Question

Giải phương trình: (x^2+4x+3).(x^2+10x+24)=72 

Answer 1

\(\left(x^2+4x+3\right)\left(x^2+10x+24\right)=72\)

\(\Leftrightarrow x^4+10x^3+24x^2+4x^3+40x^2+96x+3x^2+30x+72=72\)

\(\Leftrightarrow x^4+14x^3+67x^2+126x+72=72\)

\(\Leftrightarrow x^4+14x^3+67x^2+126x=0\)

\(\Leftrightarrow x\left(x^3+14x^2+67x+126\right)=0\)

\(\Leftrightarrow x\left(x^2+7x+18\right)\left(x+7\right)=0\)

Vì \(x^2+7x+18>0\) nên:

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-7\end{cases}}\)