a: \(x^2-x+1\)
\(=x^2-x+\frac14+\frac34\)
\(=\left(x-\frac12\right)^2+\frac34\ge\frac34>0\forall x\)
b: \(-x^2+4x-5\)
\(=-\left(x^2-4x+5\right)\)
\(=-\left(x^2-4x+4+1\right)=-\left(x-2\right)^2-1\le-1<0\forall x\)
c: \(a^2\left(a+1\right)+2a\left(a+1\right)\)
\(=\left(a+1\right)\left(a^2+2a\right)=a\left(a+2\right)\left(a+1\right)\)
Vì a;a+1;a+2 là ba số nguyên liên tiếp
nên a(a+1)(a+2)⋮3!=6
=>\(a^2\left(a+1\right)+2a\left(a+1\right)\) ⋮6
d: Đặt A=a(2a-3)-2a(a+1)
\(=2a^2-3a-2a^2-2a\)
=-5a
=>A⋮5
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