1+\(\frac{2}{3}\)
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\(1+\frac{1+\frac{1+\frac{3}{2}}{2}}{2}=1+\frac{1+\frac{\frac{5}{2}}{2}}{2}=1+\frac{1+\frac{5}{4}}{2}=1+\frac{\frac{9}{4}}{2}=1+\frac{9}{8}=\frac{17}{8}\)
\(1+\frac{2}{1+\frac{2}{1+\frac{2}{3}}}=1+\frac{2}{1+\frac{2}{\frac{5}{3}}}=1+\frac{2}{1+\frac{6}{5}}=1+\frac{2}{\frac{11}{5}}=1+\frac{10}{11}=\frac{21}{11}\)
\(1+\frac{1+\frac{1+\frac{2}{3}}{3}}{3}=1+\frac{1+\frac{\frac{5}{3}}{3}}{3}=1+\frac{1+\frac{5}{9}}{3}=1+\frac{\frac{14}{9}}{3}=1+\frac{14}{27}=\frac{41}{27}\)
\(\frac{3}{\frac{3}{\frac{3}{\frac{3}{2}+1}+1}+1}+1=1+\frac{3}{\frac{3}{\frac{3}{\frac{5}{2}}+1}+1}=1+\frac{3}{\frac{3}{\frac{6}{5}+1}+1}=1+\frac{3}{\frac{15}{11}+1}=\frac{59}{26}\)
suy ra
\(\frac{\frac{17}{18}}{\frac{21}{11}}-x=\frac{187}{378}-x=\frac{\frac{41}{27}}{\frac{59}{26}}=\frac{1066}{1593}\Rightarrow x=-\frac{1297}{7434}\)
giải:
ta có :
\(\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}}{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}}:\frac{3+\frac{3}{2}+\frac{3}{3}+\frac{3}{4}}{2-\frac{2}{2}+\frac{2}{3}-\frac{2}{4}}\)
\(\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}}{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}}.\frac{2\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}{3\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)}=\frac{2}{3}\)
\(A=3.\frac{1}{2}\left(2.\frac{1}{3}+\frac{-1}{3}\right)\)
\(A=\frac{3}{2}.\frac{1}{3}=\frac{1}{2}\)
\(B=\frac{-1}{2}\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}\right)\)
\(B=\frac{-1}{2}.\frac{1}{2}=-\frac{1}{4}\)
\(b)\) Đặt \(B=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}\) ta có :
\(B>\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}=\frac{3+3+3+3+3}{15}=\frac{3.5}{15}=\frac{15}{15}=1\)
\(\Rightarrow\)\(B>1\) \(\left(1\right)\)
Lại có :
\(B< \frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}=\frac{3+3+3+3+3}{10}=\frac{3.5}{10}=\frac{15}{10}< \frac{20}{10}=2\)
\(\Rightarrow\)\(B< 2\) \(\left(2\right)\)
Từ (1) và (2) suy ra :
\(1< B< 2\) ( đpcm )
Vậy \(1< B< 2\)
Chúc bạn học tốt ~
a) \(\frac{3}{4} + \left( {\frac{1}{2} - \frac{1}{3}} \right) = \frac{9}{{12}} + \left( {\frac{6}{{12}} - \frac{4}{{12}}} \right) = \frac{9}{{12}} + \frac{2}{{12}} = \frac{{11}}{{12}}\)
\(\frac{3}{4} + \frac{1}{2} - \frac{1}{3} = \frac{9}{{12}} + \frac{6}{{12}} - \frac{4}{{12}} = \frac{{15}}{{12}} - \frac{4}{{12}} = \frac{{11}}{{12}}\)
Vậy \(\frac{3}{4} + \left( {\frac{1}{2} - \frac{1}{3}} \right)\) = \(\frac{3}{4} + \frac{1}{2} - \frac{1}{3}\)
b)\(\frac{2}{3} - \left( {\frac{1}{2} + \frac{1}{3}} \right) = \frac{4}{6} - \left( {\frac{3}{6} + \frac{2}{6}} \right) = \frac{4}{6} - \frac{5}{6} = \frac{{ - 1}}{6}\)
\(\frac{2}{3} - \frac{1}{2} - \frac{1}{3} = \frac{4}{6} - \frac{3}{6} - \frac{2}{6} = \frac{1}{6} - \frac{2}{6} = \frac{{ - 1}}{6}\)
Vậy \(\frac{2}{3} - \left( {\frac{1}{2} + \frac{1}{3}} \right)\)=\(\frac{2}{3} - \frac{1}{2} - \frac{1}{3}\).
`#3107`
`a)`
`3/4 + (1/2 - 1/3)`
`= 3/4 + (3/6 - 2/6)`
`= 3/4 + 1/6`
`= 11/12`
`3/4 + 1/2 - 1/3`
`= 9/12 + 6/12 - 4/12`
`= (9 + 6 - 4)/12`
`= 11/12`
Vì `11/12 = 11/12`
`=> 3/4 + (1/2 - 1/3) = 3/4 + 1/2 - 1/3`
`b)`
`2/3 - (1/2 + 1/3)`
`= 2/3 - (3/6 + 2/6)`
`= 2/3 - 5/6`
`= -1/6`
`2/3 - 1/2 - 1/3`
`= 4/6 - 3/6 - 2/6`
`= (4 - 3 - 2)/6`
`= -1/6`
Vì `-1/6 = -1/6`
`=> 2/3 - (1/2 + 1/3) = 2/3 - 1/2 - 1/3`
Phần C đề thiếu
\(D=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)
\(\Rightarrow3D=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)
\(\Rightarrow3D-D=(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}})-\)\((\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}})\)
\(\Rightarrow2D=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow6D=3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(\Rightarrow6D-2D=3-\frac{101}{3^{99}}+\frac{100}{3^{100}}\)
\(\Rightarrow4D=3-\frac{203}{3^{100}}\)
\(\Rightarrow D=\frac{3}{4}-\frac{\frac{203}{3^{100}}}{4}< \frac{3}{4}\left(đpcm\right)\)
1/1+2/3=3/3+2/3=5/3
Trả lời :
= \(\frac{5}{3}\)
#_Mập
_Hok tốt