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ko ghi lại đề
ta thấy : 2019 - 1 = 2018
2020 - 2 = 2018
2021 - 3 = 2018
2022 - 4 = 2018
=> x = 2018
thử lại :
2018+1/2019 + 2018+2/2020 = 2018+3/2021 + 2018+4/2022
= 1 + 1 = 1 + 1
2 = 2
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Câu b
Ta có :x + 3 /1.3 +3/3.5 + 3/5.7+...+3/13.15=2 1/5
X + 2/3.(1-1/3+1/3-1/5+1/5-1/7+...+1/13-1/15)1=11/5
X+2/3.(1-1/15)=11/5
X+ 2/3.14/15=11/5
X + 28/45=11/5
X = 11/5 -28/45
X=71/45
Câu a gợi ý
1/2-1/3/1/6=0
1/2- 1/3 - 1/6 ) x (1/2 + 2/3 + 3/4 +4/5 + .......+ 2019 /2020 ) =0
3/4:x=9/10
X = 3/4:9/10
X = 5/6
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a) \(\left(2017\times2018+2018+2019\right)\times\left(1+\frac{1}{2}:1\frac{1}{2}-1\frac{1}{3}\right)\)
\(=\left(2017\times2018+2018+2019\right)\times\left(1+\frac{1}{2}:\frac{3}{2}-1\frac{1}{3}\right)\)
\(=\left(2017\times2018+2018+2019\right)\times\left(1+\frac{1}{3}-1\frac{1}{3}\right)\)
\(=\left(2017\times2018+2018+2019\right)\times0\)
\(=0\)
b) 10,11 + 11,12 + 12,13 + ...+ 98,99 + 99, 100
Số số hạng từ 10,11 đến 98,99 là:
( 98,99 - 10,11) : 1,01 + 1= 89
Tổng dãy số trên từ 10,11 đến 98,99 là:
( 98,99 + 10,11) x 89 : 2 = 4 854,95
=> 10,11 + 11,12+12,13 + ...+ 98,99+ 99,100 = 4 854,95 + 99, 1 = 4 954, 05
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\(x\)là dấu nhân hả bạn? Nếu vậy thì mk làm cho nhé
\(A=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot....\cdot\left(1-\frac{1}{20}\right)\)
\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot.......\cdot\frac{17}{18}\cdot\frac{18}{19}\cdot\frac{19}{20}=\frac{1}{20}\)
Vậy \(A=\frac{1}{20}\)
\(B=1\frac{1}{2}\cdot1\frac{1}{3}\cdot1\frac{1}{4}\cdot........\cdot1\frac{1}{2005}\cdot1\frac{1}{2006}\cdot1\frac{1}{2007}\)
\(B=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot......\cdot\frac{2006}{2005}\cdot\frac{2007}{2006}\cdot\frac{2008}{2007}=\frac{2008}{2}=1004\)
Vậy \(B=1004\)
DẤU CHẤM LÀ DẤU NHÂN
a,
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{19}{20}=\frac{1}{20}\)
b, \(1\frac{1}{2}.1\frac{1}{3}....1\frac{1}{2017}=\frac{3}{2}.\frac{4}{3}....\frac{2018}{2017}=\frac{2018}{2}=1009\)
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\(A=\frac{2015+2016+2017}{2014+2015+2016+2017+2018}x1000\)
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\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{7\cdot8}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{8}\)
\(=\frac{1}{1}-\frac{1}{8}=\frac{7}{8}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{3.7}+\frac{1}{7.8}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}\)
\(=1-\frac{1}{8}+0+0+...+0\)
\(=\frac{7}{8}\)
\(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot...\cdot\left(1-\frac{1}{2019}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot......\cdot\frac{2018}{2019}\)
\(=\frac{1\cdot2\cdot...\cdot2018}{2\cdot3\cdot...\cdot2019}\)
\(=\frac{1}{2019}\)