1, Tính
\(\frac{15^{2020}.11^{2021}}{3^{2020}.55^{2021}}\)
K
Khách
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KT
1
B
12 tháng 7 2019
Ta có: \(2021^2=\left(2020+1\right)^2=2020^2+2.2020.1+1^2\)
\(\Rightarrow1+2020^2=2021^2-2.2020\)
\(\Rightarrow\sqrt{1+2020^2+\frac{2020^2}{2021}}+\frac{2020}{2021}\)
\(=\sqrt{2021^2-2.2020+\frac{2020^2}{2021}}+\frac{2020}{2021}\)
\(=\sqrt{2021^2-2.2021.\frac{2020}{2021}+\left(\frac{2020}{2021}\right)^2}+\frac{2020}{2021}\)
\(=\sqrt{\left(2021-\frac{2020}{2021}\right)^2}+\frac{2020}{2021}\)
\(=2021-\frac{2020}{2021}+\frac{2020}{2021}=2021\)
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:
\(S=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+.........+\frac{1}{\sqrt{2020}}-\frac{1}{\sqrt{2021}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2021}}=1-\frac{1}{\sqrt{2021}}\)
HB
1
XO
30 tháng 7 2020
Ta có : A = \(\frac{10^{2020}+1}{10^{2021}+1}\)
=> 10A = \(\frac{10^{2021}+10}{10^{2021}+1}=1+\frac{9}{10^{2021}+1}\)
Lại có : \(B=\frac{10^{2021}+1}{10^{2022}+1}\)
=> \(10B=\frac{10^{2022}+10}{10^{2022}+1}=1+\frac{9}{10^{2022}+1}\)
Vì \(\frac{9}{10^{2022}+1}< \frac{9}{10^{2021}+1}\)
=> \(1+\frac{9}{10^{2022}+1}< 1+\frac{9}{10^{2022}+1}\)
=> 10B < 10A
=> B < A
b) Ta có : \(\frac{2019}{2020+2021}< \frac{2019}{2020}\)
Lại có : \(\frac{2020}{2020+2021}< \frac{2020}{2021}\)
=> \(\frac{2019}{2020+2021}+\frac{2020}{2020+2021}< \frac{2019}{2020}+\frac{2020}{2021}\)
=> \(\frac{2019+2020}{2020+2021}< \frac{2019}{2020}+\frac{2020}{2021}\)
=> B < A
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}
DT
1
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 8 2020
Áp dụng bài vừa chứng minh bên dưới :D
\(\Rightarrow P=2021\)
31 tháng 12 2022
a)= 2021.2021-2020.(2021+1)
= 2021.(2020+1)-2020.(2021+1)
= (2021.2020)+2021-(2020.2021)-2020
= 1
31 tháng 12 2022
b) B= (1+2-3-4)+(5+6-7-8)+(9+10-11-12)...........+(2017+2018-2019-2020)+2021
B= -4+(-4)+....................(-4)+2021
B= -4x505+2021
B= -2020 + 2021
B = 1
NK
1
24 tháng 5 2020
Ta có: \(\frac{2019}{2020}>\frac{2019}{2020+2021};\frac{2020}{2021}>\frac{2020}{2020+2021}\)
=> \(\frac{2019}{2020}+\frac{2020}{2021}>\frac{2019}{2020+2021}+\frac{2020}{2020+2021}=\frac{2019+2020}{2020+2021}\)
=> A > B.
\(\frac{15^{2020}.11^{2020}}{3^{2020}.55^{2020}}=\frac{\left(15.11\right)^{2020}}{\left(3.55\right)^{2020}}=\frac{165^{2020}}{165^{2020}}=1\)