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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow\dfrac{3\left(2x-1\right)-36}{12}=\dfrac{2\left(1+x\right)}{12}\)
\(\Leftrightarrow\dfrac{6x-3-36}{12}=\dfrac{2+2x}{12}\)
\(\Leftrightarrow6x-39=2+2x\)
\(\Leftrightarrow4x=41\Leftrightarrow x=\dfrac{41}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(2-x\right)^2=\left(2x-1\right)\left(2x+1\right)\\ \Leftrightarrow4-4x+x^2=4x^2-1\\ \Leftrightarrow4x^2-1-x^2+4x-4=0\\ \Leftrightarrow3x^2+4x-5=0\)
Đến đây mik thấy nghiệm rất xấu bạn xem đề đúng chx nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
\(Đk:x\ne0;3\)
\(\Leftrightarrow\dfrac{x+3}{x-3}=\dfrac{18}{x\left(x-3\right)}+\dfrac{8}{x}\)
\(\Leftrightarrow\dfrac{x\left(x+3\right)}{x\left(x-3\right)}=\dfrac{18+8\left(x-3\right)}{x\left(x-3\right)}\)
\(\Leftrightarrow x^2+3x=18+8x-24\)
\(\Leftrightarrow x^2-5x+6=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(ktm\right)\\x=2\left(tm\right)\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow3\left(2x-1\right)-2\left(3-x\right)=-1\)
=>6x-3-6+2x=-1
=>8x-9=-1
=>8x=8
hay x=1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow x\left(7-x\right)+\left(x+2\right)\left(x-4\right)=x+4\)
\(\Leftrightarrow7x-x^2+x^2-2x-8-x-4=0\)
=>4x-12=0
hay x=3(nhận)
ĐKXĐ:\(x\ne\pm4\)
\(\dfrac{x\left(7-x\right)}{x^2-16}+\dfrac{2+x}{x+4}=\dfrac{1}{x-4}\\ \Leftrightarrow\dfrac{x\left(7-x\right)}{\left(x-4\right)\left(x+4\right)}+\dfrac{\left(x-4\right)\left(2+x\right)}{\left(x-4\right)\left(x+4\right)}-\dfrac{x+4}{\left(x-4\right)\left(x+4\right)}=0\\ \Leftrightarrow\dfrac{7x-x^2+2x-8+x^2-4x-x-4}{\left(x-4\right)\left(x+4\right)}=0\\ \Rightarrow4x-12=0\\ \Leftrightarrow x=3\left(tm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow2x\left(x+5\right)=3\left(x+5\right)\)
\(\Leftrightarrow2x\left(x+5\right)-3\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{3}{2}\end{matrix}\right.\)
⇔2x(x+5)= 3(x+5)⇔ 2x(x+5)= 3(x+5)
⇔2x(x+5)−3(x+5)= 0⇔2x(x+5)−3(x+5)= 0
⇔(x+5)(2x−3)= 0⇔(x+5)(2x−3)= 0
⇔x= -5 hoặc x= 3/2.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(ĐK:x\ne0;2\)
\(\Leftrightarrow\dfrac{x+2}{x}-\dfrac{2x+3}{2\left(x-2\right)}=0\)
\(\Leftrightarrow\dfrac{2\left(x-2\right)\left(x+2\right)-x\left(2x+3\right)}{2x\left(x-2\right)}=0\)
\(\Leftrightarrow2\left(x^2-4\right)-x\left(2x+3\right)=0\)
\(\Leftrightarrow2x^2-8-2x^2-3x=0\)
\(\Leftrightarrow-3x=8\Leftrightarrow x=-\dfrac{8}{3}\left(tm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left|x+15\right|-2021=0\\ \Rightarrow\left|x+15\right|=2021\\ \Rightarrow\left[{}\begin{matrix}x+15=2021\\x+15=-2021\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2006\\x=-2036\end{matrix}\right.\)
\(\left|x+15\right|-2021=0\)
\(\Leftrightarrow\left|x+15\right|=2021\)
\(\Leftrightarrow\left[{}\begin{matrix}x+15=2021\\x+15=-2021\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2021-15\\x=-2021-15\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2006\\x=-2036\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left|x-2\right|+7=2x\\ \Leftrightarrow\left|x-2\right|=2x-7\left(x\ge\dfrac{7}{2}\right)\\ \Leftrightarrow\left[{}\begin{matrix}x-2=2x-7\\x-2=7-2x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=3\left(ktm\right)\end{matrix}\right.\)
\(\left(2x-5\right)\left(x-7\right)=\left(x-7\right)\left(5+x\right)\\ \Leftrightarrow\left(2x-5\right)\left(x-7\right)-\left(x-7\right)\left(5+x\right)=0\\ \Leftrightarrow\left(x-7\right)\left(2x-5-5-x\right)=0\\ \Leftrightarrow\left(x-7\right)\left(x-10\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-7=0\\x-10=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=7\\x=10\end{matrix}\right.\)
\(\left(2x-5\right)\left(x-7\right)-\left(x-7\right)\left(5+x\right)=0\\ \left(x-7\right)\left(2x-5-5-x\right)=0\\ \left(x-7\right)\left(x-10\right)=0\\ \left\{{}\begin{matrix}x-7=0\\x-10=0\end{matrix}\right.\left\{{}\begin{matrix}x=7\\x=10\end{matrix}\right.\)