\(A=\frac{9}{10}+\frac{39}{40}+\frac{87}{88}+\frac{153}{154}+\frac{237}{238}+\frac{339}{340}=\frac{117}{20}\)

\(suyra:A<1\)

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\(\frac{-7}{72}\) và \(\frac{9}{-40}\)

* \(\frac{9}{-40}=\frac{-9}{40}\)

Ta có : \(MSC=BCNN(72,40)=360\)

Quy đồng : \(\frac{-7}{72}=\frac{-7\cdot5}{72\cdot5}=\frac{-35}{360}\)

\(\frac{9}{-40}=\frac{-9}{40}=\frac{-9\cdot9}{40\cdot9}=\frac{-81}{360}\)

Mà \(\frac{-35}{360}>\frac{-81}{360}\Rightarrow\frac{-7}{72}>\frac{9}{-40}\)

Vậy : \(\frac{-7}{72}>\frac{9}{-40}\)

a>\(16^{19}=2^{4^{19}}=2^{76}\\ 8^{25}=2^{3^{25}}=2^{75}\)

16^19>8^25

\(S=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{184}+\frac{1}{238}+\frac{1}{340}=\frac{1}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\right)\)

\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\right)\)

\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)=\frac{1}{3}.\frac{9}{20}=\frac{3}{20}>\frac{2}{20}=\frac{1}{10}=0,1\)

vậy S>0,1

S = \(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)

S = \(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\)

S = \(\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+....+\frac{1}{17}-\frac{1}{20}\right)\)

S = \(\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)=\frac{1}{3}.\frac{9}{20}\)

S = \(\frac{3}{20}\)

S = 0,15 > 0,1

S=\(\frac{3}{20}\)>0,1

li ke cho mình nha