ĐK: \(3\le x\le5\)

\(\begin{align} & VT=\left( \sqrt{x-3}+\sqrt{5-x} \right)\le 2\left( x-3+5-x \right) \\ & \Leftrightarrow {{\left( \sqrt{x-3}+\sqrt{5-x} \right)}^{2}}\le 4 \\ & \Rightarrow \sqrt{x-3}+\sqrt{5-x}\le 2 \\ & VP={{\left( y+\sqrt{2013} \right)}^{2}}+2\ge 2 \\ \end{align}\)

Vậy phương trình chỉ tồn tại khi $VT=VP=2$

\(\Leftrightarrow\left\{{}\begin{matrix}\left(\sqrt{x-3}+\sqrt{5-x}\right)^2=2^2\\\left(y+\sqrt{2013}\right)^2+2=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=\sqrt{2013}\end{matrix}\right.\)

\(\left(\sqrt{x-1}+\sqrt{3-x}\right)^2\le\left(1^2+1^2\right)\left(x-1+3-x\right)=4\\ \Leftrightarrow\sqrt{x-1}+\sqrt{3-x}\le2\\ y^2+2\sqrt{2020}y+2022=\left(y^2+2y\sqrt{2020}+2020\right)+2\\ =\left(y+\sqrt{2020}\right)^2+2\ge2\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x-1=3-x\\y+\sqrt{2020}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-\sqrt{2020}\end{matrix}\right.\)

Vậy ...

ĐKXĐ: \(3\ge x\ge1\)

Áp dụng BĐT Bunhiacopski:

\(1\sqrt{x-1}+1\sqrt{3-x}\le\sqrt{\left(1^2+1^2\right)\left(x-1+3-x\right)}=\sqrt{2.2}=2\)

Mặt khác: \(y^2+2\sqrt{2020}y+2022=\left(y+\sqrt{2020}\right)^2+2\ge2\)

Nên để thõa mãn yêu cầu bài toán thì

\(\left\{{}\begin{matrix}\sqrt{x-1}=\sqrt{3-x}\\y+\sqrt{2020}=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\left(tm\right)\\y=-\sqrt{2020}\end{matrix}\right.\)

Lời giải:

Ta có:\(y^2+2\sqrt{2020}y+2022=(y^2+2\sqrt{2020}y+2020)+2=(y+\sqrt{2020})^2+2\geq 2(1)\)

Áp dụng BĐT Bunhiacopxky:

$(\sqrt{x-1}+\sqrt{3-x})^2\leq (x-1+3-x)(1+1)=4$

$\Rightarrow \sqrt{x-1}+\sqrt{3-x}\leq 2(2)$

Từ $(1); (2)\Rightarrow \sqrt{x-1}+\sqrt{3-x}\leq 2\leq y^2+2\sqrt{2020}y+2022$

Dấu "=" xảy ra khi mà: \(\left\{\begin{matrix} \frac{x-1}{1}=\frac{3-x}{1}\\ y+\sqrt{2020}=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=2\\ y=-\sqrt{2020}\end{matrix}\right.\)

ĐKXĐ: x,y >1

\(\sqrt{x^2+5}+\sqrt{x-1}+x^2=\sqrt{y^2+5}+\sqrt{y-1}+y^2\\ \)

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

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

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

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

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

\(\Rightarrow x-y=0\Leftrightarrow x=y\)

Giả sử x=y

Khi đó:

\(\sqrt{x^2+5}+\sqrt{x-1}+x^2\)

\(=\sqrt{y^2+5}+\sqrt{x-1}+y^2\)

Luôn đúng 

Vậy ta suy ra đpcm

Ta có\(x\sqrt{\frac{\left(2015+y^2\right)\left(2015+z^2\right)}{2015+x^2}}=x\sqrt{\frac{\left(xy+yz+zx+y^2\right)\left(xy+yz+zx+z^2\right)}{xy+yz+zx+x^2}}\)

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

Tương tự:\(y\sqrt{\frac{\left(2015+x^2\right)\left(2015+z^2\right)}{2015+y^2}}=yx+yz\)

               \(z\sqrt{\frac{\left(2015+x^2\right)\left(2015+y^2\right)}{2015+z^2}}=zx+zy\)

Ta có :\(P=xy+xz+yx+yz+zx+zy=2\left(xy+yz+zx\right)=4030\)

=>P không phải là số chính phương

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

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

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

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

\(\Leftrightarrow x=y\)

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

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

Theo đề bài, ta có:

\(x^3+y^3=x^2-xy+y^2\)

hay \(\left(x^2-xy+y^2\right)\left(x+y-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x^2-xy+y^2=0\\x+y=1\end{cases}}\)

+ Với \(x^2-xy+y^2=0\Rightarrow x=y=0\Rightarrow P=\frac{5}{2}\)

+ với \(x+y=1\Rightarrow0\le x,y\le1\Rightarrow P\le\frac{1+\sqrt{1}}{2+\sqrt{0}}+\frac{2+\sqrt{1}}{1+\sqrt{0}}=4\)

Dấu đẳng thức xảy ra <=> x=1;y=0 và \(P\ge\frac{1+\sqrt{0}}{2+\sqrt{1}}+\frac{2+\sqrt{0}}{1+\sqrt{1}}=\frac{4}{3}\)

Dấu đẳng thức xảy ra <=> x=0;y=1

Vậy max P=4 và min P =4/3