(x^2+3x+2)(x^2+7X+12)=24
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=24\)
\(\Leftrightarrow\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]=24\)
\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)=24\)
đặt \(x^2+5x+5=a\)=> ta có phương trình \(\Leftrightarrow\left(a-1\right)\left(a+1\right)=24\)
\(\Leftrightarrow a^2-1=24\)\(\Leftrightarrow a^2=25\Leftrightarrow a=\orbr{\begin{cases}5\\-5\end{cases}}\)
+)\(x^2+5x+5=5\)\(\Leftrightarrow x^2+5x=0\Leftrightarrow x\left(x+5\right)=0\)
\(\Leftrightarrow x=\orbr{\begin{cases}5\\0\end{cases}}\)
+) \(x^2+5x+5=-5\)\(\Leftrightarrow x^2+5x+10=0\)\(\Rightarrowđenta=5^2-4.10=-15< 0\Rightarrow ptvonghiem\)
vậy \(x=\orbr{\begin{cases}0\\5\end{cases}}\)
( x^2 + 3x + 2 )( x^2 + 7x + 12 ) = 24
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=24\)
\(\Leftrightarrow\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)=24\)
\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)=24\)
Đặt x2 + 5x + 5 = a = ta có : \(\Leftrightarrow\left(a-1\right)\left(a+1\right)=24\)
\(\Leftrightarrow a^2-1=24\Leftrightarrow a^2=25\Leftrightarrow a=\orbr{\begin{cases}5\\-5\end{cases}}\)
+)\(x^2+5x+5=5\Leftrightarrow x^2+5x=0\Leftrightarrow x\left(x+5\right)=0\)
\(\Leftrightarrow x=\orbr{\begin{cases}5\\0\end{cases}}\)
+)\(x^2+5x+5=-5\Leftrightarrow x^2+5x+10=0\)
\(\Rightarrowđenta=5^2-4.10=-15< 0\Rightarrow ptvonghiem\)
\(Vay.x=\orbr{\begin{cases}5\\0\end{cases}}\)