a, \(4x=-2\Leftrightarrow x=-\dfrac{1}{2}\)
b, \(\left(x+5\right)\left(x-2\right)=0\Leftrightarrow x=-5;x=2\)
c, \(\Rightarrow2x-4=6x+3+3x\Leftrightarrow9x+3=2x-4\Leftrightarrow7x=-7\Leftrightarrow x=-1\)
d, đk : x khác -1 ; 1
\(\Rightarrow x^2-x+x^2+2x+1=x^2+1\Leftrightarrow x^2+x=0\Leftrightarrow x=0\left(tm\right);x=-1\left(ktm\right)\)
c.\(\dfrac{x-2}{3}=x+\dfrac{1+x}{2}\)
\(\Leftrightarrow\dfrac{2\left(x-2\right)}{6}=\dfrac{6x+3\left(1+x\right)}{6}\)
\(\Leftrightarrow2\left(x-2\right)=6x+3\left(1+x\right)\)
\(\Leftrightarrow2x-4=6x+3+3x\)
\(\Leftrightarrow7x=-7\)
\(\Leftrightarrow x=-1\)
d.\(\dfrac{x}{x+1}+\dfrac{x+1}{x-1}=\dfrac{x^2+1}{x^2-1}\)
\(ĐK:x\ne\pm1\)
\(\Rightarrow\dfrac{x}{x+1}+\dfrac{x+1}{x-1}=\dfrac{x^2+1}{\left(x+1\right)\left(x-1\right)}\)
\(\Leftrightarrow\dfrac{x\left(x-1\right)+\left(x+1\right)\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}=\dfrac{x^2+1}{\left(x+1\right)\left(x-1\right)}\)
\(\Leftrightarrow x\left(x-1\right)+\left(x+1\right)^2=x^2+1\)
\(\Leftrightarrow x^2-x+x^2+2x+1-x^2-1=0\)
\(\Leftrightarrow x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)