2: \(M=-4x^2-4x+1\)
\(=-4x^2-4x-1+2\)
\(=-\left(4x^2+4x+1\right)+2\)
\(=-\left(2x+1\right)^2+2< =2\forall x\)
Dấu '=' xảy ra khi 2x+1=0
=>2x=-1
=>x=-1/2
3:
Đặt \(A=n^4+2n^3-n^2-2n\)
\(=\left(n^4+2n^3\right)-\left(n^2+2n\right)\)
\(=n^3\left(n+2\right)-n\left(n+2\right)\)
\(=\left(n+2\right)\left(n^3-n\right)\)
\(=n\cdot\left(n+2\right)\left(n^2-1\right)\)
\(=\left(n-1\right)\cdot n\cdot\left(n+1\right)\left(n+2\right)\)
Vì n-1;n;n+1;n+2 là bốn số nguyên liên tiếp
nên \(A=\left(n-1\right)\cdot n\cdot\left(n+1\right)\left(n+2\right)⋮4!=24\)
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