bài 1 ta có 

\(\left(\frac{1}{a}+\frac{1}{b}\right)\left(2020a+2021b\right)\ge\left(\sqrt{2020}+\sqrt{2021}\right)^2\)  ( BDT Bunhia )

do đó

\(a+b=ab.\left(\frac{1}{a}+\frac{1}{b}\right)\ge\left(\frac{1}{a}+\frac{1}{b}\right)\left(2020a+2021b\right)\ge\left(\sqrt{2020}+\sqrt{2021}\right)^2\)

vậy ta có đpcm.

bài 2.

ta có \(VT=\sqrt{x-3}+\sqrt{5-x}\le2\)( BDT Bunhia )

\(VP=y^2+2.\sqrt{2019}y+2021=\left(y+\sqrt{2019}\right)^2+2\ge2\)

suy ra PT có nghiệm \(\hept{\begin{cases}x-3=5-x\\y+\sqrt{2019}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=4\\y=-\sqrt{2019}\end{cases}}}\)

xét x=y,x>y và x<y chú ý tới điều kiện x,y thuộc -1;1 nữa 

ĐKXĐ : \(\left\{{}\begin{matrix}x>2019\\y>2020\\z>2021\end{matrix}\right.\)

Đặt \(\sqrt{x-2019}=a,......\)

Ta được PT : \(\dfrac{1-a}{a^2}+\dfrac{1-b}{b^2}+\dfrac{1-c}{c^2}+\dfrac{3}{4}=0\)

\(\Leftrightarrow\dfrac{1}{a^2}-\dfrac{1}{a}+\dfrac{1}{4}+\dfrac{1}{b^2}-\dfrac{1}{b}+\dfrac{1}{4}+\dfrac{1}{c^2}-\dfrac{1}{c}+\dfrac{1}{4}=0\)

\(\Leftrightarrow\left(\dfrac{1}{a}-\dfrac{1}{2}\right)^2+\left(\dfrac{1}{b}-\dfrac{1}{2}\right)^2+\left(\dfrac{1}{c}-\dfrac{1}{2}\right)^2=0\)

- Thấy : \(\left(\dfrac{1}{a}-\dfrac{1}{2}\right)^2\ge0,......\)

\(\Rightarrow\left(\dfrac{1}{a}-\dfrac{1}{2}\right)^2+\left(\dfrac{1}{b}-\dfrac{1}{2}\right)^2+\left(\dfrac{1}{c}-\dfrac{1}{2}\right)^2\ge0\)

- Dấu " = " xảy ra <=> \(\left\{{}\begin{matrix}\dfrac{1}{a}=\dfrac{1}{2}\\\dfrac{1}{b}=\dfrac{1}{2}\\\dfrac{1}{c}=\dfrac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=2\\c=2\end{matrix}\right.\)

- Thay lại a. b. c ta được : \(\left\{{}\begin{matrix}\sqrt{x-2019}=2\\\sqrt{y-2020}=2\\\sqrt{z-2021}=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2019=4\\y-2020=4\\z-2021=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2023\\y=2024\\z=2025\end{matrix}\right.\) ( TM )

Vậy ...

\(x+y+z-6046=2\sqrt{x-2019}+4\sqrt{y-2020}+6\sqrt{z-2021}\)

\(\left(x-2019\right)+\left(x-2020\right)+\left(x-2021\right)+1+4+9\)\(=2\sqrt{x-2019}+4\sqrt{y-2020}+6\sqrt{z-2021}\)

đặt :\(\hept{\begin{cases}\sqrt{x-2019}=a\\\sqrt{y-2020}=b\\\sqrt{z-2021}=c\end{cases}\left(đk:a,b,c\ge0\right)}\)

PT <=>  \(a^2+b^2+c^2+1+4+9=2a+4b+6c\)

\(\Leftrightarrow\left(a-1\right)^2+\left(b-2\right)^2+\left(c-6\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}a-1=0\\b-2=0\\c-3=0\end{cases}\Leftrightarrow\hept{\begin{cases}a=1\\b=2\\c=3\end{cases}\left(tm\right)}}\)

\(\Rightarrow\hept{\begin{cases}x=2020\\y=2024\\z=2030\end{cases}}\)

\(x=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\)

\(\Leftrightarrow x^3=18+3x\)

Tương tự co:

\(y^3=6+3y\)

\(\Rightarrow P=18+3x+6+3y-3\left(x+y\right)+2019=2043\)

Nhân 2 vế của giả thiết với \(\sqrt{x^2+2021}-x>0\):

\(\left(\sqrt{x^2+2021}-x\right)\left(x+\sqrt{x^2+2021}\right)\left(y+\sqrt{y^2+2021}\right)=2021\left(\sqrt{x^2+2021}-x\right)\)

\(\Leftrightarrow2021\left(y+\sqrt{y^2+2021}\right)=2021\left(\sqrt{x^2+2021}-x\right)\)

\(\Leftrightarrow y+\sqrt{y^2+2021}=\sqrt{x^2+2021}-x\) (1)

Tương tự, nhân 2 vế giả thiết với \(\sqrt{y^2+2021}-y\) và rút gọn ta được:

\(x+\sqrt{x^2+2021}=\sqrt{y^2+2021}-y\) (2)

Cộng vế với vế (1) và (2):

\(x+y+\sqrt{x^2+2021}+\sqrt{y^2+2021}=\sqrt{x^2+2021}+\sqrt{y^2+2021}-x-y\)

\(\Leftrightarrow2x+2y=0\Rightarrow A=0\)

:v e xin kiếu

Em tham khảo nhé

https://hoc24.vn/cau-hoi/cho-xsqrtx22021ysqrty220212021tinh-axy.332667728355