2017 = 2016 + 1 = x + 1

suy ra 2017x¹⁵ = x¹⁶ + x¹⁵

2017x¹⁴ = x¹⁵ + x¹⁴

.... 

từ đó ta dễ tính ra A

\(A=x^4\left(y-z\right)+y^4\left(z-x\right)+z^4\left(x-y\right)\)

\(A=x^4\left(y-z\right)+y^4\left(z-x\right)-z^4\left[\left(y-z\right)+\left(z-x\right)\right]\)

\(A=x^4\left(y-z\right)-z^4\left(y-z\right)+y^4\left(z-x\right)-z^4\left(z-x\right)\)

\(A=\left(y-z\right)\left(x^4-z^4\right)+\left(z-x\right)\left(y^4-z^4\right)\)

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

\(A=\left(y-z\right)\left(x-z\right)\left(x^3+xz^2+x^2z+z^3-y^3-yz^2-y^2z-z^3\right)\)

\(A=\left(y-z\right)\left(x-z\right)\left(x-y\right)\left(x^2+xy+y^2+z^2+zx+yz\right)\)

\(A=\frac{1}{2}\left(x-y\right)\left(y-z\right)\left(x-z\right)\left[\left(x+y\right)^2+\left(y+z\right)^2+\left(z+x\right)^2\right]\)

Vì \(x>y>z\Rightarrow A>0\)

Đặt \(2n+2017=a^2;n+2019=b^2\)

\(\Rightarrow2n+4038=2b^2\)

\(\Rightarrow2b^2-a^2=2021\)

\(\Leftrightarrow\left(\sqrt{2b}-a\right)\left(\sqrt{2b}+a\right)=2021=1\cdot2021=47\cdot43\)

Tự xét nốt nha

\(\frac{1}{a}+\frac{1}{b}=\frac{1}{2019}\)

\(\Leftrightarrow\frac{a+b}{ab}=\frac{1}{2019}\)

\(\Leftrightarrow2019a+2019b-ab=0\)

\(\Leftrightarrow ab-2019a-2019b=0\)

\(\sqrt{a+b}=\sqrt{a-2019}+\sqrt{b-2019}\)

\(\Leftrightarrow a+b=a-2019+b-2019+2\sqrt{\left(a-2019\right)\left(b-2019\right)}\)

\(\Leftrightarrow2\sqrt{ab-2019a-2019b+2019^2}=2\cdot2019\)

\(\Leftrightarrow2\cdot2019=2\cdot2019\) ( LUÔN OK THEO COOL KID ĐZ )

P/S:SORRY NHA.LÚC CHIỀU BẬN VÀI VIỆC NÊN KO ONL DC:(((

CÁC BẠN GIÚP MÌNH VỚI

\(x^3-3x\)

\(=x\left(x^2-3\right)\)

\(=x\left(x+\sqrt{3}\right)\left(x-\sqrt{3}\right)\)