CMR (x-1)(x-2)(x-3)(x-4) ≥-1
(x-1)(x-2)(x-3)(x-4)=[(x-1)(x-4)][(x-2)(x-3)=(x^2-5x+4)(x^2-5x+6)
đặt x^2-5x+5=y ta đc
(y-1)(y+1)=y^2-1
vì y^2> hoặc =0
=>x^2-1> hoặc bằng -1
Có :
\(\text{(x-1)(x-2)(x-3)(x-4) ≥ -1}\)
\(\Rightarrow\left(x-1\right)\left(x-4\right)\left(x-2\right)\left(x-3\right)+1\ge0\)
\(\Rightarrow\left(x^2-5x+4\right)\left(x^2-5x+6\right)+1\ge0\)
Thay x2-5x+4 = a
\(\Rightarrow a\left(a+2\right)+1\) \(\ge0\)
\(\Rightarrow a^2+2a+1\) \(\ge0\)
\(\Rightarrow\left(a+1\right)^2\) \(\ge0\)
\(\Rightarrow\left(x^2+5x+5\right)^2\ge0\) ( luôn đúng)
\(\Rightarrow\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)\ge-1\)
Vậy \(\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)\ge-1\)
