Ta có : \(n^3+2018n=n\left(n^2-1+2019\right)=\left(n-1\right)n\left(n+1\right)+2019n⋮3\forall n\inℤ\) (*)

Lại có : \(2020\equiv1\left(mod3\right)\)

\(\Rightarrow2020^{2019}\equiv1\left(mod3\right)\)

Và : \(4\equiv1\left(mod3\right)\)

Do đó : \(2020^{2019}+4\equiv2\left(mod3\right)\)

hay \(2020^{2019}+4⋮̸3\) . Điều này mâu thuẫn với (*)

Do đó, không tồn tại số nguyên n thỏa mãn đề.

Cho hỏi D là điểm nào vậy

Ta có:\(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)=a^2-ab+b^2\)(vì a+b=1)

Lại có \(2\left(a-b\right)^2\ge0\Leftrightarrow2a^2-4ab+2b^2\ge2a^2+2b^2\)

\(\Leftrightarrow4\left(a^2-ab+b^2\right)\ge2\left(a^2+b^2\right)\ge\left(a+b\right)^2=1\)

\(\Leftrightarrow4\left(a^2-ab+b^2\right)\ge1\Leftrightarrow a^2-ab+b^2\ge\frac{1}{4}\Rightarrow a^3+b^3\ge\frac{1}{4}\)

\(\Rightarrow a^3+b^3\ge\frac{1}{4}\)(đpcm)

1) \(P=x^2+3x+3=\left(x^2+2.x\cdot\frac{3}{2}+\frac{9}{4}\right)+\frac{3}{4}\)

\(=\left(x+\frac{3}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Dấu "=" xảy ra khi \(x=-\frac{3}{2}\)

2) \(Q=\left(x+y\right)^2+y^2-2\ge-2\)

Dấu "=" xảy ra khi x=0,y=0

\(P=x^2+3x+3\)

\(=x^2+2.x.\frac{3}{2}+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2+3\)

\(=\left(x+\frac{3}{2}\right)^2+\frac{3}{4}\)

Vì \(\left(x+\frac{3}{2}\right)^2\ge0\forall x\)

\(\Rightarrow\left(x+\frac{3}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\)

\(\Rightarrow P_{min}=\frac{3}{4}\Leftrightarrow x=-\frac{3}{2}\)

\(Q=x^2+2y^2+2xy-2\)

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

\(=\left(x^2+2xy+y^2\right)+y^2-2\)

\(=\left(x+y\right)^2+y^2-2\)

Vì \(\hept{\begin{cases}\left(x+y\right)^2\ge0\forall x,y\\y^2\ge0\forall y\end{cases}}\)

\(\Rightarrow\left(x+y\right)^2+y^2\ge0\forall x,y\)

\(\Rightarrow\left(x+y\right)^2+y^2-2\ge-2\forall x,y\)

\(\Rightarrow Q_{min}=-2\Leftrightarrow x=y=0\)

dai vcl

\(1,\frac{x^6+2x^3y^3+y^6}{x^7-xy^6}=\frac{\left(x^3+y^3\right)^2}{x\left(x^6-y^6\right)}=\frac{\left(x^3+y^3\right)^2}{x\left(x^3-y^3\right)\left(x^3+y^3\right)}=\frac{x^3+y^3}{x\left(x^3-y^3\right)}\)

\(2,=\frac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}=\frac{\left(a+b+c\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(a+c-b\right)}=\frac{a+b-c}{a+c-b}\)

pt thành nhân tử là ra

Ta có :\(x^7+y^7\ge x^3y^3\left(x+y\right)\)

\(\Leftrightarrow x^7-x^4y^3+y^7-x^3y^4\ge0\)

\(\Leftrightarrow x^4\left(x^3-y^3\right)-y^4\left(x^3-y^3\right)\ge0\)

\(\Leftrightarrow\left(x^3-y^3\right)\left(x^4-y^4\right)\ge0\)

\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2\right)\left(x^2-y^2\right)\left(x^2+y^2\right)\ge0\)

\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2\right)\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\ge0\)

\(\Leftrightarrow\left(x-y\right)^2\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2+y^2\right)\ge0\) ( Luôn đúng với x,y dương )

Do đó : \(x^7+y^7\ge x^3y^3\left(x+y\right)\)