\(a.x^3+1=x\left(x+1\right)\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)-x\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-1\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
Vậy ,...
\(b.1-\dfrac{3x-5}{x-1}=\dfrac{5-2x}{x-2}\left(x\ne1;x\ne2\right)\)
\(\Leftrightarrow\dfrac{\left(x-1-3x+5\right)\left(x-2\right)}{\left(x-1\right)\left(x-2\right)}=\dfrac{\left(5-2x\right)\left(x-1\right)}{\left(x-1\right)\left(x-2\right)}\)
\(\Leftrightarrow-2\left(x^2-4x+4\right)=5x-5-2x^2+2x\)
\(\Leftrightarrow-2x^2+8x-8+2x^2-7x+5=0\)
\(\Leftrightarrow x=3\) ( TM ĐKXĐ )
Vậy ,....
\(c.x^2-x-12=0\)
\(\Leftrightarrow x^2+3x-4x-12=0\)
\(\Leftrightarrow x\left(x+3\right)-4\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=4\end{matrix}\right.\)
Vậy ,....
