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Vì \(\frac{1}{33}>\frac{1}{34}>\frac{1}{35}>\frac{1}{36}\)
\(\Rightarrow M>\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}\)\(\)
\(\Rightarrow M>\frac{4}{36}=\frac{1}{9}\)
Mà \(\frac{1}{9}>\frac{1}{10}\)
\(\Rightarrow\)\(M>\frac{1}{9}>\frac{1}{10}\)
Vậy : M > N
a,64 x 25 + 35 x 25 +25=25.(64+35+1)=25.100=2500
b) 58 x 42 + 32 x 8 + 5=2436+256+5=2697
c) A =1 +2+3+4+...+100=(100+1)x[(100-1):1+1]:2=5050
\(64\times25+35\times25+25\)
\(=25\times\left(64+35+1\right)\)
\(=25\times100\)
\(=2500\)
\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
Mình chỉnh lại đề B nha:
\(B=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{9999}\)
\(=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{101}\right)\)
\(=\frac{1}{2}.\frac{100}{101}=\frac{50}{101}\)
\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
a) 1+2+3+...+99+100
giải
Từ 1 đến 100 có 100 số.Như vậy,số cặp số là:
100:2=50(cặp)
Mỗi cặp số có tổng bằng:
1+100(2+99)(3+98)...=11
Kết quả của phép tính là:
101x50=5050
Đáp số:5050
=1/3*5+1/5*7+1/7*9+...+1/99*101
=1/3-1/5+1/5-1/7+...+1/99-1/101
=1/3-1/101
=98/303
a) 890:x=35 dư 15
X= ( 890-15) :35
X = 875 : 35
X= 35
b) 648-34. x= 444
34 .x = 648-444
34.x = 204
x = 204:34
x= 6
c) 1482:x+23= 80
1482:x= 80-23
1482: x= 57
x= 1482 : 57
x = 26
+ \(\frac{1}{n\times\left(n+2\right)}=\frac{\left(n+2\right)-n}{n\times\left(n+2\right)}\)
\(=\frac{n+2}{n\times\left(n+2\right)}-\frac{n}{n\times\left(n+2\right)}=\frac{1}{n}-\frac{1}{n+2}\)
+ \(\frac{2}{3}+\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+\frac{98}{99}\)
\(=1-\frac{1}{3}+1-\frac{1}{15}+1-\frac{1}{35}+1-\frac{1}{63}+1-\frac{1}{99}\)
\(=5-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\)
\(=5-\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+\frac{2}{9\times11}\right)\)
\(=5-\frac{1}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\right)\)
\(=5-\frac{1}{2}\times\left(1-\frac{1}{11}\right)\)
\(=5-\frac{1}{2}+\frac{1}{22}=\frac{50}{11}\)
=00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
=00000000000000000000000000000000000000000000