Câu a:
\((x^2+x)^2+4(x^2+x)=12\)
\(\Leftrightarrow (x^2+x)^2+4(x^2+x)+4=16\)
\(\Leftrightarrow (x^2+x+2)^2=16\)
\(\Rightarrow \left[\begin{matrix} x^2+x+2=4\\ x^2+x+2=-4\end{matrix}\right.\Rightarrow \left[\begin{matrix} x^2+x-2=0\\ x^2+x+6=0\end{matrix}\right.\)
Với \(x^2+x-2=0\Leftrightarrow (x-1)(x+2)=0\Rightarrow \left[\begin{matrix} x=1\\ x=-2\end{matrix}\right.\)
Với \(x^2+x+6=0\Leftrightarrow (x^2+x+\frac{1}{4})+\frac{23}{4}=0\)
\(\Leftrightarrow (x+\frac{1}{2})^2=\frac{-23}{4}<0\) (vô lý- loại)
Vậy \(x\in \left\{-2;1\right\}\)
Câu b:
\(x(x-1)(x+1)(x+2)=24\)
\(\Leftrightarrow [x(x+1)][(x-1)(x+2)]=24\)
\(\Leftrightarrow (x^2+x)(x^2+x-2)=24\)
\(\Leftrightarrow a(a-2)=24\) (đặt \(x^2+x=a\) )
\(\Leftrightarrow a^2-2a-24=0\)
\(\Leftrightarrow (a-6)(a+4)=0\Rightarrow \left[\begin{matrix} a-6=0\\ a+4=0\end{matrix}\right.\)
Nếu \(a-6=0\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow (x-2)(x+3)=0\Rightarrow \left[\begin{matrix} x=2\\ x=-3\end{matrix}\right.\)
Nếu \(a+4=0\Leftrightarrow x^2+x+4=0\Leftrightarrow (x+\frac{1}{2})^2=\frac{-15}{4}<0\) (vô lý)
Vậy............
Câu c:
\((x-7)(x-5)(x-4)(x-2)=72\)
\(\Leftrightarrow [(x-7)(x-2)][(x-5)(x-4)]=72\)
\(\Leftrightarrow (x^2-9x+14)(x^2-9x+20)=72\)
\(\Leftrightarrow a(a+6)=72\) (đặt \(x^2-9x+14=a\))
\(\Leftrightarrow a^2+6a-72=0\)
\(\Leftrightarrow (a-6)(a+12)=0\Rightarrow \left[\begin{matrix} a-6=0\\ a+12=0\end{matrix}\right.\)
Nếu \(a-6=0\Leftrightarrow x^2-9x+8=0\)
\(\Leftrightarrow (x-1)(x-8)=0\Rightarrow \left[\begin{matrix} x=1\\ x=8\end{matrix}\right.\)
Nếu \(a+12=0\Leftrightarrow x^2-9x+26=0\)
\(\Leftrightarrow (x-\frac{9}{2})^2=\frac{-23}{4}<0\) (vô lý- loại)
Vậy.............
