8}{x}\)

\(\Leftrightarrow\dfrac{x\left(x+3\right)}{x\left(x-3\right)}=\dfrac{18+8\left(x-3\right)}{x\left(x-3\right)}\)

\(\Leftrightarrow x^2+3x=18+8x-24\)

\(\Leftrightarrow x^2-5x+6=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(ktm\right)\\x=2\left(tm\right)\end{matrix}\right.\)

x+3/x-3=18/x²-3x+8/x

x(x+3)/x(x-3)=18/x(x-3)+8(x-3)/x(x-3)

=>x(x+3)=18+8(x-3)

x²+3x=18+8x-24

x²+3x-8x=18-24

x²-5x=-6

x²-5x+6=0

x²-2x-3x+6=0

x(x-2)-3(x-2)=0

(x-3)(x-2)=0

=>x-3=0 hoặc x-2=0

=>x=3 hoặc x=2

Bài 1: 

a: ĐKXĐ: \(x\ge\dfrac{3}{2}\)

Bài 4:

\(a,A=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\\ P=A:B=\dfrac{\sqrt{x}+1}{\sqrt{x}}\cdot\dfrac{x-1}{\sqrt{x}+1}=\dfrac{x-1}{\sqrt{x}}\\ b,P\sqrt{x}=m-\sqrt{x}+x\\ \Leftrightarrow x-1=m-\sqrt{x}+x\\ \Leftrightarrow m=\sqrt{x}-1\)

\(a,ĐK:x\le2\\ PT\Leftrightarrow x^2-x-8=4-2x\Leftrightarrow x^2+x-12=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\left(ktm\right)\\x=-4\left(tm\right)\end{matrix}\right.\Leftrightarrow x=-4\\ b,ĐK:5x^2+10x+1\ge0\\ PT\Leftrightarrow5x^2+10x+1=\left(7-x^2-2x\right)^2\\ \Leftrightarrow5x^2+10x+1=x^4+4x^2+49-14x^2+4x^3-28x\\ \Leftrightarrow x^4+4x^3-15x^2-38x+48=0\\ \Leftrightarrow x^4-x^3+5x^3-5x^2-10x^2+10x-48x+48=0\\ \Leftrightarrow\left(x-1\right)\left(x^3+5x^2-10x-48\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x^3+3x^2+2x^2+6x-16x-48\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+3\right)\left(x^2+2x-16\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\\x^2+2x-16=0\left(1\right)\end{matrix}\right.\)

\(\Delta\left(1\right)=4+64=68\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2-2\sqrt{17}}{2}=-1-\sqrt{17}\\x=\dfrac{-2+2\sqrt{17}}{2}=-1+\sqrt{17}\end{matrix}\right.\)

Vậy pt có nghiệm ...