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Ta có :
\(\dfrac{1}{11}>\dfrac{1}{20}\\ \dfrac{1}{12}>\dfrac{1}{20}\\ ..........\\ \dfrac{1}{20}=\dfrac{1}{20}\)
\(\Rightarrow\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+...+\dfrac{1}{20}>\dfrac{1}{20}+\dfrac{1}{20}+...+\dfrac{1}{20}\\ \Rightarrow S>\dfrac{10}{20}\\ \Rightarrow S>\dfrac{1}{2}\)
Ta có: \(\dfrac{1}{11}>\dfrac{1}{20}\)
\(\dfrac{1}{12}>\dfrac{1}{20}\)
\(\dfrac{1}{13}>\dfrac{1}{20}\)
\(\dfrac{1}{14}>\dfrac{1}{20}\)
\(\dfrac{1}{15}>\dfrac{1}{20}\)
\(\dfrac{1}{16}>\dfrac{1}{20}\)
\(\dfrac{1}{17}>\dfrac{1}{20}\)
\(\dfrac{1}{18}>\dfrac{1}{20}\)
\(\dfrac{1}{19}>\dfrac{1}{20}\)
\(\dfrac{1}{20}=\dfrac{1}{20}\)
=> \(\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{20}>\dfrac{1}{20}.10\)
hay S > \(\dfrac{1}{2}\)
Ta có :
\(\dfrac{1}{11}>\dfrac{1}{20}\) ( vì 1 > 0 , 0 < 11 < 20 )
\(\dfrac{1}{12}>\dfrac{1}{20}\) ( vì 1 > 0 , 0 < 12 < 20 )
...
\(\dfrac{1}{20}=\dfrac{1}{20}\)
\(\Rightarrow\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+...+\dfrac{1}{20}>\dfrac{1}{20}+\dfrac{1}{20}+...+\dfrac{1}{20}\)( 10 số hạng )
\(\Rightarrow S>\dfrac{1}{20}.10\Rightarrow S>\dfrac{10}{20}\Rightarrow S>\dfrac{1}{2}\)
Vậy ...
\(\dfrac{1}{13}A=\dfrac{13^{19}+1}{13^{19}+\dfrac{1}{13}}=1+\dfrac{\dfrac{12}{13}}{13^{19}+\dfrac{1}{13}}\)
\(\dfrac{1}{13}B=\dfrac{13^{20}+1}{13^{20}+\dfrac{1}{13}}=1+\dfrac{\dfrac{12}{13}}{13^{20}+\dfrac{1}{13}}\)
Vì \(\dfrac{\dfrac{12}{13}}{13^{20}+\dfrac{1}{13}}< \dfrac{\dfrac{12}{13}}{13^{20}+\dfrac{1}{13}}\Rightarrow1+\dfrac{\dfrac{12}{13}}{13^{20}+\dfrac{1}{13}}< 1+\dfrac{\dfrac{12}{13}}{13^{20}+\dfrac{1}{13}}\)
\(\Rightarrow\dfrac{1}{13}A>\dfrac{1}{13}B\Rightarrow A>B\)
Vậy...
Ta xét hiệu:
\(A-1=\dfrac{3^{19}+1}{3^{18}+1}-1=\dfrac{3^{19}-3^{18}}{3^{18}+1}=\dfrac{3^{18}.2}{3^{18}+1}\)
\(B-1=\dfrac{3^{20}+1}{3^{19}+1}-1=\dfrac{3^{20}-3^{19}}{3^{19}+1}=\dfrac{3^{19}.2}{3^{19}+1}\)
Xét: \(\dfrac{A-1}{B-1}=\dfrac{3^{18}.2}{3^{18}+1}\cdot\dfrac{3^{19}+1}{3^{19}.2}=\dfrac{3^{19}+1}{\left(3^{18}+1\right).3}=\dfrac{3^{19}+1}{3^{19}+3}< 1\)
=> A-1<B-1
=>A<B
\(P=\dfrac{1+19+\dfrac{19}{13}+\dfrac{19}{101}}{7+\dfrac{7}{13}+\dfrac{7}{19}+\dfrac{7}{101}}\)
\(=\dfrac{19\left(1+\dfrac{1}{3}+\dfrac{1}{19}+\dfrac{1}{101}\right)}{7\left(1+\dfrac{1}{13}+\dfrac{1}{19}+\dfrac{1}{101}\right)}=\dfrac{19}{7}\)
A = 6/19 . -7/11 + 6/19 . -4/11 + -13/19
A = 6/19 . [-7/11 + (-4/11)] + (-13/19)
A = 6/19 . -11/11 + (-13/19)
A = 6/19 . (-1) + (-13/19)
A = -6/19 + (-13/19)
A = -19/19
A = -1
Vậy A = -1
a)
\(3\dfrac{14}{19}+\dfrac{13}{17}+\dfrac{35}{43}+6\dfrac{5}{19}+\dfrac{8}{43}\\ =\left(3\dfrac{14}{19}+6\dfrac{5}{19}\right)+\left(\dfrac{35}{43}+\dfrac{8}{43}\right)+\dfrac{13}{17}\\ =10+1+\dfrac{13}{17}\\ =11\dfrac{13}{17}\)
b)
\(\dfrac{-5}{7}\cdot\dfrac{2}{11}+\dfrac{-5}{7}\cdot\dfrac{9}{11}+1\dfrac{5}{7}\\ =\dfrac{-5}{7}\cdot\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+1\dfrac{5}{7}\\ =\dfrac{-5}{7}\cdot1+1\dfrac{5}{7}\\ =\dfrac{-5}{7}+1\dfrac{5}{7}\\ =1\)
a) \(3\dfrac{14}{19}+\dfrac{13}{17}+\dfrac{35}{43}+6\dfrac{5}{19}+\dfrac{8}{43}\)
\(=\left(3\dfrac{14}{19}+6\dfrac{5}{19}\right)+\left(\dfrac{35}{43}+\dfrac{8}{43}\right)+\dfrac{13}{17}\)
\(=\left[\left(3+6\right)+\left(\dfrac{14}{19}+\dfrac{5}{19}\right)\right]+1+\dfrac{13}{17}\)
\(=\left[9+1\right]+1+\dfrac{13}{17}\)
\(=10+1+\dfrac{13}{17}\)
\(=11+\dfrac{13}{17}\)
\(=\dfrac{187}{17}+\dfrac{13}{17}\)
\(=\dfrac{200}{17}\)
b) \(\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}.\dfrac{9}{11}+1\dfrac{5}{7}\)
\(=\dfrac{-5}{7}.\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{12}{7}\)
\(=\dfrac{-5}{7}.1+\dfrac{12}{7}\)
\(=\dfrac{-5}{7}+\dfrac{12}{7}\)
\(=\dfrac{7}{7}\)
\(=1\)
c) \(11\dfrac{3}{13}-\left(2\dfrac{4}{7}+5\dfrac{3}{13}\right)\)
= \(11\dfrac{3}{13}-2\dfrac{4}{7}-5\dfrac{3}{13}\)
\(=\left(11\dfrac{3}{13}-5\dfrac{3}{13}\right)-2\dfrac{4}{7}\)
\(=\left[\left(11-5\right)+\left(\dfrac{3}{13}-\dfrac{3}{13}\right)\right]-\dfrac{18}{7}\)
\(=\left[6+0\right]-\dfrac{18}{7}\)
\(=6-\dfrac{18}{7}\)
\(=\dfrac{42}{7}-\dfrac{18}{7}\)
\(=\dfrac{24}{7}\)
d) \(\dfrac{2}{7}.5\dfrac{1}{4}-\dfrac{2}{7}.3\dfrac{1}{4}\)
\(=\dfrac{2}{7}.\left(5\dfrac{1}{4}-3\dfrac{1}{4}\right)\)
\(=\dfrac{2}{7}.\left[\left(5-3\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)\right]\)
\(=\dfrac{2}{7}.\left[2+0\right]\)
\(=\dfrac{2}{7}.2\)
= \(\dfrac{4}{7}\)
c,
= \(\dfrac{5}{9}.\left(\dfrac{7}{13}+\dfrac{9}{13}+\dfrac{-3}{13}\right)\)
= \(\dfrac{5}{9}.1\)
= \(\dfrac{5}{9}\)
ta thấy : \(\dfrac{1}{11},\dfrac{1}{12},\dfrac{1}{13},...\dfrac{1}{19}\)đều lớn hơn\(\dfrac{1}{20}\)
=>\(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+...+\dfrac{1}{20}>\dfrac{1}{20}+\dfrac{1}{20}+...+\dfrac{1}{20}\)(20 số hạng \(\dfrac{1}{20}\))
=>\(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+..+\dfrac{1}{20}>1\) mà 1 > \(\dfrac{1}{2}\) =>\(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+..+\dfrac{1}{20}>\dfrac{1}{2}\)
\(\dfrac{14}{13}+\left(\dfrac{-1}{13}-\dfrac{19}{20}\right)=\dfrac{14}{13}-\dfrac{1}{13}-\dfrac{19}{20}\)
\(=\dfrac{13}{13}-\dfrac{19}{20}=1-\dfrac{19}{20}=\dfrac{20}{20}-\dfrac{19}{20}\)
\(=\dfrac{1}{20}\)
=\(\dfrac{14}{13}\) + ( \(\dfrac{-20}{260}\) - \(\dfrac{247}{260}\) )
=\(\dfrac{14}{13}\) + \(\dfrac{-267}{260}\)
=\(\dfrac{280}{260}\) + \(\dfrac{-267}{260}\)
=\(\dfrac{13}{260}\)
=\(\dfrac{1}{20}\)