Xét :\(VT^2=2020-x+x-2018+2\sqrt{\left(2012-x\right)\left(x-2018\right)}\)

\(=2+2\sqrt{\left(2012-x\right)\left(x-2018\right)}\)

Áp dụng bđt AM - GM ta có : \(2\sqrt{\left(2012-x\right)\left(x-2018\right)}\le2012-x+x-2018=2\)

\(\Rightarrow VT^2\le4\Rightarrow VT\le2\)(1)

Xét \(VP=x^2-4038x+4076363=\left(x^2-4038x+4076361\right)+2\)

\(=\left(x-2019\right)^2+2\ge2\) (2)

Từ (1);(2) \(\Rightarrow VT\le2\le VP\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}2020-x=x-2018\\\left(x-2019\right)^2=0\end{cases}\Rightarrow x=2019\left(TM\right)}\)

Vậy nghiệm của PT là \(S=\left\{2019\right\}\)

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

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

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

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

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

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

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

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

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

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

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

Chúc bạn học tốt ~ 

PS : sai thì thui nhá 

Bài của bạn Quân làm đúng ùi nhưng mà căn thì không ra số âm nhé!

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

1.

đk: \(x\ge2\)

Đặt y = \(\sqrt{x+2}\) ta biến pt về dạng pt thuần nhất bậc 3 đối vs x và y:

ta có : \(x^3-3x^2+2y^3-6x=0\)

\(\Leftrightarrow x^3-3xy^2+2y^3=0\)

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

ta sẽ có nghiệm : \(x=2;x=2-2\sqrt{3}\)

\(1.đk:\left(x+2\right)^3\ge0\Leftrightarrow x\ge-2\)

\(pt\Leftrightarrow x^3-3x\left(x+2\right)+2\sqrt{\left(x+2\right)^3}=0\)

\(\Leftrightarrow x^3-x\left(x+2\right)+2\sqrt{\left(x+3\right)^2}-2x\left(x+2\right)=0\)

\(\Leftrightarrow x\left[x^2-\left(x+2\right)\right]+2\left(x+2\right)\left(\sqrt{x+2}-x\right)=0\)

\(\Leftrightarrow x\left[\left(x-\sqrt{x+2}\right)\left(x+\sqrt{x+2}\right)\right]+2\left(x+2\right)\left(\sqrt{x+2}-x\right)=0\)

\(\Leftrightarrow\left(\sqrt{x+2}-x\right)\left[-x\left(\sqrt{x+2}+x\right)+2\left(x+2\right)\right]=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+2}=x\left(2\right)\\2\sqrt{x+2}=-x\left(3\right)\end{matrix}\right.\)

\(\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^2=x+2\end{matrix}\right.\)\(\Leftrightarrow x=2\left(tm\right)\)

\(\left(3\right)\Leftrightarrow\left\{{}\begin{matrix}-x\ge0\Leftrightarrow x\le0\\x^2=4\left(x+2\right)\end{matrix}\right.\)\(\Leftrightarrow x=2-2\sqrt{3}\left(tm\right)\)

\(2.đk:x^2;y^2\ge2018\Leftrightarrow\left[{}\begin{matrix}x;y\le-\sqrt{2018}\\x;y\ge\sqrt{2018}\end{matrix}\right.\)

\(pt\Leftrightarrow\sqrt{x^2+11}-\sqrt{y^2+11}+\sqrt{x^2-2018}-\sqrt{y^2-2018}+x^2-y^2=0\)

\(\Leftrightarrow\left(x+y\right)\left(x-y\right)+\dfrac{x^2+11-y^2-11}{\sqrt{x^2+11}+\sqrt{y^2+11}}+\dfrac{x^2-2018-y^2+2018}{\sqrt{x^2-2018}+\sqrt{y^2-2018}}=0\)

\(\Leftrightarrow\left(x-y\right)\left(x+y\right)\left[1+\dfrac{1}{\sqrt{x^2+11}+\sqrt{y^2+11}}+\dfrac{1}{\sqrt{x^2-2018}+\sqrt{y^2+2018}}>0\right]=0\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-y\end{matrix}\right.\)

\(x=y\Rightarrow M=x^{11}-x^{2018}\)

\(x=-y\Rightarrow M=-y^{11}-y^{2018}=:vvv\) (đến đây chịu)

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 ...

Chiều mk lm cho

Đang dùng đt

Ta có:

\(\sqrt{x^2-2018x+2018}+\sqrt{x^2-1009x+1009}=2x\)

\(\Leftrightarrow x-\sqrt{\left(2018x-2018\right)}+x-\sqrt{\left(1009x-1009\right)}=2x\)

\(\Leftrightarrow2x-\sqrt{\left(2018x-2018\right)}-\sqrt{\left(1009x-1009\right)}=2x\)

\(\Leftrightarrow\sqrt{\left(2018x\right)-2018}+\sqrt{\left(1009x-1009\right)}=0\)

\(\Leftrightarrow\sqrt{\left(2018x-2018\right)}=\sqrt{\left(1009x-1009\right)}=0\)

\(\Leftrightarrow2018x-2018=1009x-1009=0\Leftrightarrow x=1\)