ĐKXĐ: \(x\ge2020\)

- Với \(x=2020\Rightarrow A=\frac{1}{2022}\)

- Với \(x>2020\)

\(A=\frac{\sqrt{x-2019}}{x-2019+2021}+\frac{\sqrt{x-2020}}{x-2020+2020}\)

\(A=\frac{1}{\sqrt{x-2019}+\frac{2021}{\sqrt{x-2019}}}+\frac{1}{\sqrt{x-2020}+\frac{2020}{\sqrt{x-2020}}}\)

\(A\le\frac{1}{2\sqrt{2021}}+\frac{1}{2\sqrt{2020}}\)

So sánh với \(\frac{1}{2022}\Rightarrow A_{max}=\frac{1}{2\sqrt{2019}}+\frac{1}{2\sqrt{2020}}\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}x-2019=2021\\x-2020=2020\end{matrix}\right.\) \(\Rightarrow x=4040\)

Thay 2020=x+y+z vao mẫu đc

\(\frac{xy}{\sqrt{xy+zx+zy+z^2}}=\frac{xy}{\sqrt{\left(x+z\right)\left(y+z\right)}}\le\frac{xy}{2}\left(\frac{1}{x+z}+\frac{1}{y+z}\right)\)(Cauchy)

Làm tương tự mấy cái kia sau đó ghép mấy cái cũng mẫu lại là ra

\(\Sigma\left(\frac{xy}{\sqrt{xy+2020z}}\right)=\Sigma\left[\frac{xy}{\sqrt{xy+z\left(x+y+z\right)}}\right]=\Sigma\left[\frac{xy}{\sqrt{\left(y+z\right)\left(z+x\right)}}\right]\)

\(=\Sigma\left[\sqrt{\frac{xy}{y+z}\cdot\frac{xy}{z+x}}\right]\le\Sigma\left[\frac{1}{2}\cdot\left(\frac{xy}{y+z}+\frac{xy}{z+x}\right)\right]\)

\(=\frac{1}{2}\cdot\left(\frac{xy}{y+z}+\frac{xy}{z+x}+\frac{yz}{x+y}+\frac{yz}{z+x}+\frac{zx}{x+y}+\frac{zx}{y+z}\right)\)

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

\(=\frac{1}{2}\cdot\left(x+y+z\right)=\frac{2020}{2}=1010\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=z=\frac{2020}{3}\)

Ta có:

\(\frac{x}{2}=\frac{y}{3}=k\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\end{matrix}\right.\)

\(\Rightarrow A=\frac{2019x+2020y}{2019x-2020y}=\frac{2019.2k+2020.3k}{2019.2k-2020.3k}=\frac{10098k}{-2022k}=\frac{10098}{-2022}=\frac{-1683}{337}\)

Ta có:

\(\frac{x}{2}=\frac{y}{3}.\)

Đặt \(\frac{x}{2}=\frac{y}{3}=k\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\end{matrix}\right.\)

Lại có: \(A=\frac{2019x+2020y}{2019x-2020y}.\)

+ Thay \(x=2k\) và \(y=3k\) vào A ta được:

\(A=\frac{2019.2k+2020.3k}{2019.2k-2020.3k}\)

\(\Rightarrow A=\frac{4038k+6060k}{4038k-6060k}\)

\(\Rightarrow A=\frac{k.\left(4038+6060\right)}{k.\left(4038-6060\right)}\)

\(\Rightarrow A=\frac{4038+6060}{4038-6060}\)

\(\Rightarrow A=\frac{10098}{-2022}\)

\(\Rightarrow A=\frac{-1683}{337}.\)

Vậy \(A=\frac{-1683}{337}.\)

Chúc bạn học tốt!

ĐKXĐ: \(x\ge\dfrac{2020}{2019}>0\)

\(\Leftrightarrow\sqrt{2020x-2019}+\sqrt{2019x-2020}+2019\left(x+1\right)=0\)

\(\Leftrightarrow\dfrac{x+1}{\sqrt{2020x-2019}+\sqrt{2019x-2020}}+2019\left(x+1\right)=0\)

Do \(x>0\) nên hiển nhiên vế trái dương.

Pt vô nghiệm

ĐKXĐ: 

Do  nên hiển nhiên vế trái dương.

Pt vô nghiệm

mình nghĩ ra 2 cách bn thik cách nào thì làm nhé

Đặt P = ...

Ta có: \(P=\sum\sqrt{x+\frac{yz}{x+y+z}}=\sum\sqrt{\frac{\left(x+y\right)\left(x+z\right)}{x+y+z}}=\frac{\sum\sqrt{\left(x+y\right)\left(x+z\right)}}{\sqrt{2020}}\)

\(\le\frac{\sum\left(x+y+x+z\right)}{2\sqrt{2020}}=\frac{4.\left(x+y+z\right)}{2\sqrt{2020}}=2\sqrt{2020}=4\sqrt{505}\)

Dấu "=" xảy ra khi và chỉ khi x = y = z = 2020/3