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\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=24=1.2.3.4=\left(-1\right)\left(-2\right)\left(-3\right)\left(-4\right)\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=1\\x+1=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)
\(S=\left\{-2;0\right\}\)
\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24=0\)
\(\Leftrightarrow\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]-24=0\)
\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=0\)
Đặt \(x^2+5x+4=a\)
\(pt\Leftrightarrow a\left(a+2\right)-24=0\)
\(\Leftrightarrow a^2+2a-24=0\)
\(\Leftrightarrow a^2+6a-4a-24=0\)
\(\Leftrightarrow a\left(a+6\right)-4\left(a+6\right)=0\)
\(\Leftrightarrow\left(a+6\right)\left(a-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=-6\\a=4\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2+5x+4=-6\\x^2+5x+4=4\end{cases}}}\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+5x+10=0\\x^2+5x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2+2\cdot x\cdot\frac{5}{2}+\frac{25}{4}+\frac{15}{4}=0\\x\left(x+5\right)=0\end{cases}}}\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x+\frac{5}{2}\right)^2=\frac{-15}{4}\left(loai\right)\\x\in\left\{0;-5\right\}\end{cases}}\)
Vậy....
Cho mình sửa xíu, x+1 = -4 <=> x = -5 lúc nãy vội quá nên lộn.
