ĐKXĐ:\(x\ne\pm1\)

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

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

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

cái này cơ bản lắm mà b vẫn chx bt làm luôn?

đkxđ: x khác 3 ; x khác -3 

\(\Leftrightarrow\dfrac{x^2+6x+9}{\left(x-3\right)\left(x+3\right)}-\dfrac{x^2-6x+9}{\left(x+3\right)\left(x-3\right)}=\dfrac{3x-1}{\left(3-x\right)\left(3+x\right)}\)

\(\Leftrightarrow\dfrac{..}{..}-\dfrac{...}{...}=-\dfrac{3x-1}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow\dfrac{...}{...}-\dfrac{....}{....}+\dfrac{3x-1}{....}=0\)

\(\Leftrightarrow x^2+6x+9-x^2+6x-9+3x-1=0\)

\(\Leftrightarrow15x-1=0\)

=> x = 1/15 ( thỏa mãn)

\(ĐK:x\ne\pm3\)

\(\Rightarrow\dfrac{x+3}{x-3}-\dfrac{x-3}{x+3}=-\dfrac{3x-1}{\left(x-3\right)\left(x+3\right)}\)

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

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

\(\Leftrightarrow x^2+6x+9-x^2+6x-9+3x-1=0\)

\(\Leftrightarrow15x-1=0\)

\(\Leftrightarrow15x=1\)

\(\Leftrightarrow x=\dfrac{1}{15}\)

\(\dfrac{-3x-18+5x-10}{-15}-2\ge0\)

\(\Leftrightarrow\dfrac{2x-28+30}{-15}\ge0\Leftrightarrow\dfrac{2x+2}{-15}\ge0\)

\(\left\{{}\begin{matrix}2x+2< 0\\2x+2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< -1\\x\ne-1\end{matrix}\right.\)

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

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

=>8x-9=-1

=>8x=8

hay x=1