Question

[Shu Sakamaki]

Tìm x, biết:

\(\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)=24\)

Answer 1

[lý canh hy]

\(\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)=24\)

\(\Rightarrow\left(x-1\right)\left(x-4\right)\left(x-2\right)\left(x-3\right)=24\)

\(\left(x^2-5x+4\right)\left(x^2-5x+6\right)=24\)

Đặt \(x^2-5x+5=a,\)ta có

\(\left(a-1\right)\left(a+1\right)=24\Rightarrow a^2=25\Rightarrow a=\pm5\)

Theo cánh đặt,ta có

+,\(x^2-5x+5=5\Rightarrow x\left(x-5\right)=0\Rightarrow\orbr{\begin{cases}x=0\\x=5\end{cases}}\)

+\(x^2-5x+5=-5\Rightarrow x^2-2\cdot\frac{5}{2}+\frac{25}{4}+\frac{15}{4}=0\)

\(\Rightarrow\left(x-\frac{5}{2}\right)^2+\frac{15}{4}=0\)(vô lí)

Vậy