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Ta có A = \(\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}+\dfrac{1}{132}\)
= \(\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}+\dfrac{1}{10\cdot11}+\dfrac{1}{11\cdot12}\)
= \(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}\)
= \(\dfrac{1}{6}-\dfrac{1}{12}=\dfrac{1}{12}\)
B = \(\dfrac{\dfrac{2}{29}-\dfrac{2}{39}+\dfrac{2}{49}}{\dfrac{23}{29}-\dfrac{23}{39}+\dfrac{23}{49}}=\dfrac{2\left(\dfrac{1}{29}-\dfrac{1}{39}+\dfrac{1}{49}\right)}{23\left(\dfrac{1}{29}-\dfrac{1}{39}+\dfrac{1}{49}\right)}=\dfrac{2}{23}\)
Lại có \(\dfrac{2}{23}>\dfrac{2}{24}=\dfrac{1}{12}\) hay A < B
Vậy A < B
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Ta có :
\(\frac{1}{31}>\frac{1}{40};\frac{1}{32}>\frac{1}{40};...;\frac{1}{40}=\frac{1}{40}\)
\(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{39}+\frac{1}{40}\)
\(>\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}+\frac{1}{40}=\frac{1}{40}.10=\frac{1}{4}\)
\(\frac{1}{31}+\frac{1}{32}+..................+\frac{1}{39}+\frac{1}{40}>\frac{1}{4}\)
KẾT BẠN NHA
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A = \(\frac{5^{30}-2}{5^{31-2}}\) = 5
B = \(\frac{5^{31}-2}{5^{32}-2}\) = \(\frac{1}{5}\) = 0.2
Mà 5 > 0.2
Nên: A > B
\(5A=\frac{5^{31}-10}{5^{31}-2}=\frac{5^{31}-2-8}{5^{31}-2}=\frac{5^{31}-2}{5^{31}-2}-\frac{8}{5^{31}-2}=1-\frac{8}{5^{31}-2}\left(1\right)\)
\(5B=\frac{5^{32}-10}{5^{32}-2}=\frac{5^{32}-8}{5^{32}-2}=\frac{5^{32}-2}{5^{32}-2}-\frac{8}{5^{32}-2}=1-\frac{8}{5^{32}-2}\left(2\right)\)
từ (1) và (2)
=>A>B
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32 . 53 - 31 = 53.(31 + 1) - 31 = 53.31 < 53.31 + 32
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Ta có \(\dfrac{{31}}{{32}} > 0\) và \(\dfrac{{ - 5}}{{57}} < 0\) nên \(\dfrac{{31}}{{32}} > \dfrac{{ - 5}}{{57}}\).
-31/42