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= \(\frac{17}{8}:\frac{25}{14}-\left(15-\frac{40}{3}\right):\frac{25}{6}\)
= \(\frac{17}{8}.\frac{14}{25}-\left(\frac{45}{3}-\frac{40}{3}\right).\frac{6}{25}\)
= \(\frac{119}{100}-\frac{5}{3}.\frac{6}{25}\) = \(\frac{119}{100}-\frac{2}{5}\)
= \(\frac{119}{100}-\frac{40}{100}=\frac{79}{100}\)
Chúc bạn Hk tốt!!!!!
Bài 1 :
36/1212 = 3/101
13/1313 = 1/101
3/101 + 1/101 = 4/101
Vậy 36/1212 + 13/1313 = 4/101.
Bài 2 :
A = 5/13 + 1/2 + -5/9 + -3/6 + 4/-9
A = 5/13 + 1/2 + -5/9 + -1/2 + -4/9
A = (1/2 + -1/2) + (-5/9 + -4/9) + 5/13
A = 0 + (-1) + 5/13
A = (-1) + 5/13 = -13/13 + 5/13 = 8/13.
Chúc bạn học giỏi nhé.
quên tớ đang tính thì dí nhầm tớ làm lại :
= \(\dfrac{17}{8}\) : \(\dfrac{51}{40}\) - 15 + \(\dfrac{40}{3}\) : \(\dfrac{25}{6}\)
= \(\dfrac{17.5.8}{8.17.3}\) - 15 + \(\dfrac{5.8.3.2}{3.5.5}\)
= \(\dfrac{5}{3}\) - 15 + \(\dfrac{16}{5}\)
= \(\dfrac{25-225+48}{15}\)
= \(\dfrac{-152}{15}\)
Giải:
\(2\dfrac{1}{8}:1\dfrac{11}{40}-\left(15-13\dfrac{1}{3}\right):4\dfrac{1}{6}\)
\(=2\dfrac{1}{8}:1\dfrac{11}{40}-2\dfrac{1}{3}:4\dfrac{1}{6}\)
\(=\dfrac{17}{8}:\dfrac{51}{40}-\dfrac{7}{3}:\dfrac{25}{6}\)
\(=\dfrac{17}{8}.\dfrac{40}{51}-\dfrac{7}{3}.\dfrac{6}{25}\)
\(=\dfrac{17}{8}.\dfrac{8.5}{17.3}-\dfrac{7}{3}.\dfrac{3.2}{25}\)
\(=\dfrac{5}{3}-\dfrac{14}{25}\)
\(=\dfrac{83}{75}\)
Vậy ...
\(\dfrac{\left(12:13\right)+\left(12:131\right)-\left(12:1313\right)+\left(12:13131\right)}{\left(15:13\right)+\left(15:131\right)-\left(15:1313\right)+\left(15:13131\right)}\\ =\dfrac{12:\left(13+131-1313+13131\right)}{15:\left(13+131-1313+13131\right)}=\dfrac{12}{15}=\dfrac{4}{5}\)
29 . ( 85 -47) + 85. ( 47-29 )
=29.38+85.18
=1102+1530
=1632
2,7 . 10,5 - 7,3 . 10,5 - 7,3 . 15 + 2,7 . 15
=(2,7+7,3).10,5-(2,7+7,3).15
=10.10,5-10.15
=105-150
=-45
\(B=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+\frac{1}{11\cdot13}\)
\(=\frac{1}{2}\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+\frac{2}{11\cdot13}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{13}\right)=\frac{1}{2}\cdot\frac{10}{39}=\frac{5}{39}\)
\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{1.13}\)
\(=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{13}\right)=\frac{1}{2}.\frac{10}{39}=\frac{5}{39}\)
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