Tính tỉ số A/B biết:
A\(=\frac{1}{1.101}+\frac{1}{2.102}+...\frac{1}{10.110}\)
B\(=\frac{1}{1.11}+\frac{1}{2.12}+...\)
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Tính tỉ số \(\frac{A}{B}\)biết:
A= \(\frac{1}{1.101}+\frac{1}{2.102}+...+\frac{1}{10.110}\)
B=\(\frac{1}{1.11}+\frac{1}{2.12}\)\(+...+\frac{1}{100.110}\)
\(A= {1\over1.101}+{1\over2.102}+...+{1\over10.110}= \)
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\(E=\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103}+.........+\frac{1}{10.110}\)
\(F=\frac{1}{1.11}+\frac{1}{2.12}+\frac{1}{3.13}+.........+\frac{1}{100.110}\)
tính tỉ số \(\frac{E}{F}\)
\(100E\)\(=\frac{100}{1.101}+\frac{100}{2.102}+..........+\frac{100}{10.110}\)
\(=1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+........+\frac{1}{10}-\frac{1}{110}\)
\(10F=\frac{10}{1.11}+\frac{10}{2.12}+......+\frac{10}{100.110}\)
\(=1-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+......+\frac{1}{100}-\frac{1}{110}\)
\(=1+\frac{1}{2}+...+\frac{1}{10}+\frac{1}{11}+....+\frac{1}{100}-\frac{1}{11}-\frac{1}{12}-....-\frac{1}{100}-\frac{1}{101}-...-\frac{1}{110}\)
\(=1+\frac{1}{2}+...+\frac{1}{10}-\frac{1}{101}-\frac{1}{102}-...-\frac{1}{110}\)\(=100E\)
\(\Rightarrow10F=100E\Rightarrow\frac{E}{F}=\frac{1}{10}\)
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Tìm số nguyên x,biết:
\(\frac{1}{1.101}+\frac{1}{2.102}+...+\frac{1}{10.110}\left(x\right)=\frac{1}{1.11}+\frac{1}{2.12}+...+\frac{1}{100.101}\)
tìm x
\(\left(\frac{1}{1.101}+\frac{1}{2.102}+...+\frac{1}{10.110}\right).x=\frac{1}{100.110}+\frac{1}{99.109}+...+\frac{1}{2.12}+1.11\)
Vì gõ trên Hoc24 khá lâu nên mình gửi hình ảnh cho lẹ
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Tìm x bít
\(\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103}+...+\frac{1}{10.110}.x=\frac{1}{1.11}+\frac{1}{2.12}+\frac{1}{3.13}+...+\frac{1}{100.110}\)
\(A=\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103+...}+\frac{1}{10.110}\)
\(A=\frac{1}{100}(\frac{100}{1.101}+\frac{100}{2.102}+\frac{100}{3.103}+...+\frac{100}{10.110})\)
\(A=\frac{1}{100}(\frac{1}{1}-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110})\)
\(A=\frac{1}{100}((\frac{1}{1}+\frac{1}{2}+...+\frac{1}{10})-(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}))\) ok?
\(B=\frac{1}{1.11}+\frac{1}{2.12}+...+\frac{1}{100.110}\)
\(B=\frac{1}{10}(\frac{10}{1.11}+\frac{10}{2.12}+...+\frac{10}{100.110})\)
\(B=\frac{1}{10}(\frac{1}{1}-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110})\)
\(B=\frac{1}{10}((\frac{1}{1}+\frac{1}{2}+...+\frac{1}{100})-(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{110}))\)=\(\frac{1}{10}((\frac{1}{1}+\frac{1}{2}+...+\frac{1}{10})-(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}))\)
B=10A
A.x=10A suy ra x=10
gõ xong mém xỉu. :)
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Xem thêm câu trả lời
Tìm x biết: \(\left(\frac{1}{1.101}+\frac{1}{2.102}+...+\frac{1}{10.110}\right).x=\frac{1}{1.11}+\frac{1}{2.12}+...+\frac{1}{100.110}\)
Giải phương trình :\(\left(\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103}+...+\frac{1}{10.110}\right).x=\frac{1}{1.11}+\frac{1}{2.12}+...+\frac{1}{100.110}\)
Câu hỏi của Huỳnh Ngọc Cẩm Tú - Toán lớp 6 - Học toán với OnlineMath
Tìm x , bíÊt:
\(\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103}+...+\frac{1}{10.110}x=\frac{1}{1.11}+\frac{1}{2.12}+\frac{1}{3.13}+...+\frac{1}{100.110}\)
Tìm x:
( \(\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103}+...+\frac{1}{10.110}\) ) . x = \(\frac{1}{1.11}+\frac{1}{2.12}+...+\frac{1}{100.110}\)