\(E=\frac{1}{1\times101}+\frac{1}{2\times102}+\frac{1}{3\times103}+.........+\frac{1}{10\times110}\) và \(F=\frac{1}{1\times11}+\frac{1}{2\times12}+\frac{1}{3\times13}+.......+\frac{1}{100\times110}\)
tìm tỉ số \(\frac{E}{F}\)
Cho:
\(A=\frac{1}{1\times101}+\frac{1}{2\times102}+\frac{1}{3\times103}+...+\frac{1}{25\times125}\)
\(B=\frac{1}{1\times26}+\frac{1}{2\times27}+\frac{1}{3\times28}+...+\frac{1}{100\times125}\)
Trong đó A có 25 số hạng, B có 100 số hạng. Tìm thương A:B
\(A=\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103}+...+\frac{1}{25.125}\)
\(A=\frac{1}{100}.\left(1-\frac{1}{101}\right)+\frac{1}{100}.\left(\frac{1}{2}-\frac{1}{102}\right)+\frac{1}{100}.\left(\frac{1}{3}-\frac{1}{103}\right)+...+\frac{1}{100}.\left(\frac{1}{25}-\frac{1}{125}\right)\)
\(A=\frac{1}{100}.\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+\frac{1}{3}-\frac{1}{103}+...+\frac{1}{25}-\frac{1}{125}\right)\)
\(A=\frac{1}{100}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}-\frac{1}{101}-\frac{1}{102}-\frac{1}{103}-...-\frac{1}{125}\right)\)
\(B=\frac{1}{1.26}+\frac{1}{2.27}+\frac{1}{3.28}+...+\frac{1}{100.125}\)
\(B=\frac{1}{25}.\left(1-\frac{1}{26}\right)+\frac{1}{25}.\left(\frac{1}{2}-\frac{1}{27}\right)+\frac{1}{25}.\left(\frac{1}{3}-\frac{1}{28}\right)+...+\frac{1}{25}.\left(\frac{1}{100}-\frac{1}{125}\right)\)
\(B=\frac{1}{25}.\left(1-\frac{1}{26}+\frac{1}{2}-\frac{1}{27}+\frac{1}{3}-\frac{1}{28}+...+\frac{1}{100}-\frac{1}{125}\right)\)
\(B=\frac{1}{25}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-\frac{1}{26}-\frac{1}{27}-\frac{1}{28}-...-\frac{1}{125}\right)\)
\(B=\frac{1}{25}.\left(1+\frac{1}{2}+...+\frac{1}{25}+\frac{1}{26}+\frac{1}{27}+...+\frac{1}{100}-\frac{1}{26}-\frac{1}{27}-...-\frac{1}{100}-\frac{1}{101}-...-\frac{1}{125}\right)\)\(B=\frac{1}{25}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}-\frac{1}{101}-\frac{1}{102}-\frac{1}{103}-...-\frac{1}{125}\right)\)
Ta thấy biểu thức trong ngoặc của hai vế A và B giống nhau
Vậy A : B = \(\frac{1}{100}:\frac{1}{25}=\frac{1}{4}\)
\(A=\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103}+...+\frac{1}{25.125}\)
\(\Rightarrow A=\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{24.25}\right)+\left(\frac{1}{101.102}+\frac{1}{102.103}+...+\frac{1}{124.125}\right)\)
\(A=\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{24}-\frac{1}{25}\right)+\left(\frac{1}{101}-\frac{1}{102}+\frac{1}{102}-\frac{1}{103}+...+\frac{1}{124}-\frac{1}{125}\right)\)
\(A=\left(1-\frac{1}{25}\right)+\left(\frac{1}{101}-\frac{1}{125}\right)\)
\(A=\frac{24}{25}+\frac{24}{12625}\)
Bạn tự tính luôn nha trog máy tính của mình là : 0,961... ( k làm thành phân số được )
\(B=\frac{1}{1.26}+\frac{1}{2.27}+\frac{1}{3.28}+...+\frac{1}{100.125}\)
\(\Rightarrow B=\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)+\left(\frac{1}{26.27}+\frac{1}{27.28}+...+\frac{1}{124.125}\right)\)
\(B=\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)+\left(\frac{1}{26}-\frac{1}{27}+\frac{1}{27}-\frac{1}{28}+...+\frac{1}{124}-\frac{1}{125}\right)\)
\(B=\left(1-\frac{1}{100}\right)+\left(\frac{1}{26}-\frac{1}{125}\right)\)
\(B=\frac{99}{100}+\frac{99}{3250}\)
\(B=\frac{6633}{6500}\)
Vậy từ bài trên mình làm ta có : => A:B = ...
\(\frac{1}{1\times5}+\frac{1}{5\times9}+\frac{1}{9\times13}+...+\frac{1}{97\times101}\)
\(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+.....+\frac{1}{97.101}\)
\(=\frac{1}{4}\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+......+\frac{4}{97.101}\right)\)
\(=\frac{1}{4}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+.....+\frac{1}{97}-\frac{1}{101}\right)\)
\(=\frac{1}{4}\left(1-\frac{1}{101}\right)\)
\(=\frac{1}{4}.\frac{100}{101}=\frac{25}{101}\)
\(A=\frac{1}{1\times300}+\frac{1}{2\times301}+...+\frac{1}{101\times400}\)
\(B=\frac{1}{1\times102}+\frac{1}{2\times103}+....+\frac{1}{298\times399}+\frac{1}{299\times400}\)
\(Tính:\frac{A}{B}\)
Giups mk voi mk can gap
A=\(\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+...+\frac{1}{101.400}\)
299.A= 299.(\(\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+...+\frac{1}{101.400}\))
299.A=\(\frac{299}{1.300}+\frac{299}{2.301}+\frac{299}{3.302}+...+\frac{299}{101.400}=\frac{1}{1}-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+...+\frac{1}{101}-\frac{1}{400}\)
A= \(=\frac{1}{299}\left(1+\frac{1}{2}+...+\frac{1}{101}-\frac{1}{300}-\frac{1}{301}-...-\frac{1}{400}\right)\)
Tương tự
B=\(\frac{1}{101}.\left(\frac{1}{1}-\frac{1}{102}+\frac{1}{2}-\frac{1}{103}+...+\frac{1}{299}-\frac{1}{400}\right)\)
B= \(\frac{1}{101}.\left(\frac{1}{1}+\frac{1}{2}...+\frac{1}{299}-\frac{1}{102}-\frac{1}{103}-...-\frac{1}{400}\right)\)
B= \(\frac{1}{101}.\left(\frac{1}{1}+\frac{1}{2}...+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{299}-\frac{1}{102}-\frac{1}{103}-...-\frac{1}{400}\right)\)
B= \(\frac{1}{101}.\left(\frac{1}{1}+\frac{1}{2}...+\frac{1}{101}-\frac{1}{300}-\frac{1}{301}-...-\frac{1}{400}\right)\)
Hai dấu ngoặc ở biểu thức A và biểu thức B như nhau
Vậy \(A:B=\frac{1}{299}:\frac{1}{101}=\frac{101}{299}\)
Bài 1 :
a, \(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{101\times102}\)
b, \(\frac{5}{5\times10}+\frac{5}{10\times15}+...+\frac{5}{115\times120}\)
c, \(\frac{2}{5\times7}+\frac{2}{7\times9}+\frac{2}{9\times11}+...+\frac{2}{997\times999}\)
A. = 1/2-1/3+1/3-1/4+1/4-1/5...+1/101-1/102=1/2-1/102=25/51.
B. =1/5-1/10+1/10-1/15+...+1/115-1/120=1/5-1/120=23/120.
C. = 1/5-1/7+1/7-1/9+1/9-1/11+...+1/997-1/999=1/5-1/999=994/4995.
Minh kiem tra bang may tinh roi do.
\(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{101\times102}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{101}-\frac{1}{102}+\frac{1}{102}\)
\(=1-\frac{1}{102}\)
\(=\frac{101}{102}\)
Tính \(\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+\frac{1}{9\times11}+\frac{1}{11\times13}\)
SAI HẾT RỒI.........CẦN THÌ TỚ GIẢI LẠI CHO !!
thế này :
= \(\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{11.13}\right)\)
= \(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{11}-\frac{1}{13}\right)\)
= \(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{13}\right)\)
= \(\frac{1}{2}.\frac{10}{39}\)
= \(\frac{5}{39}\)
Vậy kq = \(\frac{5}{39}\)
a,A=\(\frac{1}{10\times11}+\frac{1}{11\times12}+...+\frac{1}{99\times100}\)
\(A=\frac{1}{10\times11}+\frac{1}{11\times12}+...+\frac{1}{99\times100}\)
\(A=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{10}-\frac{1}{100}\)
\(A=\frac{9}{100}\)
\(A=\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{99.100}\)
\(=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{10}-\frac{1}{100}\)
\(=\frac{9}{100}\)
9/100 nha bn
chuc bn hoc tot
happy new year
1\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+............+\frac{1}{99\times100}+\frac{1}{100\times101}\)
\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101}.\)
\(=\frac{1}{1}-\frac{1}{101}\)
\(=\frac{101}{101}-\frac{1}{101}=\frac{100}{101}\)
TÍNH
\(\frac{\left(1+2+3+...+100\right)\times\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right)\times\left(6,3\times12-21\times3,6\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
Dễ thấy 6,3 . 12 - 21 . 3,6 = 63 . 1,2 - 63 . 1,2 = 0
Do đó biểu thức trên bằng 0
Tính \(A=\frac{M}{N}\)biết :
\(M=\frac{1}{1\times51}+\frac{1}{2\times52}+...+\frac{1}{12\times62}\)
\(N=\frac{1}{1\times11}+\frac{1}{2\times12}+...+\frac{1}{52\times62}\)