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

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

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

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

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

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}

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

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

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

\(\Leftrightarrow3\left(2x-1\right)-2\left(3-x\right)=-1\)

=>6x-3-6+2x=-1

=>8x-9=-1

=>8x=8

hay x=1

\(\Leftrightarrow x\left(7-x\right)+\left(x+2\right)\left(x-4\right)=x+4\)

\(\Leftrightarrow7x-x^2+x^2-2x-8-x-4=0\)

=>4x-12=0

hay x=3(nhận)

ĐKXĐ:\(x\ne\pm4\)

\(\dfrac{x\left(7-x\right)}{x^2-16}+\dfrac{2+x}{x+4}=\dfrac{1}{x-4}\\ \Leftrightarrow\dfrac{x\left(7-x\right)}{\left(x-4\right)\left(x+4\right)}+\dfrac{\left(x-4\right)\left(2+x\right)}{\left(x-4\right)\left(x+4\right)}-\dfrac{x+4}{\left(x-4\right)\left(x+4\right)}=0\\ \Leftrightarrow\dfrac{7x-x^2+2x-8+x^2-4x-x-4}{\left(x-4\right)\left(x+4\right)}=0\\ \Rightarrow4x-12=0\\ \Leftrightarrow x=3\left(tm\right)\)

\(ĐK:x\ne\pm4\)

\(\Leftrightarrow\dfrac{x\left(7-x\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{x+4-\left(2+x\right)\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}\)

\(\Leftrightarrow x\left(7-x\right)=x+4-\left(2+x\right)\left(x-4\right)\)

\(\Leftrightarrow7x-x^2=x+4-2x+8-x^2+4x\)

\(\Leftrightarrow4x-12=0\)

\(\Leftrightarrow x=3\left(tm\right)\)

\(\Leftrightarrow2x\left(x+5\right)=3\left(x+5\right)\)

\(\Leftrightarrow2x\left(x+5\right)-3\left(x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(2x-3\right)=0\)

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

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 = \(\left\{-\dfrac{8}{3}\right\}\)

\(\left|x+15\right|-2021=0\\ \Rightarrow\left|x+15\right|=2021\\ \Rightarrow\left[{}\begin{matrix}x+15=2021\\x+15=-2021\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2006\\x=-2036\end{matrix}\right.\)

\(\left|x+15\right|-2021=0\)

\(\Leftrightarrow\left|x+15\right|=2021\)

\(\Leftrightarrow\left[{}\begin{matrix}x+15=2021\\x+15=-2021\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2021-15\\x=-2021-15\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2006\\x=-2036\end{matrix}\right.\)

=−2036

\(\left|x-2\right|+7=2x\\ \Leftrightarrow\left|x-2\right|=2x-7\left(x\ge\dfrac{7}{2}\right)\\ \Leftrightarrow\left[{}\begin{matrix}x-2=2x-7\\x-2=7-2x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=3\left(ktm\right)\end{matrix}\right.\)