cho E = \(\frac{1}{1.100}+\frac{1}{2.102}+\frac{1}{3.103}+...+\frac{1}{10.110}\)
F = \(\frac{1}{1.11}+\frac{1}{2.12}+\frac{1}{3.13}+...+\frac{1}{100.110}\)
Tính tỉ số \(\frac{E}{F}\)
\(E=\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103}+.........+\frac{1}{10.110}\)
\(F=\frac{1}{1.11}+\frac{1}{2.12}+\frac{1}{3.13}+.........+\frac{1}{100.110}\)
tính tỉ số \(\frac{E}{F}\)
\(100E\)\(=\frac{100}{1.101}+\frac{100}{2.102}+..........+\frac{100}{10.110}\)
\(=1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+........+\frac{1}{10}-\frac{1}{110}\)
\(10F=\frac{10}{1.11}+\frac{10}{2.12}+......+\frac{10}{100.110}\)
\(=1-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+......+\frac{1}{100}-\frac{1}{110}\)
\(=1+\frac{1}{2}+...+\frac{1}{10}+\frac{1}{11}+....+\frac{1}{100}-\frac{1}{11}-\frac{1}{12}-....-\frac{1}{100}-\frac{1}{101}-...-\frac{1}{110}\)
\(=1+\frac{1}{2}+...+\frac{1}{10}-\frac{1}{101}-\frac{1}{102}-...-\frac{1}{110}\)\(=100E\)
\(\Rightarrow10F=100E\Rightarrow\frac{E}{F}=\frac{1}{10}\)
Help me do my homework !
Cho \(E=\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103}+....+\frac{1}{10.110}\) và \(\frac{1}{1.11}+\frac{1}{2.12}+\frac{1}{3.13}+.....+\frac{1}{100.110}\)
Tính tỉ số \(\frac{E}{F}\)
E = 1/1.101+1/2.102+...+1/10.110
E = 1/100[100/1.101+100/2.102+...+100/10.110]
E = 1/100[1/1-1/101+1/2-1/102+...+1/10-1/110]
E = 1/100[[1/1+1/2+1/3...+1/10]-[1/101+1/102+...+1/110] - xg cái E
F = 1/1.11+1/2.12+...+1/100.110
F = 1/10[10/1.11+10/2.12+...+10/100.110]
F = 1/10[1/1-1/11+1/2-1/12+...+1/100-1/110]
F = 1/10[[1/1+1/2+...+1/100]-[1/11+1/12...+1/110]]
F = 1/10[[1/1+1/2+...+1/10]-[1/101+1/102+...+1/110]
\(\Rightarrow\frac{E}{F}=\frac{\frac{1}{100}\left[\left[\frac{1}{1}+\frac{1}{2}+...+\frac{1}{10}\right]-\left[\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right]\right]}{\frac{1}{10}\left[\left[\frac{1}{1}+\frac{1}{2}+...+\frac{1}{10}\right]-\left[\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right]\right]}=\frac{1}{10}\)
Xỉu... vì đuối sau khi bấm
<br class="Apple-interchange-newline"><div id="inner-editor"></div>⇒EF =1100 [[11 +12 +...+110 ]−[1101 +1102 +...+1110 ]]110 [[11 +12 +...+110 ]−[1101 +1102 +...+1110 ]] =110
<br class="Apple-interchange-newline"><div id="inner-editor"></div>⇒EF =1100 [[11 +12 +...+110 ]−[1101 +1102 +...+1110 ]]110 [[11 +12 +...+110 ]−[1101 +1102 +...+1110 ]] =110
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. :)
Tìm x , bíÊt:
\(\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103}+...+\frac{1}{10.110}x=\frac{1}{1.11}+\frac{1}{2.12}+\frac{1}{3.13}+...+\frac{1}{100.110}\)
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:
( \(\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103}+...+\frac{1}{10.110}\) ) . x = \(\frac{1}{1.11}+\frac{1}{2.12}+...+\frac{1}{100.110}\)
Cho E=\(\frac{1}{1.101}\)+\(\frac{1}{2.102}\)+\(\frac{1}{3.103}\)+...+\(\frac{1}{10.110}\)va F=\(\frac{1}{1.11}\)+\(\frac{1}{2.12}\)+\(\frac{1}{3.13}\)+...+\(\frac{1}{100.110}\)
Tinh \(\frac{E}{F}\)
Cho \(E=\frac{1}{1.101}+\frac{1}{2.102}+...+\frac{1}{10.101}\)
\(F=\frac{1}{1.11}+\frac{1}{2.12}+...+\frac{1}{100.110}\)
Tính tỉ số \(\frac{E}{F}\) và \(\frac{F}{E}\)
Cho
E=\(\frac{1}{1.101}\) + \(\frac{1}{2.102}\) +...+\(\frac{1}{10.110}\)
F=\(\frac{1}{1.11}\)+\(\frac{1}{2.12}\) +...+\(\frac{1}{100.110}\)
Tính \(\frac{E}{F}\)