a: Ta có: \(\dfrac{3\left(2x-1\right)}{4}-\dfrac{3x+1}{10}+1=\dfrac{2\left(3x+2\right)}{5}\)
\(\Leftrightarrow15\left(2x-1\right)-2\left(3x+1\right)+20=8\left(3x+2\right)\)
\(\Leftrightarrow30x-15-6x-2+20=24x+16\)
\(\Leftrightarrow24x+3=24x+16\)(vô lý)
b: Ta có: \(\dfrac{x+1}{x-2}-\dfrac{x-1}{x+2}=\dfrac{2\left(x^2+2\right)}{x^2-4}\)
Suy ra: \(x^2+3x+2-x^2+3x-2=2x^2+4\)
\(\Leftrightarrow2x^2-6x+4=0\)
\(\Leftrightarrow x=1\)
a, \(\dfrac{3\left(2x-1\right)}{4}-\dfrac{3x+1}{10}+1=\dfrac{2\left(3x+2\right)}{5}\)
=> \(\dfrac{15\left(2x-1\right)-2\left(3x+1\right)+20}{20}=\dfrac{8\left(3x+2\right)}{20}\)
=> \(15\left(2x-1\right)-2\left(3x+1\right)+20=8\left(3x+2\right)\)
=> \(30x-15-6x-2+20=24x+16\)
=> \(30x-6x-24x=16+15+2-20\)
=> 0 = 13 (Vô lí)
=> x không tồn tại
b, \(\dfrac{x+1}{x-2}-\dfrac{x-1}{x+2}=\dfrac{2\left(x^2+2\right)}{x^2-4}\)
=> \(\dfrac{\left(x+1\right)\left(x+2\right)-\left(x-1\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{2\left(x^2+2\right)}{\left(x-2\right)\left(x+2\right)}\)
=> \(\left(x+1\right)\left(x+2\right)-\left(x-1\right)\left(x-2\right)=2\left(x^2+2\right)\)
=> \(x^2+3x+2-\left(x^2-3x+2\right)-2x^2-4=0\)
=> \(x^2+3x+2-x^2+3x-2-2x^2-4=0\)
=> \(6x-4=0\)
=> 6x = 4
=> \(x=\dfrac{2}{3}\)
c, \(x^3+1=x\left(x+1\right)\)
=> \(x^3+1=x^2+x\)
=> \(x^3-x^2-x+1=0\)
=> \(x^3+x^2-2x^2-2x+x+1=0\)
=> \(x^2\left(x+1\right)-2x\left(x+1\right)+\left(x+1\right)=0\)
=> \(\left(x^2-2x+1\right)\left(x+1\right)=0\)
=> \(\left(x-1\right)^2\left(x+1\right)=0\)
=> \(\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
d, \(\left|x+6\right|=3x+2\)
=> \(\left[{}\begin{matrix}x+6=3x+2\\-x-6=3x+2\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x-3x=2-6\\-x-3x=2+6\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}-2x=-4\\-4x=8\end{matrix}\right.\)
=> \(x=-2\)
