\(P=\dfrac{x}{\sqrt{y}}+\dfrac{y}{\sqrt{x}}\Rightarrow P^2=\dfrac{x^2}{y}+\dfrac{y^2}{x}+2\sqrt{xy}\)

\(P^2=\left(\dfrac{x^2}{y}+\sqrt{xy}+\sqrt{xy}\right)+\left(\dfrac{y^2}{x}+\sqrt{xy}+\sqrt{xy}\right)-2\sqrt{xy}\)

\(P^2\ge3x+3y-2\sqrt{xy}\ge3\left(x+y\right)-\left(x+y\right)=2\left(x+y\right)=4038\)

\(\Rightarrow P\ge\sqrt{4038}\)

Dấu "=" xảy ra khi \(x=y=\dfrac{2019}{2}\)

Ta có:

\(P=\dfrac{x}{\sqrt{2019-x}}+\dfrac{y}{\sqrt{y-2019}}=\dfrac{x}{\sqrt{y}}+\dfrac{y}{\sqrt{x}}\ge\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^2}{\sqrt{x}+\sqrt{y}}=\sqrt{x}+\sqrt{y}\)

Lại có:

\(P=\dfrac{x}{\sqrt{2019-x}}+\dfrac{y}{\sqrt{2019-y}}=\dfrac{2019-y}{\sqrt{y}}+\dfrac{2019-x}{\sqrt{x}}\\ =\dfrac{2019}{\sqrt{x}}+\dfrac{2019}{\sqrt{y}}-\sqrt{x}-\sqrt{y}\)

\(\Rightarrow2P=\dfrac{2019}{\sqrt{x}}+\dfrac{2019}{\sqrt{y}}=2019\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\right)\ge2019\cdot\dfrac{2}{\sqrt[4]{xy}}\\ \ge2019\dfrac{2}{\sqrt[2]{\dfrac{x+y}{2}}}=2019\cdot\dfrac{2}{\sqrt{\dfrac{2019}{2}}}=2\sqrt{2}\sqrt{2019}\)

\(\Rightarrow P\ge\sqrt{2}\sqrt{2019}\)

Dấu = khi \(x=y=\dfrac{2019}{2}\)

Có \(y^2+2019=y^2+xy+yz+zx=y\left(x+y\right)+z\left(x+y\right)=\left(y+z\right)\left(x+y\right)\)

\(x^2+2019=x^2+xy+yz+zx=x\left(x+y\right)+z\left(x+y\right)=\left(x+z\right)\left(x+y\right)\)

\(z^2+2019=z^2+xy+yz+xz=z\left(z+y\right)+x\left(y+z\right)=\left(z+x\right)\left(y+z\right)\)

Có \(P=x\sqrt{\frac{\left(y^2+2019\right)\left(z^2+2019\right)}{x^2+2019}}+y\sqrt{\frac{\left(z^2+2019\right)\left(x^2+2019\right)}{y^2+2019}}+z\sqrt{\frac{\left(x^2+2019\right)\left(y^2+2019\right)}{z^2+2019}}\)

=\(x\sqrt{\frac{\left(y+z\right)\left(x+y\right)\left(x+z\right)\left(z+y\right)}{\left(x+z\right)\left(y+x\right)}}+y\sqrt{\frac{\left(z+x\right)\left(y+z\right)\left(x+z\right)\left(x+y\right)}{\left(y+z\right)\left(x+y\right)}}+z\sqrt{\frac{\left(x+z\right)\left(x+y\right)\left(y+z\right)\left(x+y\right)}{\left(z+x\right)\left(y+z\right)}}\)

=\(x\sqrt{\left(y+z\right)^2}+y\sqrt{\left(x+z\right)^2}+z\sqrt{\left(x+y\right)^2}\)

=\(x\left|y+z\right|+y\left|x+z\right|+z\left|x+y\right|\)

=\(x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\) (vì x,y,z >0)

= xy+xz+xy+yz+xz+yz

=2(xy+xz+yz)=2.2019(vì xy+xz+yz=2019)

=4038

Vậy P=4038

Từ gt suy ra: \(x+\sqrt{x^2+2019}=\dfrac{2019}{y+\sqrt{y^2+2019}}=\sqrt{y^2+2019}-y\).

Tương tự: \(y+\sqrt{y^2+2019}=\sqrt{x^2+2019}-x\).

Do đó dễ dàng suy ra được: \(x+y=0\).

\(\Rightarrow x=-y\Rightarrow x^{2019}+y^{2019}=x^{2019}+\left(-x\right)^{2019}=0\left(đpcm\right)\).

bạn lm ra 2 hướng

hướng 1 ) liên hợp với (x - căn (x²+2019)) ( nhân vào 2 vế)

biến đổi nhân ra => ....(1)

hướng 2) liên hợp với (y-căn (y² + 2019)) ( nhân vào 2 vế)

biến đổi  nhân ra=>....(2)

từ (1) và (2) => x=-y hay x=y gì đó

r tính A

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học tốt

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\(\left(x+\sqrt{x^2+2019}\right)\left(y+\sqrt{y^2+2019}\right)=2019\)

\(\Rightarrow\left(x+\sqrt{x^2+2019}\right)\left(y+\sqrt{y^2+2019}\right)\left(\sqrt{x^2+2019}-x\right)=2019\left(\sqrt{x^2+2019}-x\right)\)

\(\Leftrightarrow\left(y+\sqrt{y^2+2019}\right)\left(\left(\sqrt{x^2+2019}\right)^2-x^2\right)=2019\left(\sqrt{x^2+2019}-x\right)\)

\(\Leftrightarrow\left(y+\sqrt{y^2+2019}\right)\left(x^2+2019-x^2\right)=2019\left(\sqrt{x^2+2019}-x\right)\)

\(\Leftrightarrow2019\left(y+\sqrt{y^2+2019}\right)=2019\left(\sqrt{x^2+2019}-x\right)\)

\(\Leftrightarrow y+\sqrt{y^2+2019}=\sqrt{x^2+2019}-x\)(1)

Tương tự ta có:

\(\Leftrightarrow x+\sqrt{x^2+2019}=\sqrt{y^2+2019}-y\)(2)

Cộng vế theo vế (1) vả (2) ta có:

\(x+y+\sqrt{x^2+2019}+\sqrt{y^2+2019}=\sqrt{x^2+2019}+\sqrt{y^2+2019}-x-y\)

\(\Leftrightarrow2x+2y=0\)\(\Leftrightarrow x+y=0\)

\(\Rightarrow A=2019\left(x+y\right)=2019.0=0\)

\(\left(x+\sqrt{x^2+2019}\right)\left(\sqrt{x^2+2019}-x\right)=x^2+2019-x^2=2019\)

\(\Rightarrow\sqrt{x^2+2019}-x=y+\sqrt{y^2+2019}\left(2\right)\)

Tương tự \(\sqrt{y^2+2019}-y=x+\sqrt{x^2+2019}\left(1\right)\)

Lấy (2) - (1) được: -2x = 2y

                       <=> -x = y

                       <=> x + y = 0