a: \(\Leftrightarrow\left(x-1\right)\left(x^2+5x-2\right)-\left(x-1\right)\left(x^2+x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+5x-2-x^2-x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x-3\right)=0\)
=>x=1 hoặc x=3/4
b: \(\Leftrightarrow\left(x+5\right)\left(x-3\right)=x^2-1-8\)
\(\Leftrightarrow x^2+2x-12-x^2+9=0\)
=>2x-3=0
hay x=3/2(nhận)
\(\left(x-1\right)\left(x^2+5x-2\right)=x^3-1\)
\(\left(x-1\right)\left(x^2+5x-2\right)-\left(x-1\right)\left(x^2+x+1\right)=0\)
\(\left(x-1\right)\left(x^2+5x-2-x^2-x-1\right)=0\)
\(\left(x-1\right)\left(4x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{3}{4}\end{matrix}\right.\)
<=>\(4x^2+7x+3=0\)
<=>\(4x^2-4x-3x+3=0 \)
<=>\((4x-3)(x-1)=0 \)
<=>x=3/4 hoặc x=1