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tìm x biết:
(1/1.101 + 1/2.102 + 1/3.103+....+1/10.110) .x = 1/1.11 + 1/2.12 + 1/3.13 +....+1/100.110
⇒(1−1101 +12 −1102 +13 −1103 +...+110 −1110 ).x=10.(1−111 +12 −112 +...+1100 −1110 )
⇒((1+12 +13 +...+110 )−(1101 +1102 +...+1110 )).x=10.((1+12 +..+110 +111 +112 +...+1100 )−(111 +112 +...+1110 ))
\(100E\)\(=\frac{100}{1.101}+\frac{100}{2.102}+..........+\frac{100}{10.110}\)
\(=1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+........+\frac{1}{10}-\frac{1}{110}\)
\(10F=\frac{10}{1.11}+\frac{10}{2.12}+......+\frac{10}{100.110}\)
\(=1-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+......+\frac{1}{100}-\frac{1}{110}\)
\(=1+\frac{1}{2}+...+\frac{1}{10}+\frac{1}{11}+....+\frac{1}{100}-\frac{1}{11}-\frac{1}{12}-....-\frac{1}{100}-\frac{1}{101}-...-\frac{1}{110}\)
\(=1+\frac{1}{2}+...+\frac{1}{10}-\frac{1}{101}-\frac{1}{102}-...-\frac{1}{110}\)\(=100E\)
\(\Rightarrow10F=100E\Rightarrow\frac{E}{F}=\frac{1}{10}\)
E100E=1001.101+1002.102+..........+10010.110=1.101100+2.102100+..........+10.110100
=1−1101+12−1102+........+110−1110=1−1011+21−1021+........+101−1101
10F=101.11+102.12+......+10100.11010F=1.1110+2.1210+......+100.11010
=1−111+12−112+......+1100−1110=1−111+21−121+......+1001−1101
=1+12+...+110+111+....+1100−111−112−....−1100−1101−...−1110=1+21+...+101+111+....+1001−111−121−....−1001−1011−...−1101
=1+12+...+110−1101−1102−...−1110=1+21+...+101−1011−1021−...−1101=100E=100E lm như bn này nha bn
⇒10F=100E⇒EF=110
\(\left(\frac{1}{1\cdot101}+\frac{1}{2\cdot102}+\frac{1}{3\cdot103}+...+\frac{1}{10\cdot110}\right)x=\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+...+\frac{1}{100\cdot110}\)
\(\Rightarrow\left(\frac{100}{1\cdot101}+\frac{100}{2\cdot102}+\frac{100}{3\cdot103}+...+\frac{100}{100\cdot110}\right)x=10\left(\frac{10}{1\cdot11}+\frac{10}{2\cdot12}+...+\frac{10}{100\cdot110}\right)\)
\(\Rightarrow\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+\frac{1}{3}-\frac{1}{103}+....+\frac{1}{10}-\frac{1}{110}\right)x=10\)\(\left(1-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110}\right)\)
\(\Rightarrow\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)-\left(\frac{1}{101}+\frac{1}{102}+....+\frac{1}{110}\right)\right]x=10\)\(\left[\left(1+\frac{1}{2}+....+\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+....+\frac{1}{100}\right)-\left(\frac{1}{11}+\frac{1}{12}+....+\frac{1}{110}\right)\right]\)
\(\Rightarrow\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)-\left(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{110}\right)\right]x=10\)
\(\left[\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{10}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right)\right]\)
\(\Rightarrow x=10\)
\(A=\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103+...}+\frac{1}{10.110}\)
\(A=\frac{1}{100}(\frac{100}{1.101}+\frac{100}{2.102}+\frac{100}{3.103}+...+\frac{100}{10.110})\)
\(A=\frac{1}{100}(\frac{1}{1}-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110})\)
\(A=\frac{1}{100}((\frac{1}{1}+\frac{1}{2}+...+\frac{1}{10})-(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}))\) 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. :)
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
(100/1.101 + 100/2.102 + 100/3.103 +....+100/10.110) . x
= (10/1.11 + 10/2.12 + 10/100.110 )10
=>(1+1/2+1/3+...+1/10-1/101-...-1/110)x
=(1+1/2+1/3+...+1/10+1/11+...+1/100-1/11-...-1/100-1/101-...-1/110)10 =>(1+1/2+1/3+...+1/10-1/101-...-1/110)x
=(1+1/2+1/3+...+1/10-1/101-...-1/110)10 =>x=10
\(\dfrac{1}{10}nhébạn\)
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]
⇒EF=1100[[11+12+...+110]−[1101+1102+...+1110]]110[[11+12+...+110]−[1101+1102+...+1110]]=110