Lời giải:
a)
\(x^2-2x=24\)
\(\Leftrightarrow x^2-6x+4x-24=0\)
\(\Leftrightarrow x(x-6)+4(x-6)=0\Leftrightarrow (x+4)(x-6)=0\)
\(\Rightarrow \left[\begin{matrix} x+4=0\\ x-6=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=-4\\ x=6\end{matrix}\right.\)
b)
\(x^3-7x+6=0\Leftrightarrow (x^3-x)-(6x-6)=0\)
\(\Leftrightarrow x(x^2-1)-6(x-1)=0\)
\(\Leftrightarrow x(x-1)(x+1)-6(x-1)=0\)
\(\Leftrightarrow (x-1)(x^2+x-6)=0\)
\(\Leftrightarrow (x-1)(x^2-2x+3x-6)=0\)
\(\Leftrightarrow (x-1)[x(x-2)+3(x-2)]=0\)
\(\Leftrightarrow (x-1)(x-2)(x+3)=0\)
\(\Rightarrow \left[\begin{matrix} x-1=0\\ x-2=0\\ x+3=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=1\\ x=2\\ x=-3\end{matrix}\right.\)
c) Xem lại đề.
d) Đặt \(x^2+x+4=a\) thì pt trở thành:
\(a^2+8ax+16x^2=0\)
\(\Leftrightarrow a^2+2.a.4x+(4x)^2=0\)
\(\Leftrightarrow (a+4x)^2=0\Rightarrow a+4x=0\)
\(\Rightarrow x^2+x+4+4x=0\)
\(\Rightarrow x(x+1)+4(x+1)=0\Leftrightarrow (x+1)(x+4)=0\)
\(\Rightarrow \left[\begin{matrix} x+4=0\rightarrow x=-4\\ x+1=0\rightarrow x=-1\end{matrix}\right.\)
