`(x+sqrt{x^2+2020})(sqrt{x^2+2020}-x)=x^2+2020-x^2=2020`

`=>y+sqrt{y^2+2020}=sqrt{x^2+2020}-x`

`<=>x+y=sqrt{x^2+2020}-sqrt{y^2+2020}`

Tương tự:`x+y=sqrt{y^2+2020}-sqrt{x^2+2020}`

Cộng từng vế

`=>2(x+y)=0`

`<=>S=0+2020=2020`

Gt\(\Leftrightarrow\left(x+\sqrt{x^2+2020}\right)\left(x-\sqrt{x^2+2020}\right)\left(y+\sqrt{y^2+2020}\right)=2020\left(x-\sqrt{x^2+2020}\right)\)

\(\Leftrightarrow\left(x^2-x^2-2020\right)\left(y+\sqrt{y^2+2020}\right)=2020\left(x-\sqrt{x^2+2020}\right)\)

\(\Leftrightarrow-y-\sqrt{y^2+2020}=x-\sqrt{x^2+2020}\) (1)

Gt\(\Leftrightarrow\left(x+\sqrt{x^2+2020}\right)\left(y-\sqrt{y^2+2020}\right)\left(y+\sqrt{y^2+2020}\right)=2020\left(y-\sqrt{y^2+2020}\right)\)

\(\Leftrightarrow\left(y^2-y^2-2020\right)\left(x+\sqrt{x^2+2020}\right)=2020\left(y-\sqrt{y^2+2020}\right)\)

\(\Leftrightarrow-x-\sqrt{x^2+2020}=y-\sqrt{y^2+2020}\) (2)

Từ (1) (2) cộng vế với vế \(\Rightarrow-\left(x+y\right)-\left(\sqrt{y^2+2020}+\sqrt{x^2+2020}\right)=x+y-\left(\sqrt{y^2+2020}+\sqrt{x^2+2020}\right)\)

\(\Leftrightarrow-2\left(x+y\right)=0\)

\(\Leftrightarrow x+y=0\)

\(S=x+y+2020=2020\)

Đk: \(\forall x\in R\)

Ta có:\(\sqrt{x^2+1-2x}+\sqrt{x^2+4x+4}=\sqrt{1+2020^2+\frac{2020^2}{2021^2}}+\frac{2020}{2021}\)

<=> \(\sqrt{\left(x-1\right)^2}+\sqrt{\left(x+2\right)^2}=\sqrt{1+2020^2+2.2020+\frac{2020^2}{2021^2}-2.2020}+\frac{2020}{2021}\)

<=> \(\left|x-1\right|+\left|x+2\right|=\sqrt{\left(1+2020\right)^2+\frac{2020^2}{2021^2}-2.2020}+\frac{2020}{2021}\)

<=> \(\left|x-1\right|+\left|x+2\right|=\sqrt{\left(2021-\frac{2020}{2021}\right)^2}+\frac{2020}{2021}\)

<=> \(\left|x-1\right|+\left|x+2\right|=\frac{2021^2-2020}{2021}+\frac{2020}{2021}\)

<=> \(\left|x-1\right|+\left|x+2\right|=2021\)

Lập bảng xét dầu

x                   -2                   1 

x - 1   -         |           -          0       +

x + 2   -        0         +          |            -

Xét các TH xảy ra :

TH1: x \(\le\)-2 => pt trở thành: 1 - x - x - 2 = 2021

<=> -2x = 2022 <=> x = -1011 (tm)

TH2: \(-2< x\le1\) => pt trở thành: 1 - x + x + 2 = 2021

<=> 0x = 2018 (vô lí) => pt vô nghiệm

TH3: \(x>1\) => pt trở thành: x - 1 + x + 2 = 2021

<=> 2x = 2020 <=> x = 1010 (tm)

Vậy S = {-1011; 1010}

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 ...

Ta có :

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

Tương tự :

\(\dfrac{1}{\sqrt{x+2}+\sqrt{x+3}}=-\sqrt{x+2}+\sqrt{x+3}\)

\(\dfrac{1}{\sqrt{x+3}+\sqrt{x+4}}=-\sqrt{x+3}+\sqrt{x+4}\)

....

\(\dfrac{1}{\sqrt{x+2019}+\sqrt{x+2010}}=-\sqrt{x+2019}+\sqrt{x+2010}\)

Từ những ý trên , pt trở thành :

\(-\sqrt{x+1}+\sqrt{x+2}-\sqrt{x+2}+\sqrt{x+3}-\sqrt{x+3}+\sqrt{x+4}-.....-\sqrt{x+2019}+\sqrt{x+2020}=11\)

\(\Leftrightarrow\sqrt{x+2020}-\sqrt{x+1}=11\)

\(\Leftrightarrow x+2020-2\sqrt{\left(x+2020\right)\left(x+1\right)}+x+1=121\)

\(\Leftrightarrow2x+1900=2\sqrt{\left(x+1\right)\left(x+2020\right)}\)

\(\Leftrightarrow x+950=\sqrt{\left(x+1\right)\left(x+2020\right)}\)

\(\Leftrightarrow x^2+1900x+902500=x^2+2021x+2020\)

\(\Leftrightarrow121x-900480=0\)

\(\Leftrightarrow x=\dfrac{900480}{121}\)

\(\frac{1}{\sqrt{x+1}+\sqrt{x+2}}+\frac{1}{\sqrt{x+2}+\sqrt{x+3}}+...+\frac{1}{\sqrt{x+2019}+\sqrt{x+2020}}=11\)

\(\Leftrightarrow\)\(\frac{\sqrt{x+2}-\sqrt{x+1}}{\left(\sqrt{x+1}+\sqrt{x+2}\right)\left(\sqrt{x+2}-\sqrt{x+1}\right)}+\frac{\sqrt{x+3}-\sqrt{x+2}}{\left(\sqrt{x+2}+\sqrt{x+3}\right)\left(\sqrt{x+3}-\sqrt{x+2}\right)}\)

\(+...+\frac{\sqrt{x+2020}-\sqrt{x+2019}}{\left(\sqrt{x+2019}+\sqrt{x+2020}\right)\left(\sqrt{x+2020}-\sqrt{x+2019}\right)}=11\)

\(\Leftrightarrow\)\(\frac{\sqrt{x+2}-\sqrt{x+1}}{x+2-x-1}+\frac{\sqrt{x+3}-\sqrt{x+2}}{x+3-x-2}+...+\frac{\sqrt{x+2020}-\sqrt{x+2019}}{x+2020-x-2019}=11\)

\(\Leftrightarrow\)\(\sqrt{x+2}-\sqrt{x+1}+\sqrt{x+3}-\sqrt{x+2}+...+\sqrt{x+2020}-\sqrt{x+2019}=11\)

\(\Leftrightarrow\)\(\sqrt{x+2020}-\sqrt{x+1}=11\)

\(\Leftrightarrow\)\(\sqrt{x+2020}=11+\sqrt{x+1}\)

\(\Leftrightarrow\)\(x+2020=121+22\sqrt{x+1}+x+1\)

\(\Leftrightarrow\)\(22\sqrt{x+1}=1898\)

\(\Leftrightarrow\)\(\sqrt{x+1}=\frac{949}{11}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x+1=\frac{900601}{121}\\x+1=\frac{-900601}{121}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{900480}{121}\\x=\frac{-900722}{121}\end{cases}}\)

Chúc bạn học tốt ~ 

PS : sai thì thui nhá 

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