\(1=x+y+3xy\le x+y+\dfrac{3}{4}\left(x+y\right)^2\)

\(\Rightarrow3\left(x+y\right)^2+4\left(x+y\right)-4\ge0\)

\(\Rightarrow3\left(x+y+2\right)\left(x+y-\dfrac{2}{3}\right)\ge0\)

\(\Rightarrow x+y\ge\dfrac{2}{3}\) \(\Rightarrow\dfrac{1}{x+y}\le\dfrac{3}{2}\)

Đồng thời: \(x^2+y^2\ge\dfrac{1}{2}\left(x+y\right)^2\ge\dfrac{1}{2}.\left(\dfrac{2}{3}\right)^2=\dfrac{2}{9}\)

\(\Rightarrow-\left(x^2+y^2\right)\le-\dfrac{2}{9}\)

Từ đó ta có:

\(A=\sqrt{1-x^2}+\sqrt{1-y^2}+\dfrac{1-\left(x+y\right)}{x+y}=\sqrt{1-x^2}+\sqrt{1-y^2}+\dfrac{1}{x+y}-1\)

\(A\le\sqrt{2\left[2-\left(x^2+y^2\right)\right]}+\dfrac{1}{x+y}-1\le\sqrt{2\left(2-\dfrac{2}{9}\right)}+\dfrac{3}{2}-1=\dfrac{3+8\sqrt{2}}{6}\)

Dấu "=" xảy ra khi \(x=y=\dfrac{1}{3}\)

\(\sqrt{x+2017}-y^3=\sqrt{y+2017}-x^3\)

\(\Leftrightarrow\left(\sqrt{x+2017}-\sqrt{y+2017}\right)+\left(x^3-y^3\right)=0\)

\(\Leftrightarrow\dfrac{x-y}{\sqrt{x+2017}+\sqrt{y+2017}}+\left(x-y\right)\left(x^2+xy+y^2\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(\dfrac{1}{\sqrt{x+2017}+\sqrt{y+2017}}+\left(x^2+xy+y^2\right)\right)=0\)

\(\Leftrightarrow x=y\)

\(\Rightarrow P=x^2-3x^2+12x-x^2+2018\)

\(=-3x^2+12x+2018=2030-3\left(x-2\right)^2\le2030\)

\(\sqrt{x+5}-y^3=\sqrt{y+5}-x^3\)

\(\Leftrightarrow\sqrt{x+5}-\sqrt{y+5}+x^3-y^3=0\)

\(\Leftrightarrow\dfrac{\left(\sqrt{x+5}-\sqrt{y+5}\right)\left(\sqrt{x+5}+\sqrt{y+5}\right)}{\sqrt{x+5}+\sqrt{y+5}}+\left(x-y\right)\left(x^2+xy+y^2\right)=0\)

\(\Leftrightarrow\dfrac{\left(x+5\right)-\left(y+5\right)}{\sqrt{x+5}+\sqrt{y+5}}+\left(x-y\right)\left(x^2+xy+y^2\right)=0\)

\(\Leftrightarrow\dfrac{x-y}{\sqrt{x+5}+\sqrt{y+5}}+\left(x-y\right)\left(x^2+xy+y^2\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(\dfrac{1}{\sqrt{x+5}+\sqrt{y+5}}+x^2+xy+y^2\right)=0\)

\(\Leftrightarrow x-y=0\) và\(\dfrac{1}{\sqrt{x+5}+\sqrt{y+5}}+x^2+xy+y^2>0\)

\(\Leftrightarrow x=y\)

Khi đó: \(P=x^2-3xy+12-y^2+2018\)

\(\Leftrightarrow P=x^2-3x^2+12-x^2+2018\)

\(\Leftrightarrow P=2030-3x^2\le2030\)

dấu "=" xảy ra khi và chỉ khi x=y=0

vậy maxP=2030 khi và chỉ khi x=y=0

giải bằng Bunhiaskopki nha bạn, search gg

Ta có P \(\le\dfrac{1^2+\left(\sqrt{x-1}\right)^2}{2}+\dfrac{2^2+\left(\sqrt{y-4}\right)^2}{2}+\dfrac{3^2+\left(\sqrt{z-9}\right)^2}{2}\)

\(=\dfrac{1+x-1+4+y-4+9+z-9}{2}=\dfrac{x+y+z}{2}=\dfrac{28}{2}=14\)

Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}1=\sqrt{x-1}\\2=\sqrt{y-4}\\3=\sqrt{z-9}\end{matrix}\right.\Leftrightarrow x=2;y=8;z=18\)(tm) 

\(\sqrt{x-1}-y\sqrt{y}=\sqrt{y-1}-x\sqrt{x}\)

\(\Leftrightarrow\left(\sqrt{x-1}-\sqrt{y-1}\right)+\left(x\sqrt{x}-y\sqrt{y}\right)=0\)

\(\Leftrightarrow\frac{x-y}{\sqrt{x-1}+\sqrt{y-1}}+\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)=0\)

\(\Leftrightarrow\left(\sqrt{x}-\sqrt{y}\right)\left(\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x-1}+\sqrt{y-1}}+x+\sqrt{xy}+y\right)=0\)

\(\Leftrightarrow x=y\)

\(\Rightarrow S=2x^2-8x+5=2\left(x-2\right)^2-3\ge-3\)

Tại sao từ:\(\left(\sqrt{x-1}-\sqrt{y-1}\right)\)  lại => đc: \(\frac{x-y}{\sqrt{x-1}+\sqrt{y-1}}\)??????????

Từ giả thiết ta có:

\(x+y=3\left(\sqrt{x+1}+\sqrt{y+2}\right)\le3\sqrt{2\left(x+y+3\right)}\)

\(\Leftrightarrow P\le3\sqrt{2\left(P+3\right)}\)

\(\Leftrightarrow\left\{{}\begin{matrix}P\ge0\\18P+54\ge P^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}P\ge0\\P^2-18P-54\le0\end{matrix}\right.\)

\(\Leftrightarrow0\le P\le9+3\sqrt{15}\)

\(\Rightarrow maxP=9+3\sqrt{15}\Leftrightarrow\left(x;y\right)=\left(\dfrac{10+3\sqrt{15}}{2};\dfrac{8+3\sqrt{15}}{2}\right)\)

\(P=\sqrt{y}\left(\sqrt{x}+2\sqrt{z}\right)+3\sqrt{zx}=\left(6-\sqrt{x}-\sqrt{z}\right)\left(\sqrt{x}+2\sqrt{z}\right)+3\sqrt{zx}\)

\(P=-x+6\sqrt{x}-2z+12z=-\left(\sqrt{x}-3\right)^2-2\left(\sqrt{z}-3\right)^2+27\le27\)

\(P_{max}=27\) khi \(\left(x;y;z\right)=\left(9;0;9\right)\)