so sánh \(\sqrt{2021}-\sqrt{2020}\&\sqrt{2020}-\sqrt{2019}\)
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NV
Nguyễn Việt Lâm
Giáo viên
2 tháng 8 2021
\(\sqrt{2021}-\sqrt{2020}=\dfrac{\left(\sqrt{2021}-\sqrt{2020}\right)\left(\sqrt{2021}+\sqrt{2020}\right)}{\sqrt{2021}+\sqrt{2020}}=\dfrac{1}{\sqrt{2021}+\sqrt{2020}}\) là nghịch đảo của \(\sqrt{2021}+\sqrt{2020}\) (đpcm)
2 tháng 8 2021
\(\sqrt{2021}-\sqrt{2020}=\dfrac{\left(\sqrt{2021}-\sqrt{2020}\right)\left(\sqrt{2021}+\sqrt{2020}\right)}{\sqrt{2021}+\sqrt{2020}}\)
\(=\dfrac{1}{\sqrt{2021}+\sqrt{2020}}\)(đpcm)
N
1
10 tháng 10 2020
Ta có: \(\sqrt{2020}-\sqrt{2019}=\frac{\left(\sqrt{2020}-\sqrt{2019}\right)\left(\sqrt{2020}+\sqrt{2019}\right)}{\sqrt{2020}+\sqrt{2019}}\)
\(=\frac{2020-2019}{\sqrt{2020}+\sqrt{2019}}=\frac{1}{\sqrt{2020}+\sqrt{2019}}\)
\(\sqrt{2021}-\sqrt{2020}=\frac{\left(\sqrt{2021}-\sqrt{2020}\right)\left(\sqrt{2021}+\sqrt{2020}\right)}{\sqrt{2021}+\sqrt{2020}}\)
\(=\frac{2021-2020}{\sqrt{2021}+\sqrt{2020}}=\frac{1}{\sqrt{2021}+\sqrt{2020}}\)
Vì \(\sqrt{2020}+\sqrt{2019}< \sqrt{2021}+\sqrt{2020}\)
\(\Rightarrow\) \(\frac{1}{\sqrt{2020}+\sqrt{2019}}>\frac{1}{\sqrt{2021}+\sqrt{2020}}\)
Hay \(\sqrt{2020}-\sqrt{2019}>\sqrt{2021}-\sqrt{2020}\)
Chúc bn học tốt!
22 tháng 8 2020
TA XÉT PHÂN THỨC TỔNG QUÁT SAU:
\(A=\frac{1}{n\sqrt{n+1}+\left(n+1\right)\sqrt{n}}\)
\(A=\frac{1}{\sqrt{n\left(n+1\right)}.\left(\sqrt{n}+\sqrt{n+1}\right)}\)
\(A=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)}{\sqrt{n\left(n+1\right)}.\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}\)
\(A=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}\left(n+1-n\right)}\)
\(A=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}\)
\(A=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
THAY LẦN LƯỢT CÁC GIÁ TRỊ n từ 1 => 2021 vào ta được:
=> \(A=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2020}}-\frac{1}{\sqrt{2021}}\)
=> \(A=1-\frac{1}{\sqrt{2021}}=\frac{\sqrt{2021}-1}{\sqrt{2021}}\)
VẬY \(A=\frac{\sqrt{2021}-1}{\sqrt{2021}}.\)
NN
22 tháng 8 2020
Ta có: \(\frac{1}{\left(a-1\right)\sqrt{a}+a.\sqrt{a-1}}=\frac{a-\left(a-1\right)}{\sqrt{a}.\sqrt{a-1}.\left(\sqrt{a}+\sqrt{a-1}\right)}\)
\(=\frac{\left(\sqrt{a}-\sqrt{a-1}\right)\left(\sqrt{a}+\sqrt{a-1}\right)}{\sqrt{a}.\sqrt{a-1}.\left(\sqrt{a}+\sqrt{a-1}\right)}=\frac{\sqrt{a}-\sqrt{a-1}}{\sqrt{a}.\sqrt{a-1}}\)
\(=\frac{\sqrt{a}}{\sqrt{a}.\sqrt{a-1}}-\frac{\sqrt{a-1}}{\sqrt{a}.\sqrt{a-1}}=\frac{1}{\sqrt{a-1}}-\frac{1}{\sqrt{a}}\)
Thay lần lượt các giá trị của a bằng \(2;3;4;........;2021\)ta được:
\(S=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+.........+\frac{1}{\sqrt{2020}}-\frac{1}{\sqrt{2021}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2021}}=1-\frac{1}{\sqrt{2021}}\)
21 tháng 10 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}
5 tháng 4 2020
Ta có : VT2 = \(\sqrt{2019}^2+2\sqrt{2019.2021}+\sqrt{2021}^2\)
\(=2.2020+2\sqrt{\left(2020-1\right).\left(2020+1\right)}\)
\(=2.2020+2\sqrt{2020^2-1}\)
Ta thấy : \(2\sqrt{2020^2-1}< 2.2020\)
=> \(2.2020+2\sqrt{2020^2-1}< 4.2020\)
=> \(2.2020+2\sqrt{2020^2-1}< \left(2\sqrt{2020}\right)^2\)
-> \(\sqrt{VT^2}< \sqrt{\left(2\sqrt{2020}\right)^2}\)
-> \(VT< 2\sqrt{2020}\)
Vậy \(2\sqrt{2020}>\sqrt{2019}+\sqrt{2021}\)
9 tháng 5 2021
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: \(\sqrt{2021}-\sqrt{2020}=\frac{\left(\sqrt{2021}-\sqrt{2020}\right)\left(\sqrt{2021}+\sqrt{2020}\right)}{\sqrt{2021}+\sqrt{2020}}=\frac{1}{\sqrt{2021}+\sqrt{2020}}\)
\(\sqrt{2020}-\sqrt{2019}=\frac{\left(\sqrt{2020}+\sqrt{2019}\right)\left(\sqrt{2020}-\sqrt{2019}\right)}{\sqrt{2020}+\sqrt{2019}}=\frac{1}{\sqrt{2020}+\sqrt{2019}}\)
Do \(\frac{1}{\sqrt{2021}+\sqrt{2020}}< \frac{1}{\sqrt{2020}+\sqrt{2019}}\) => \(\sqrt{2021}-\sqrt{2020}< \sqrt{2020}-\sqrt{2019}\)