tính giá trị biểu thức:
\(B=\frac{3^{2021}-3^{2019}}{3^{2021}+3^{2020}}\)
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B/A
\(=\dfrac{1+\dfrac{2020}{2}+1+\dfrac{2019}{3}+...+1+\dfrac{1}{2021}+1}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}}\)
\(=\dfrac{2022\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}}=2022\)
bài 1:
ssh của A là:
(151-3):2+1=75
A=(151+3)x75:2=5775
đáp số: 5775
\(x=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3+2\sqrt{2}}\)
Ta có: Đặt \(A=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}\)=> \(A^2=\frac{\sqrt{5}+2+\sqrt{5}-2+2\sqrt{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}}{\sqrt{5}+1}\)
=> \(A^2=\frac{2\sqrt{5}+2\sqrt{5-4}}{\sqrt{5}+1}=\frac{2\left(\sqrt{5}+1\right)}{\sqrt{5}+1}=2\)=> \(A=\sqrt{2}\)
\(\sqrt{3+2\sqrt{2}}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\)
==> \(x=\sqrt{2}-\left(\sqrt{2}+1\right)=-1\)
Do đó: N = (-1)2019 + 3.(-1)2020 - 2.(-1)2021 = -1 + 3 + 2 = 4
a) \(2021^{2020}-2021^{2019}=2021^{2019}.\left(2021-1\right)=2021^{2019}.2020\)
b) Ta có :\(7x-140=3.7^2\)
\(\implies\) \(7x-140=3.49\)
\(\implies\) \(7x-140=147\)
\(\implies\) \(7x=287\)
\(\implies\) \(x=41\)
Ta có: \(C=\frac{\left|x-2019\right|+2020}{\left|x-2019\right|+2021}=\frac{\left|x-2019\right|+2021-1}{\left|x-2019\right|+2021}=1-\frac{1}{\left|x-2019\right|+2021}\)
=> C đạt giá trị nhỏ nhất khi \(\frac{1}{\left|x-2019\right|+2021}\) lớn nhất
=> |x - 2019| + 2021 nhỏ nhất
Ta có: \(\left|x-2019\right|\ge0\)
\(\Rightarrow\left|x-2019\right|+2021\ge2021\)
Dấu "=" xảy ra khi x - 2019 = 0
=> x = 2019
\(\Rightarrow C=\frac{\left|2019-2019\right|+2020}{\left|2019-2019\right|+2021}=\frac{2020}{2021}\)
Vậy \(MinC=\frac{2020}{2021}\Leftrightarrow x=2019\).
Ta có x = 2020
=> x + 1 = 2021
A = x2021 - 2021x2020 + .... + 2021x - 2021
= x2021 - (x + 1)x2020 + .... + (x + 1)x - (x + 1)
= x2021 - x2021 - x2020 + .... + x2 + x - x + 1
= 1
Vậy A = 1
Ta có : \(x=2020\Rightarrow x+1=2021\)
\(A=x^{2021}-\left(x+1\right)x^{2020}+\left(x+1\right)x^{2019}-\left(x+1\right)x^{2018}+...-\left(x+1\right)x^2+\left(x+1\right)x-2021\)
= x2021 - x2021 - x2020 + x2020 + x2019 - x2019 - x2018 + ... - x3 - x2 + x2 + x - 2021 = x - 2021
mà x = 2020 hay 2020 - 2021 = -1
Vậy với x = 2020 thì A = -1
\(\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times...\times\left(1-\dfrac{1}{2023}\right)\\ =\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times...\times\dfrac{2022}{2023}\\ =\dfrac{1}{2023}\)
Ta có: \(B=\frac{3^{2021}-3^{2019}}{3^{2021}+3^{2020}}\)
\(\Leftrightarrow B=\frac{3^{2019}.\left(3^2-1\right)}{3^{2020}.\left(3+1\right)}\)
\(\Leftrightarrow B=\frac{8}{3.4}\)
\(\Leftrightarrow B=\frac{2}{3}\)
Vậy \(B=\frac{2}{3}\)
Ta có : \(B=\frac{3^{2021}-3^{2019}}{3^{2021}+3^{2020}}\)
\(=\frac{3^{2019}\left(3^2-1\right)}{3^{2020}\left(3+1\right)}\)
\(=\frac{3^{2019}.8}{3^{2020}.4}\)
\(=\frac{2}{3}\)