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\(\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{2019}\Rightarrow\dfrac{a+b}{ab}=\dfrac{1}{2019}\Rightarrow2019=\dfrac{ab}{a+b}\)
\(\dfrac{1}{a}=\dfrac{1}{2019}-\dfrac{1}{b}=\dfrac{b-2019}{2019b}\Rightarrow b-2019=\dfrac{2019b}{a}\)
\(\dfrac{1}{b}=\dfrac{1}{2019}-\dfrac{1}{a}=\dfrac{a-2019}{2019a}\Rightarrow a-2019=\dfrac{2019a}{b}\)
\(\Rightarrow\sqrt{a-2019}+\sqrt{b-2019}=\sqrt{\dfrac{2019a}{b}}+\sqrt{\dfrac{2019b}{a}}=\dfrac{\sqrt{2019}\left(a+b\right)}{\sqrt{ab}}=\sqrt{\dfrac{ab}{a+b}}.\dfrac{a+b}{\sqrt{ab}}=\sqrt{a+b}\)
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 ...
Bài 1: Ta có: \(\sqrt{2020}-\sqrt{2019}=\frac{1}{\sqrt{2020}+\sqrt{2019}};\)\(\sqrt{2018}-\sqrt{2017}=\frac{1}{\sqrt{2018}+\sqrt{2017}}\)
Dễ thấy \(\sqrt{2020}+\sqrt{2019}>\sqrt{2018}+\sqrt{2017}\)nên \(\frac{1}{\sqrt{2020}+\sqrt{2019}}< \frac{1}{\sqrt{2018}+\sqrt{2017}}\)
Suy ra\(\sqrt{2020}-\sqrt{2019}< \sqrt{2018}-\sqrt{2017}\)
Bài 2: Xét biểu thức \(\sqrt{a^2+a^2\left(a+1\right)^2+\left(a+1\right)^2}=\sqrt{a^2\left(a^2+2a+1+1\right)+\left(a+1\right)^2}=\sqrt{a^4+2a^2\left(a+1\right)+\left(a+1\right)^2}=\sqrt{\left(a^2+a+1\right)^2}=a^2+a+1\)(Vì \(a^2+a+1>0\forall a\inℝ\))
Áp dụng công thức tổng quát trên, ta được: \(\sqrt{2019^2+2019^2.2020^2+2020^2}=2019^2+2019+1\)(là số tự nhiên) (đpcm)
\(\sqrt{2019^2+2019^2.2020^2+2020^2}=\sqrt{2019^2+\left(2020-1\right)^2.2020^2+2020^2}=\sqrt{2019^2+2020^4-2.2020.2020^2+2020^2+2020^2}=\sqrt{2020^4+2.2020^2-2.\left(2019+1\right).2020^2+2019^2}=\sqrt{2020^4+2.2020^2-2.2019.2020^2-2.2020^2+2019^2}=\sqrt{2020^4-2.2019.2020^2+2019^2}=\sqrt{\left(2020^2-2019\right)^2}=\left|2020^2-2019\right|=2020^2-2019\)
Vì 20202-2019\(\in N\)
Vậy \(\sqrt{2019^2+2019^2.2020^2+2020^2}\)\(\in N\)
Đặt \(K\left(x\right)=P\left(x\right)-\left(x+1\right)\)
\(\Rightarrow K\left(2016\right)=K\left(2017\right)=K\left(2018\right)=K\left(2019\right)=0\)
Vì P(x) có hệ số của bậc cao nhất bằng 1 nên K(x) cũng có hệ số của bậc cao nhất bằng 1
Do đó K(x) có dạng \(\left(x-2016\right)\left(x-2017\right)\left(x-2018\right)\left(x-2019\right)\)
Lúc đó \(P\left(x\right)=\left(x-2016\right)\left(x-2017\right)\left(x-2018\right)\left(x-2019\right)\)
\(+\left(x+1\right)\Rightarrow P\left(2020\right)=2045⋮5\)
Vậy P(2020) là một số tự nhiên chia hết cho 5 (đpcm)
Bài toán quá hay (người ra đề quá đẳng cấp)
A = \(\dfrac{2020}{2019^2+1}\) + \(\dfrac{2020}{2019^2+2}\)+......+\(\dfrac{2020}{2019^{2^{ }}+2019}\)
A = 2020 x ( \(\dfrac{1}{2019^{2^{ }}+1}\)+ \(\dfrac{1}{2019^2+2}\)+....+\(\dfrac{1}{2019^2+2019}\))
đặtB =( \(\dfrac{1}{2019^{2^{ }}+1}\)+ \(\dfrac{1}{2019^2+2}\)+....+\(\dfrac{1}{2019^2+2019}\))⇒ A =2020.B
mặt khác ta có \(\dfrac{1}{2019^2+1}\) > \(\dfrac{1}{2019^2+2}\)>.....>\(\dfrac{1}{2019^2+2019}\)
⇔\(\dfrac{2019}{2019^2+1}\) > \(\dfrac{1}{2019^{2^{ }}+1}\)> \(\dfrac{1}{2019^{2^{ }}+2}\)+......+\(\dfrac{1}{2019^2+2019}\) > \(\dfrac{2019}{2019^{2^{ }}+2019}\)
⇔ \(\dfrac{2019}{2019^{2^{ }}+2019}\) < B < \(\dfrac{2019}{2019^{2^{ }}+1}\)
⇔ \(\dfrac{2020.2019}{2019^{2^{ }}+2019}\) <2020. B < \(\dfrac{2020.2019}{2019^{2^{ }}+1}\)
⇔ \(\dfrac{2019.2020}{2019.\left(2019+1\right)}\) < 2012.B < \(\dfrac{\left(2019+1\right).2019}{2019^{2^{ }}+1}\)
⇔ \(\dfrac{2019.2020}{2019.2020}\)< 2020.B < \(\dfrac{2019^{2^{ }}+2019}{2019^{2^{ }}+1}\) = 1 + \(\dfrac{2018}{2019^{2^{ }}+1}\)< 2
⇔ 1 < 2020 .B < 2
⇔ 1 < A < 2
⇔ A không phải là số nguyên điều phải chứng minh