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\(A=\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103+...}+\frac{1}{10.110}\)
\(A=\frac{1}{100}(\frac{100}{1.101}+\frac{100}{2.102}+\frac{100}{3.103}+...+\frac{100}{10.110})\)
\(A=\frac{1}{100}(\frac{1}{1}-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110})\)
\(A=\frac{1}{100}((\frac{1}{1}+\frac{1}{2}+...+\frac{1}{10})-(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}))\) 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. :)
VẾ TRÁI = (1/1x101 + 1/2x102 + 1/3x103 + ... + 1/10x110)xa
=1/100x(1/1 - 1/101 + 1/2 - 1/102 + 1/3 - 1/103 + ... +1/10 - 1/110)xa
=1/100x(1/1 + 1/2 + 1/3 + ... + 1/10 - 1/101 - 1/102 - 1/103 - ... - 1/110)xa(1)
VẾ PHẢI = 1/1x11 + 1/2x12 + 1/3x13 + ... +1/100x110
= 1/10x(1/1 -1/11 + 1/2 - 1/12 +1/3 - 1/13 + ...+ 1/100 - 1/110)
= 1/10x(1/1 + 1/2 + 1/3 +...+1/100 - 1/11 - 1/12 - 1/13 -...- 1/100 -1/101 -... -1/110)
= 1/10x(1/1 + 1/2 + 1/3 + ... + 1/10 - 1/101 - 1/102 - 1/103 - ... - 1/110)(2)
Từ (1) và (2) ta thấy để vế trái bằng vế phải thì a = 1/10 : 1/100 = 10.
Vậy a = 10
\(\Rightarrow\left(\frac{100}{1.101}+\frac{100}{2.102}+\frac{100}{3.103}+...+\frac{100}{10.110}\right).x=10.\left(\frac{10}{1.10}+\frac{10}{2.12}+...+\frac{10}{100.110}\right)\)
\(\Rightarrow\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+\frac{1}{3}-\frac{1}{103}+...+\frac{1}{10}-\frac{1}{110}\right).x=10\)\(\left(1-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110}\right)\)
\(\Rightarrow\left(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)- \left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right)\right).x=10\)
\(\left(\left(1+\frac{1}{2}+...+\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}\right)-\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{110}\right)\right)\)
\(\Rightarrow\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right).x=10\)
\(\left(\left(1+\frac{1}{2}+...+\frac{1}{10}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right)\right)\Rightarrow x=10\)
\(100E\)\(=\frac{100}{1.101}+\frac{100}{2.102}+..........+\frac{100}{10.110}\)
\(=1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+........+\frac{1}{10}-\frac{1}{110}\)
\(10F=\frac{10}{1.11}+\frac{10}{2.12}+......+\frac{10}{100.110}\)
\(=1-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+......+\frac{1}{100}-\frac{1}{110}\)
\(=1+\frac{1}{2}+...+\frac{1}{10}+\frac{1}{11}+....+\frac{1}{100}-\frac{1}{11}-\frac{1}{12}-....-\frac{1}{100}-\frac{1}{101}-...-\frac{1}{110}\)
\(=1+\frac{1}{2}+...+\frac{1}{10}-\frac{1}{101}-\frac{1}{102}-...-\frac{1}{110}\)\(=100E\)
\(\Rightarrow10F=100E\Rightarrow\frac{E}{F}=\frac{1}{10}\)
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
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