Question

Giaỉ phương trình sau

     x(x-1)(x+1)(x+2)= 24

Answer 1

  \(x\left(x-1\right)\left(x+1\right)\left(x+2\right)=24\)

\(\Leftrightarrow\left[x\left(x+1\right)\right]\left[\left(x-1\right)\left(x+2\right)\right]=24\)

\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)=24\)

Đặt \(x^2+x-1=a\)

Ta có : \(x^2+x-1=\left(x+\frac{1}{2}\right)^2-\frac{5}{4}\ge-\frac{5}{4}\)

\(\Rightarrow a\ge-\frac{5}{4}\)

Ta có pt : \(\left(a+1\right)\left(a-1\right)=24\)

\(\Leftrightarrow a^2-1=24\)

\(\Leftrightarrow a^2=25\)

\(\Leftrightarrow a=5\left(Do\text{ }a\ge-\frac{5}{4}\right)\)

\(\Leftrightarrow x^2+x-1=5\)

\(\Leftrightarrow x^2+x-6=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)