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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
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
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
Ta có : A = 32 . 53 - 31 = (31 + 1) . 53 - 31 = 31 . 53 + 53 -31 = 53 . 31 +22
Vì 22 < 32 Nên : 53 . 31 + 22 < 53 . 21 +32
Hay : A < B
Vậy A < B .
Ta có: A = 32 . 53 - 31
A = ( 31 + 1 ) . 53 - 31
A = 31 . 53 + 53 - 31
A= 31 . 53 + 22
Vì 31 . 53 + 22 < 53 . 31 + 32 nên A < B
Vậy A < B
-31/42