Tính nhanh:
a) A=\(\frac{2020^3+1}{2020^2-2019}\) b)B=\(\frac{2020^3-1}{2020^2+2021}\)
Những câu hỏi liên quan
Tính nhanh :
a) A = \(\frac{2020^3+1}{2020^2-2019}\)
b) B = \(\frac{2020^3-1}{2020^2+2021}\)
so sánh
a)
A=\(\frac{10^{2020}+1}{10^{2021}+1};B=\frac{10^{2021}+1}{10^{2022}+1}\)
b)
\(A=\frac{2019}{2020}+\frac{2020}{2021}\)và \(B=\frac{2019+2020}{2020+2021}\)
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
a) A=10^2020+1/10^2021+1 < 10^2020+1+9/10^2022+1+9 =
10.(10^2021+1)/10.(10^2022+1) = 10^2021+1/10^2022+1 = B
Vậy A < B.
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Tính nhanh :
a, A = 2020^3 + 1 / 2020^2 - 2019
b, B = 2020^3 - 1 / 2020^2 + 2021
Giúp mk vs ạ mk đang cần gấp
So sánh:
\(A=\frac{2019}{2020}+\frac{2020}{2021}\) và \(B=\frac{2019+2020}{2020+2021}\)
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.
Bài 1:
A,3+5+7+9+,...+151
Bài 2:So sánh 2 biểu thức
A=2019/2020+2020/2021 và
B=2019+2020/2020+2021
Không làm tính cộng
bài 1:
ssh của A là:
(151-3):2+1=75
A=(151+3)x75:2=5775
đáp số: 5775
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Cho \(A=1-\frac{2019}{2020}+\left(\frac{2019}{2020}\right)^2-\left(\frac{2019}{2020}\right)^3+...+\left(\frac{2019}{2020}\right)^{2020}\). Chứng tỏ A ko phải là 1 số nguyên.
Mk cần gấp. Mai nộp rồi!!!
sao ko có ai giúp mk vậy
tính A/B biết:
A=1/2+1/3+1/4+..+1/2021.
B=2020/1+2019/2+...+2/2019+1/2020.
Nhanh giúp mk nhé!
Cần gấp lắm!
số lượng số hạng của dãy số là
( 2021 - 2 ) : 1 + 1 = 2020
tổng của dãy số là
( 2021 + 2) x 2020 : 2 = 2043230
vậy A = \(\frac{1}{2043230}\)
xong a rồi vậy b thì sao bạn
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So sánh A và B:
\(A=\frac{2018^2}{2^{2018}+3^{2019}}+\frac{3^{2019}}{3^{2019}+5^{2020}}+\frac{5^{2020}}{5^{2020}+2^{2018}}\)
\(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2019.2020}\)
B= 1/1.2+1/2.3+...+1/2019.2020
B=1/1-1/2+1/2-1/3+...+1/2019-1/2020
B=1-1/2020=2020/2020-1/2020=2019/2020
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Rút gọn:
\(\frac{\frac{1}{2020}+\frac{2}{2019}+\frac{3}{2018}+...+\frac{2019}{2}+\frac{2020}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2021}}\)
Đặt \(A=\frac{\frac{1}{2020}+\frac{2}{2019}+\frac{3}{2018}+...+\frac{2019}{2}+\frac{2020}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2021}}\)
\(A=\frac{1+\left(\frac{1}{2020}+1\right)+\left(\frac{2}{2019}+1\right)+\left(\frac{3}{2018}+1\right)+...+\left(\frac{2019}{2}+1\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2021}}\)
\(A=\frac{\frac{2021}{2021}+\frac{2021}{2020}+\frac{2021}{2019}+...+\frac{2021}{2}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2021}}\)
\(A=\frac{2021\left(\frac{1}{2021}+\frac{1}{2020}+\frac{1}{2019}+...+\frac{1}{2}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2021}}=2021\)