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ĐKXĐ: ...
Đặt \(\left(\sqrt{x-2018};\sqrt{y-2019};\sqrt{z-2020}\right)=\left(a;b;c\right)\) \(\Rightarrow a;b;c>0\)
\(\frac{a-1}{a^2}+\frac{b-1}{b^2}+\frac{c-1}{c^2}=\frac{3}{4}\)
\(\Leftrightarrow\frac{4a-4}{a^2}+\frac{4b-4}{b^2}+\frac{4c-4}{c^2}=3\)
\(\Leftrightarrow1-\frac{4a-a}{a^2}+1-\frac{4b-4}{b^2}+1-\frac{4c-4}{c^2}=0\)
\(\Leftrightarrow\frac{a^2-4a+4}{a^2}+\frac{b^2-4b+4}{b^2}+\frac{c^2-4c+4}{c^2}=0\)
\(\Leftrightarrow\left(\frac{a-2}{a}\right)^2+\left(\frac{b-2}{b}\right)^2+\left(\frac{c-2}{c}\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-2=0\\b-2=0\\c-2=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=2\\b=2\\c=2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x-2018}=2\\\sqrt{y-2019}=2\\\sqrt{z-2020}=2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=2022\\y=2023\\z=2024\end{matrix}\right.\)
\(2x^2+4x+2=21-3y^2\)
\(\Leftrightarrow2\left(x+1\right)^2=3\left(7-y^2\right)\)
Do \(\left(x+1\right)^2\ge0\Rightarrow7-y^2\ge0\) \(\Rightarrow y^2\le7\) (1)
Mà \(2\left(x+1\right)^2\) là một số tự nhiên chẵn và 3 là số lẻ
\(\Rightarrow7-y^2\) là một số chẵn \(\Rightarrow y^2\) là một số lẻ (2)
Từ (1); (2) \(\Rightarrow y^2\) là số chính phương lẻ và nhỏ hơn 7
\(\Rightarrow y^2=1\Rightarrow y=\pm1\)
\(\Rightarrow2\left(x+1\right)^2=3\left(7-1\right)=18\)
\(\Rightarrow\left(x+1\right)^2=9\)
\(\Rightarrow\left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)
Lời giải:
Đặt mẫu số của $B$ là $M$.
Từ \(2018x^3=2019y^3=2020z^3\)
\(\Rightarrow \sqrt[3]{2018}x=\sqrt[3]{2019}y=\sqrt[3]{2020}z=\frac{\sqrt[3]{2018}}{\frac{1}{x}}=\frac{\sqrt[3]{2019}}{\frac{1}{y}}=\frac{\sqrt[3]{2020}}{\frac{1}{z}}=\frac{\sqrt[3]{2018}+\sqrt[3]{2019}+\sqrt[3]{2020}}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}\)
\(=\frac{\sqrt[3]{2018}+\sqrt[3]{2019}+\sqrt[3]{2020}}{8}=\frac{M}{8}\)
\(\Rightarrow \left\{\begin{matrix} x=\frac{M}{8\sqrt[3]{2018}}\\ y=\frac{M}{8\sqrt[3]{2019}}\\ z=\frac{M}{8\sqrt[3]{2020}}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} 2018x^2=\frac{\sqrt[3]{2018}M^2}{64}\\ 2019y^2=\frac{\sqrt[3]{2019}M^2}{64}\\ 2020z^2=\frac{\sqrt[3]{2020}M^2}{64}\end{matrix}\right.\)
\(\Rightarrow 2018x^2+2019y^2+2020z^2=\frac{M^2(\sqrt[3]{2018}+\sqrt[3]{2019}+\sqrt[3]{2020})}{64}=\frac{M^3}{64}\)
\(\Rightarrow B=\frac{\sqrt[3]{\frac{M^3}{64}}}{M}=\frac{M}{4M}=\frac{1}{4}\)
Bài 1:Áp dụng C-S dạng engel
\(\frac{3}{xy+yz+xz}+\frac{2}{x^2+y^2+z^2}=\frac{6}{2\left(xy+yz+xz\right)}+\frac{2}{x^2+y^2+z^2}\)
\(\ge\frac{\left(\sqrt{6}+\sqrt{2}\right)^2}{\left(x+y+z\right)^2}=\left(\sqrt{6}+\sqrt{2}\right)^2>14\)
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 ...
1) Ta có: \(2020^2=\left(2019+1\right)^2=2019^2+2.2019+1.\)
\(\Rightarrow1+2019^2=2020^2-2.2019\)
\(\Rightarrow M=\sqrt{1+2019^2+\frac{2019^2}{2020^2}}+\frac{2019}{2020}=\sqrt{2020^2-2.2019+\frac{2019^2}{2020^2}}+\frac{2019}{2020}\)
\(=\sqrt{2020^2-2.2020.\frac{2019}{2020}+\left(\frac{2019}{2020}\right)^2}+\frac{2019}{2020}\)
\(=\sqrt{\left(2020-\frac{2019}{2020}\right)^2}+\frac{2019}{2020}=2020-\frac{2019}{2020}+\frac{2019}{2020}\)
\(=2020\)
Vậy M=2020.
2) Xét : \(k\in N;k\ge2\)ta có:
\(\left(1+\frac{1}{k-1}-\frac{1}{k}\right)^2=1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}+\frac{2}{k-1}-\frac{2}{\left(k-1\right)k}-\frac{2}{k}\)
\(=1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}+\frac{2}{k-1}-\frac{2}{k-1}+\frac{2}{k}-\frac{2}{k}\)
\(\Rightarrow\left(1+\frac{1}{k-1}-\frac{1}{k}\right)^2=1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}\)
\(\Rightarrow\sqrt{1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}}=1+\frac{1}{k-1}+\frac{1}{k}\)
Cho \(k=3,4,...,2020.\)Ta có:
\(N=\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+...+\sqrt{1+\frac{1}{2019^2}+\frac{1}{2020^2}}\)
\(=\left(1+\frac{1}{2}-\frac{1}{3}\right)+\left(1+\frac{1}{3}-\frac{1}{4}\right)+...+\left(1+\frac{1}{2018}-\frac{1}{2019}\right)+\left(1+\frac{1}{2019}-\frac{1}{2020}\right)\)
\(=2018+\frac{1}{2}-\frac{1}{2020}=2018\frac{1009}{2020}\)
Vậy \(N=2018\frac{1009}{2020}.\)