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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
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:
\(\sqrt{x^2-2018x+2018}+\sqrt{x^2-1009x+1009}=2x\)
\(\Leftrightarrow x-\sqrt{\left(2018x-2018\right)}+x-\sqrt{\left(1009x-1009\right)}=2x\)
\(\Leftrightarrow2x-\sqrt{\left(2018x-2018\right)}-\sqrt{\left(1009x-1009\right)}=2x\)
\(\Leftrightarrow\sqrt{\left(2018x\right)-2018}+\sqrt{\left(1009x-1009\right)}=0\)
\(\Leftrightarrow\sqrt{\left(2018x-2018\right)}=\sqrt{\left(1009x-1009\right)}=0\)
\(\Leftrightarrow2018x-2018=1009x-1009=0\Leftrightarrow x=1\)
Pttđ: \(x^2-x-1=2018\left(\sqrt{x^2+x+2}-\sqrt{2x^2+1}\right)\)(1)
Đặt \(\sqrt{2x^2+1}=a;\sqrt{x^2+x+2}=b\Rightarrow x^2-x-1=a^2-b^2\)
(1) <=> a2-b2=2018(b-a)
<=>(a-b)(a+b)=-2018(a-b)
<=>a=b hoặc a+b=-2018
Tự giải tiếp nha
ĐKXĐ: ...
\(VT\le\sqrt{2\left(2020-x+x-2018\right)}=2\)
\(VP=\left(x-2019\right)^2+2\ge2\)
\(\Rightarrow VT\le VP\)
Dấu "=" xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}2020-x=x-2018\\x-2019=0\end{matrix}\right.\) \(\Rightarrow x=2019\)