\(\left(x-1\right)\cdot\left(x-2\right)\cdot\left(x-3\right)\cdot\left(x-4\right)+15\)
\(=\left(x-1\right)\cdot\left(x-4\right)\cdot\left(x-2\right)\cdot\left(x-3\right)+15\)
\(=\left(x^2-5x+4\right)\cdot\left(x^2-5x+6\right)+15\)
Đặt \(t=x^2-5x+4\), ta có:
\(t\cdot\left(t+2\right)+15=t^2+2t+15\)
\(=t^2+2t+1+14=\left(t+1\right)^2+14\ge14\)
Dấu "=" xảy ra khi \(t+1=0\Rightarrow t=-1\Leftrightarrow x^2-5x+4=-1\)
\(\Rightarrow x=\dfrac{5\pm\sqrt{5}}{2}\)
\(\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)+15\)
\(=\left[\left(x-1\right)\left(x-4\right)\right]\left[\left(x-2\right)\left(x-3\right)\right]+15\)
\(=\left(x^2-5x+4\right)\left(x^2-5x+6\right)+15\)
Đặt \(x^2-5x+4=t\) ,ta có :
\(t\left(t+2\right)+15\)
\(=t^2+2t+15\)
\(=\left(t^2+2t+1\right)+14\)
\(=\left(t+1\right)^2+14\)
\(=\left(x^2-5x+4+1\right)^2+14\)
\(=\left(x^2-5x+1\right)^2+14\)
Ta có :
\(\left(x^2-5x+5\right)^2\ge0\) \(\Rightarrow\left(x^2-5x+5\right)^2+14\ge14\)
Dấu = xảy ra khi \(x^2-5x+5=0\)
