(1-1/2020).(1-2/2020).(1-3/2020)...(1-2021/2020) Giúp tui đi, mốt thi rồi
So sánh:
A=2021^2020+2/2021^2020-1 và B=2021^2020/2021^2020-3
tính giá trị biểu thức:
P=2/2020*(1/2020+5/2020)-1/2021*(7-2/2020)-2/2020*(1/2020+6/2021)
𝑝=−2856279824648840
so sánh
C=\(\dfrac{2^{2020}-1}{2^{2021}-1}\)và D=\(\dfrac{2^{2021}-1}{2^{2022}-1}\)
Bạn nào giúp mình với
thứ 2 mình thi mất rồi😥
tham khảo:
https://hoidap247.com/cau-hoi/3987981
bài này thì mình bó tay chấm com xin lỗi bạn :(((
1/1-1/2+1/2-1/3+1/3-...-1/2020+1/2020-1/2021
1/1-1/2+1/2-1/3+1/3-...-1/2020+1/2020-1/2021
=1/1+(1/2-1/2)+(1/3-1/3)+...+(1/2020-1/2020)-1/2021
=1/1-1/2021
=1-1/2021
=2020/2021
Học tốt nha!!!
so sánh:
a)C= \(\dfrac{100^{99}+1}{100^{100}+1}\) và D= \(\dfrac{100^{100}+1}{100^{101}+1}\)
b)E=\(\dfrac{2020^{2021}+1}{2020^{2022}+1}\) và F=\(\dfrac{2020^{2020}+1}{2020^{2021}+1}\)
c: \(100C=\dfrac{100^{100}+100}{100^{100}+1}=1+\dfrac{99}{100^{100}+1}\)
\(100D=\dfrac{100^{101}+100}{100^{101}+1}=1+\dfrac{99}{100^{101}+1}\)
100^100+1<100^101+1
=>\(\dfrac{99}{100^{100}+1}>\dfrac{99}{100^{101}+1}\)
=>100C>100D
=>C>D
b: \(2020E=\dfrac{2020^{2022}+2020}{2020^{2022}+1}=1+\dfrac{2019}{2020^{2022}+1}\)
\(2020F=\dfrac{2020^{2021}+2020}{2020^{2021}+1}=1+\dfrac{2019}{2020^{2021}+1}\)
2020^2022+1>2020^2021+1(Do 2022>2021)
=>\(\dfrac{2019}{2020^{2022}+1}< \dfrac{2019}{2020^{2021}+1}\)
=>2020E<2020F
=>E<F
Câu 24: Cho biểu thức: A=1/2+1/3+1/4+.........+1/2021+1/2022 Và B=2021/1+2020/2+2019/3+.........+3/2019+2020+1/2021
B/A
\(=\dfrac{1+\dfrac{2020}{2}+1+\dfrac{2019}{3}+...+1+\dfrac{1}{2021}+1}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}}\)
\(=\dfrac{2022\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}}=2022\)
2019 x 2020 - 1/ 2019 x 2020 và 2020 x 2021 - 1/ 2020 x 2021
so sánh phân số
1+1/3+1/4+.....+1/2021:
2020/1+2019/2+2018/3+.....+1/2020
B=1*2*3*...*2021-1*2*3*...*2020-1*2*3*...*2019*2020^2 Giá trị của B là? giúp mình nốt câu này với ạ