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Đặt A = 1 + 2 + 3 + 4 + ... + 2023
Tổng có 2023 - 1 + 1 số hạng
A = (2023 + 1) × 2023 : 2
= 2047276
-----------------------
Đặt B = 20 + 21 + 22 + ... + 2024
Tổng có: 2024 - 20 + 1 = 2005 số hạng
B = (2024 + 20) × 2005 : 2
= 2049110
------------------------
Đặt C = 2 + 4 + 6 + ... + 2024
Tổng có (2024 - 2) : 2 + 1 = 1012 số hạng
C = (2024 + 2) × 1012 : 2
= 1025156
------------------------
Đặt D = 1 + 2 + 4 + 8 + 16 + ... + 8192
2 × D = 2 + 4 + 8 + 16 + 32 + ... + 16384
2 × D - D = (2 + 4 + 8 + 16 + 32 + ... + 16384) - (1 + 2 + 4 + 8 + 16 + ... + 8192)
= 16384 - 1
= 16383
Vậy D = 16383
\(a,A=1+2+3+4+5..+2023\)
Số số hạng:
\(\left(2023-1\right):1+1=2023\)
Tổng :
\(\dfrac{\left(2023+1\right).2023}{2}=2047276\)
\(b,20+21+22+..+2024\)
Số số hạng:
\(\left(2024-20\right):1+1=2005\)
Tổng:
\(\dfrac{\left(2024+20\right).2005}{2}=2049110\)
\(c,2+4+6+..+2024\)
Số số hạng:
\(\left(2024-2\right):2+1=1012\)
Tổng:
\(\dfrac{\left(2024+2\right).1012}{2}=1025156\)
\(\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times...\times\dfrac{2023}{2024}\\ =\dfrac{1\times2\times3\times...\times2023}{2\times3\times4\times...\times2024}\\ =\dfrac{1}{2024}\)
Ta có:
Mẫu số chung 2 phân số: 84
\(\dfrac{3}{7}=\dfrac{3*12}{7*12}=\dfrac{36}{84}\)
\(\dfrac{5}{12}=\dfrac{5*7}{12*7}=\dfrac{35}{84}\)
Vì \(36>35\) nên\(\dfrac{36}{84}>\dfrac{35}{84}\)
Vậy \(\dfrac{3}{7}>\dfrac{5}{12}\)
Ta có:
\(\dfrac{9}{8}>1>\dfrac{2023}{2024}\) nên \(\dfrac{9}{8}>\dfrac{2023}{2024}\)
Ta có:
\(\dfrac{1+15}{16}=1\)
\(\dfrac{1+16}{15}=\dfrac{17}{15}>1\)
\(\Rightarrow\dfrac{1+15}{16}>\dfrac{1+16}{15}\)
\(1\dfrac{1}{2}\times1\dfrac{1}{3}\times1\dfrac{1}{4}\times...\times1\dfrac{1}{2023}\times1\dfrac{1}{2024}\)
\(=\left(1+\dfrac{1}{2}\right)\times\left(1+\dfrac{1}{3}\right)\times\left(1+\dfrac{1}{4}\right)\times...\times\left(1+\dfrac{1}{2023}\right)\times\left(1+\dfrac{1}{2024}\right)\)
\(=\dfrac{3}{2}\times\dfrac{4}{3}\times\dfrac{5}{4}\times\dfrac{6}{5}\times...\times\dfrac{2024}{2023}\times\dfrac{2025}{2024}\)
\(=\dfrac{3\times4\times5\times...\times2024\times2025}{2\times3\times4\times...\times2023\times2024}\)
\(=\dfrac{2025}{2}\)
\(=1012,5\)
\(x\times\dfrac{2}{7}+x\times\dfrac{5}{7}=2024\\ x\times\left(\dfrac{2}{7}+\dfrac{5}{7}\right)=2024\\ x\times1=2024\\ x=2024\)
= (2021+2029) + (2022+2028) + (2023+2027) + (2024+2026) + 2025
= 4050 + 4050 + 4050 + 4050 + 2025
= 8100 + 4050 +4050 + 2025
= 12 150 + 4050 + 2025
= 16 200 +2025
= 18 225
nếu đúng tick dùm mik nhé
2021+2029+2022+2028+2023+2027+2024+2026+2025
Bằng bao nhiêu bạn tự tính ra nhé
a:
b: \(\dfrac{1}{2}+\dfrac{2}{8}+\dfrac{4}{77}+\dfrac{5}{176}+\dfrac{6}{352}+\dfrac{7}{638}\)
\(=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{4}{77}+\dfrac{5+3}{352}+\dfrac{7}{638}\)
\(=\dfrac{3}{4}+\dfrac{4}{77}+\dfrac{1}{44}+\dfrac{7}{638}\)
\(=\dfrac{3\cdot77+4\cdot4+7}{308}+\dfrac{7}{638}=\dfrac{127}{154}+\dfrac{7}{638}\)
\(=\dfrac{3683}{4466}+\dfrac{49}{4466}=\dfrac{3732}{4466}=\dfrac{1866}{2233}\)
c: \(\left(\dfrac{161616}{212121}+\dfrac{2022}{3033}\right)\cdot7\)
\(=\left(\dfrac{16}{21}+\dfrac{2}{3}\right)\cdot7\)
\(=\dfrac{16+14}{21}\cdot7\)
\(=\dfrac{30}{3}=10\)