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Vì \(\frac{3}{10}>\frac{3}{11}>\frac{3}{12}>\frac{3}{13}>\frac{3}{14}\)
Ta có: \(\frac{3}{14}\) x 5 < \(\frac{3}{10}+\frac{3}{11}<\frac{3}{12}<\frac{3}{13}<\frac{3}{14}\) < \(\frac{3}{10}\) x 5
\(\frac{15}{14}\) < S < \(\frac{15}{10}\)
1\(\frac{1}{14}\) < S < 1\(\frac{5}{10}\)
1 < S < 2 (đpcm)

ΔMAB đều \(\Rightarrow \hat{A M B} = 6 0^{0}\)
Theo tính chất 2 tiếp tuyến, ta có MO là phân giác \(\hat{A M B}\)
\(\Rightarrow \hat{A M O} = \frac{1}{2} \hat{A M B} = 3 0^{0}\)
Trong tam giác vuông OAM:
\(t a n \hat{A M O} = \frac{O A}{A M} \Rightarrow O A = A M . t a n \hat{A M O} = 15 \sqrt{3} . t a n 3 0^{0} = 15 \left(\right. c m \left.\right)\)
\(\Rightarrow 2 R = 2 O A = 30 \left(\right. c m \left.\right)\)

B =\(\frac{1}{1.5}\) + \(\frac{1}{5.9}\) + ...+ \(\frac{1}{\left(4n-3\right).\left(4n+1\right)}\)
B = \(\frac14\).(\(\frac{4}{1.5}+\frac{4}{5.9}+\cdots+\frac{4}{\left(4n-3\right).\left(4n+1\right)}\)
B = \(\frac14\).(\(\frac11\) - \(\frac15\) + \(\frac15\) - \(\frac19\) + ... + \(\frac{1}{4n-3}-\frac{1}{4n+1}\))
B = \(\frac14\).(\(\frac11\) - \(\frac{1}{4n+1}\))
B = \(\frac14\).\(\frac{4n}{4n+1}\)
B = \(\frac{n}{4n+1}\)

Bài 1';
a; \(\frac13\) + \(\frac54\) = \(\frac{4}{12}+\frac{15}{12}\) = \(\frac{19}{12}\)
b; 1,6.2\(\frac14\) - 1,6.3\(\frac12\)
= 1,6.(2\(\frac14\) - 3\(\frac12\))
= 1,6.(\(\frac94\) - \(\frac72\))
= 1,6.(\(\frac94\) - \(\frac{14}{4}\))
= 1,6.(- \(\frac54\))
= -2
c; \(\frac{11}{4}\) .(- \(\frac{4}{11}\)) - \(\frac54\) : \(\frac{11}{4}\)
= -1 - \(\frac54\) x \(\frac{4}{11}\)
= - 1 - \(\frac{5}{11}\)
= - \(\frac{16}{11}\)


\(-\dfrac{15x^2y^3}{18x^3y^5}=-\dfrac{15}{18}\cdot\dfrac{x^2}{x^3}\cdot\dfrac{y^3}{y^5}=\dfrac{-5}{6\cdot x\cdot y^2}\)
Bài 3:
a; \(\frac16\) + \(\frac{-5}{6}\)
= \(-\frac46\)
= \(-\frac23\)
b; \(\frac{-2}{17}\) + \(\frac{3}{19}\) + \(\frac{-15}{17}\) + \(\frac{16}{19}\) + \(\frac56\)
= (\(\frac{-2}{17}\) + \(\frac{-15}{17}\)) + (\(\frac{3}{19}\) + \(\frac{16}{19}\)) + \(\frac56\)
= - 1 + 1 + \(\frac56\)
= 0 + \(\frac56\)
= \(\frac56\)
Bài `3`
a, 1/6 + -5/6 = 1/6 - 5/6 = -4/6 = -2/3
b,
-2/17 + 3/19 + -15/17 + 16/19 + 5/6
= (-2/17 + -15/17) + (3/19 + 16/19) + 5/6
= (-1) + 1 + 5/6
= 0 + 5/6
= 5/6