Question

Tìm x: (x-1)(x-4)(x-5)(x-8)+36=0

Answer 1

\(\left(x-1\right)\left(x-4\right)\left(x-5\right)\left(x-8\right)+36=0\)

\(\Rightarrow\left(x-1\right)\left(x-8\right)\left(x-4\right)\left(x-5\right)+36=0\)

\(\Leftrightarrow\left(x^2-8x-x+8\right)\left(x^2-4x-5x+20\right)+36=0\)

\(\Rightarrow\left(x^2-9x+8\right)\left(x^2-9x+20\right)+36=0\)

\(\Rightarrow\left(x^2-9x+14-6\right)\left(x^2-9x+14+6\right)+36=0\)

\(\Rightarrow\left(x^2-9x+14\right)^2-36+36=0\)

\(\Rightarrow x^2-9x+14=0\Leftrightarrow\left(x-2\right)\left(x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=7\end{matrix}\right.\)