2020 mũ trừ 1 chia 2020 mũ 2 cộng 2021
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I don't now
sorry
...................
nha
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Ta có: \(2021^2=\left(2020+1\right)^2=2020^2+2.2020.1+1^2\)
\(\Rightarrow1+2020^2=2021^2-2.2020\)
\(\Rightarrow\sqrt{1+2020^2+\frac{2020^2}{2021}}+\frac{2020}{2021}\)
\(=\sqrt{2021^2-2.2020+\frac{2020^2}{2021}}+\frac{2020}{2021}\)
\(=\sqrt{2021^2-2.2021.\frac{2020}{2021}+\left(\frac{2020}{2021}\right)^2}+\frac{2020}{2021}\)
\(=\sqrt{\left(2021-\frac{2020}{2021}\right)^2}+\frac{2020}{2021}\)
\(=2021-\frac{2020}{2021}+\frac{2020}{2021}=2021\)
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bài 1:
ssh của A là:
(151-3):2+1=75
A=(151+3)x75:2=5775
đáp số: 5775
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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 \(A=\frac{\sqrt{2021}-1}{\sqrt{2021}}.\)
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:
\(S=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+.........+\frac{1}{\sqrt{2020}}-\frac{1}{\sqrt{2021}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2021}}=1-\frac{1}{\sqrt{2021}}\)
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B/A
\(=\dfrac{1+\dfrac{2020}{2}+1+\dfrac{2019}{3}+...+1+\dfrac{1}{2021}+1}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}}\)
\(=\dfrac{2022\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}}=2022\)
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Ta có \(\frac{a}{a^2}=\frac{a^2}{a^3}=...=\frac{a^{2020}}{a^{2021}}=\frac{a+a^2+....+a^{2020}}{a^2+a^3+...+a^{2021}}\)
=> \(\frac{a}{a^2}=\frac{a+a^2+...+a^{2020}}{a^2+a^3+...+a^{2021}}\)
=> \(\left(\frac{a}{a^2}\right)^{2020}=\left(\frac{a+a^2+...+a^{2020}}{a^2+a^3+...+a^{2021}}\right)^{2020}\)
=> \(\frac{a}{a^2}.\frac{a}{a^2}...\frac{a}{a^2}=\left(\frac{a+a^2+...+a^{2020}}{a^2+a^3+...+a^{2021}}\right)^{2020}\)(2020 thừa số \(\frac{a}{a^2}\))
=> \(\frac{a}{a^2}.\frac{a^2}{a^3}...\frac{a^{2020}}{a^{2021}}=\left(\frac{a+a^2+...+a^{2020}}{a^2+a^3+...+a^{2021}}\right)^{2020}\)(Vì \(\frac{a}{a^2}=\frac{a^2}{a^3}=...=\frac{a^{2020}}{a^{2021}}\))
=> \(\frac{a}{a^{2021}}=\left(\frac{a+a^2+...+a^{2020}}{a^2+a^3+...+a^{2021}}\right)^{2020}\)(đpcm)
Sửa đề: \(\dfrac{2020^3-1}{2020^2+2021}\)
\(=\dfrac{\left(2020-1\right)\left(2020^2+2020+1\right)}{2020^2+2020+1}\)
=2020-1=2019