\(A=n^3+\left(n^3+3n^2+3n+1\right)+\left(n^3+6n^2+12n+8\right)\)

\(=3n^3+9n^2+15n+9=3\left(n^3+3n^2+5n+3\right)\)

Đặt \(B=n^3+3n^2+5n+1=n^3+n^2+2n^2+2n+3n+3\)

\(=n^2\left(n+1\right)+2n\left(n+1\right)+3\left(n+1\right)=n\left(n+1\right)\left(n+2\right)+3\left(n+1\right)\)

Ta thấy \(n\left(n+1\right)\left(n+2\right)⋮3\)( vì là tích của 3 số tự nhiên liên tiếp)

\(3\left(n+1\right)⋮3\Rightarrow B⋮3\Rightarrow A=3B⋮9\)

TA CÓ: \(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

Do đó: \(\frac{a^3+b^3+c^3}{4abc}=\frac{3}{4}+\frac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)}{4abc}\)

\(=\frac{3}{4}+\frac{1}{4}\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

\(\ge\frac{3}{4}+\frac{1}{4}.\frac{9}{ab+bc+ca}\left(a^2+b^2+c^2-ab-bc-ca\right)\)

\(=\frac{3}{4}+\frac{9\left(a^2+b^2+c^2\right)}{4\left(ab+bc+ca\right)}-\frac{9}{4}=\frac{9\left(a^2+b^2+c^2\right)}{4\left(ab+bc+ca\right)}-\frac{3}{2}\)

\(\Rightarrow P\ge\frac{1}{30}+\frac{ab+bc+ca}{15\left(a^2+b^2+c^2\right)}+\frac{9\left(a^2+b^2+c^2\right)}{4\left(ab+bc+ca\right)}-\frac{131\left(a^2+b^2+c^2\right)}{60\left(ab+bc+ca\right)}-\frac{3}{2}\)

\(=\frac{-22}{15}+\frac{ab+bc+ca}{15\left(a^2+b^2+c^2\right)}+\frac{a^2+b^2+c^2}{15\left(ab+bc+ca\right)}\)

\(\ge\frac{-22}{15}+2\sqrt{\left[\frac{ab+bc+ca}{15\left(a^2+b^2+c^2\right)}\right]\left[\frac{a^2+b^2+c^2}{15\left(ab+bc+ca\right)}\right]}=\frac{-22}{15}+\frac{2}{15}=\frac{-4}{3}\)

Dấu '=' xảy ra <=> a=b=c

Vậy GTNN của P là -4/3 khi a=b=c

trả lời hộ cái

2xy-x²-y²+16

=16-(x^2-2xy+y^2)

=16-(x-y)^2

=(4+x-y)(4(-x+y)

\(2x\left(x-1\right)-\left(x-1\right)\left(x-2\right)\)

\(=\left(x-1\right)\left(2x-x-2\right)\)

\(=\left(x-1\right)\left(x-2\right)\)