2022 kg = ... tấn ... kg
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75g=.0,075 kg
750g=.0,75..kg
75g=.0,075...kg
65dag=.0,65...kg
0,5=...kg ???
0,085=....kg ???
3tan 25kg=.3025....kg
74,5 tấn=.7500...kg
0,5 tấn=.500...kg
10 tấn 5 tạ=..10500...kg
3kg 75g=..3,075...kg
7hg=..0,7...kg
9g=..0,009...kg
0,032 tấn= 32 kg 15 lạng= 1,5 kg
0,016 tấn= 16 kg 1,25 giờ= 75 giây
0,56 km= 560 m 25 lạng= 2,5 kg
1260 phút= 21 giờ 1,15 giờ= 69 giây
0,75 km= 750 m
1560 phút= 26 giờ
\(\dfrac{-3}{7}\)(\(\dfrac{5}{9}\)+\(\dfrac{4}{9}\)) + 0
=\(\dfrac{-3}{7}\)
*Lần sau bạn viết rõ đề ra, để thế này nhiều ng sẽ không hiểu!
\(\dfrac{-3}{7}.\dfrac{5}{9}+\dfrac{4}{9}.\dfrac{-3}{7}+\left(2022\right)^0\)
\(=\dfrac{-3}{7}.\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+1\)
\(=\dfrac{-3}{7}.1+1\)
\(=\dfrac{-3}{7}+1\)
\(=\dfrac{-3+7}{7}\)
\(=\dfrac{4}{7}\)
a. 2,5 tấn = 1500… kg b/ 38 kg = 0,38… tạ
c. 4,572 tấn = 4572… kg d/ 1543 g = 1,543… kg
93kg=0,93 tạ = 0,093 tấn
950kg = 9,5 tạ = 0,95 tấn
0,5 tấn = 5 tạ = 500kg
69,45 tấn = 694,5 tạ = 69450kg
93 kg=0,93=0,093 tấn
950kg=9,50 tạ=0,950 tấn
0,5 tấn=5 tạ=5000kg
69,45 tấn=694,5 tạ=69450kg
Nhớ kb và cho mình nhé
Ta có \(A=\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{2022}{2^{2022}}+\dfrac{2023}{2^{2023}}\)
\(2A=1+\dfrac{2}{2}+\dfrac{3}{2^2}+...+\dfrac{2022}{2^{2021}}+\dfrac{2023}{2^{2022}}\)
\(2A-A=\left(1+\dfrac{2}{2}+\dfrac{3}{2^2}+...+\dfrac{2022}{2^{2021}}+\dfrac{2023}{2^{2022}}\right)-\left(\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{2022}{2^{2022}}+\dfrac{2023}{2^{2023}}\right)\)\(A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2021}}+\dfrac{1}{2^{2022}}\) - \(\dfrac{2023}{2^{2023}}\)
Đặt B = \(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2021}}+\dfrac{1}{2^{2022}}\)
2B = \(2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{2020}}+\dfrac{1}{2^{2021}}\)
2B - B = \(\left(2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{2020}}+\dfrac{1}{2^{2021}}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2021}}+\dfrac{1}{2^{2022}}\right)\)B = 2 - \(\dfrac{1}{2^{2022}}\)
Suy ra A = 2 - \(\dfrac{1}{2^{2022}}\) - \(\dfrac{2023}{2^{2023}}\) < 2
Vậy A < 2
\(A=\dfrac{1}{2}+\dfrac{2}{2^{2}}+\dfrac{3}{2^{3}}+...+\dfrac{2022}{2^{2022}}+\dfrac{2023}{2^{2023}}\)
\(2A=1+\dfrac22+\dfrac3{2^2}\ +\,.\!.\!.+\ \dfrac{2022}{2^{2021}}+\dfrac{2023}{2^{2022}}\\2A-A=\left(1+\dfrac22+\dfrac3{2^2}\ +\,.\!.\!.+\ \dfrac{2022}{2^{2021}}+\dfrac{2023}{2^{2022}}\right)-\left(\dfrac12+\dfrac2{2^2}+\dfrac3{2^3}\ +\,.\!.\!.+\ \dfrac{2022}{2^{2022}}+\dfrac{2023}{2^{2023}}\right)\\A=1+\dfrac12+\dfrac1{2^3}\ +\,.\!.\!.+\ \dfrac1{2^{2021}}+\dfrac1{2^{2022}}-\dfrac{2023}{2^{2023}}\\2\left(A+\dfrac{2023}{2^{2023}}\right)=2+1+\dfrac12+\dfrac1{2^2}\ +\,.\!.\!.+\ \dfrac1{2^{2020}}+\dfrac1{2^{2021}}\\A+\dfrac{2023}{2^{2023}}=2-\dfrac1{2^{2022}}\\A=2-\dfrac1{2^{2022}}+\dfrac{2023}{2^{2023}}<2\)
2 tấn 22 kg
chưa chắc lắm
2022 kg = 2 tấn 22 kg nhá