D=\(\dfrac{4}{8x13}\)+\(\dfrac{4}{13x18}\)+....+\(\dfrac{4}{253x258}\)
K
Khách
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D
1
ML
5 tháng 10 2017
\(B=\frac{4}{8.13}+\frac{4}{13.18}+\frac{4}{18.23}+...+\frac{4}{253.258}\)
\(B=\frac{4}{5}\left(\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+\frac{1}{18}-\frac{1}{23}+...+\frac{1}{253}-\frac{1}{258}\right)\)
\(B=\frac{4}{5}\left(\frac{1}{8}-\frac{1}{258}\right)\)
\(B=\frac{4}{5}.\frac{125}{1032}\)
\(B=\frac{25}{258}\)
C
15 tháng 5 2022
8/9 x 5/4 + 3/5 = 10/9 + 3/5 = 77/45
5/8 + 1/4 = 5/9 + 2/8 = 7/8
1/2 + 5/9 = 9/18 + 5/9 = 14/9
9/8 x 3/4 - 4/7 = 27/32 - 4/7 = 61/224
O
15 tháng 5 2022
a) \(\dfrac{8}{9}:\dfrac{4}{5}+\dfrac{3}{5}=\dfrac{8}{9}\times\dfrac{5}{4}+\dfrac{2}{5}=\dfrac{10}{9}+\dfrac{2}{5}=\dfrac{50}{45}+\dfrac{18}{45}=\dfrac{68}{45}\)
b) \(\dfrac{5}{8}+\dfrac{1}{4}=\dfrac{5}{8}+\dfrac{2}{8}=\dfrac{6}{8}=\dfrac{3}{4}\)
c) \(\dfrac{7}{8}\times\dfrac{4}{7}+\dfrac{5}{9}=\dfrac{1}{2}+\dfrac{5}{9}=\dfrac{9}{18}+\dfrac{10}{18}=\dfrac{19}{18}\)
d) \(\dfrac{9}{8}:\dfrac{4}{3}-\dfrac{4}{7}=\dfrac{9}{8}\times\dfrac{3}{4}-\dfrac{4}{7}=\dfrac{27}{32}-\dfrac{4}{7}=\dfrac{61}{244}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
5 tháng 8 2023
a, \(\dfrac{7}{8}\) \(\times\) \(\dfrac{3}{13}\) + \(\dfrac{4}{9}\) \(\times\) \(\dfrac{4}{13}\)
= \(\dfrac{1}{13}\) \(\times\)( \(\dfrac{21}{8}\) + \(\dfrac{16}{9}\))
= \(\dfrac{1}{13}\) \(\times\)( \(\dfrac{189}{72}\) + \(\dfrac{128}{72}\))
= \(\dfrac{1}{13}\) \(\times\) \(\dfrac{317}{73}\)
= \(\dfrac{317}{949}\)
b, \(\dfrac{6}{5}\) + \(\dfrac{7}{3}\) + \(\dfrac{8}{9}\)
= \(\dfrac{54}{45}\) + \(\dfrac{105}{45}\) + \(\dfrac{40}{45}\)
= \(\dfrac{199}{45}\)
c, 23 : \(\dfrac{5}{14}\) + \(\dfrac{6}{7}\) + \(\dfrac{4}{9}\)
= \(\dfrac{322}{5}\) + \(\dfrac{6}{7}\) + \(\dfrac{4}{9}\)
= \(\dfrac{20286}{315}\) + \(\dfrac{270}{315}\) + \(\dfrac{140}{315}\)
= \(\dfrac{20696}{315}\)
d, 4\(\dfrac{1}{4}\) + 7\(\dfrac{3}{7}\) - 2\(\dfrac{4}{17}\)
= 4 + \(\dfrac{1}{4}\) + 7 + \(\dfrac{3}{7}\) - 2 - \(\dfrac{4}{17}\)
= (4+7-2) + (\(\dfrac{1}{4}\) + \(\dfrac{3}{7}\) - \(\dfrac{4}{17}\))
= 9 + \(\dfrac{119}{476}\) + \(\dfrac{204}{476}\) - \(\dfrac{112}{476}\)
= 9\(\dfrac{211}{476}\) = \(\dfrac{4495}{476}\)
e, 8 - (9\(\dfrac{2}{11}\) + \(\dfrac{8}{33}\))
= 8 - 9 - \(\dfrac{2}{11}\) - \(\dfrac{8}{33}\)
= -1 - \(\dfrac{2}{11}\) - \(\dfrac{8}{33}\)
= \(\dfrac{-33}{33}\) - \(\dfrac{-6}{33}\) - \(\dfrac{8}{33}\)
= - \(\dfrac{47}{33}\)
14 tháng 12 2021
Đặt \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}=k\Rightarrow a=bk;b=ck;c=dk;d=ek\)
\(\Rightarrow a=bk=ck^2=dk^3=ek^4;b=ek^3\)
\(\Rightarrow\dfrac{a}{e}=\dfrac{ek^4}{e}=k^4\left(1\right)\)
Ta có \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}\Rightarrow\dfrac{a^4}{b^4}=\dfrac{b^4}{c^4}=\dfrac{c^4}{d^4}=\dfrac{d^4}{e^4}=\dfrac{2a^4+3b^4+4c^4+5d^4}{2b^4+3c^4+4d^4+5e^4}\left(2\right)\)
Lại có \(\dfrac{a^4}{b^4}=\left(\dfrac{a}{b}\right)^4=\left(\dfrac{ek^4}{ek^3}\right)^4=k^4\left(3\right)\)
\(\left(1\right)\left(2\right)\left(3\right)\RightarrowĐpcm\)
29 tháng 12 2021
a: \(=\dfrac{-8}{9}-\dfrac{6}{5}+\dfrac{8}{9}=-\dfrac{6}{5}\)
c: \(=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\)
25 tháng 1 2022
a: \(A=\dfrac{19}{9}+\dfrac{4}{11}+\dfrac{2}{3}=\dfrac{209}{99}+\dfrac{44}{99}+\dfrac{66}{99}=\dfrac{319}{99}\)
b: \(B=\dfrac{-50}{60}+\dfrac{-35}{60}+\dfrac{12}{60}=\dfrac{-73}{60}\)
c: \(C=\dfrac{-27}{36}+\dfrac{132}{36}+\dfrac{10}{36}=\dfrac{115}{36}\)
d: \(D=\dfrac{-19}{3}+\dfrac{2}{3}-\dfrac{4}{5}=\dfrac{-17}{3}-\dfrac{4}{5}=\dfrac{-85-12}{15}=-\dfrac{97}{15}\)
1 tháng 4 2022
a) \(\dfrac{3}{5}+\dfrac{11}{20}=\dfrac{12}{20}+\dfrac{11}{20}=\dfrac{23}{20}\)
b) \(\dfrac{5}{8}-\dfrac{4}{9}=\dfrac{45}{72}-\dfrac{32}{72}=\dfrac{13}{72}\)
c) \(\dfrac{9}{16}\times\dfrac{4}{3}=\dfrac{3}{4}\)
d) \(\dfrac{4}{7}:\dfrac{8}{11}=\dfrac{4}{7}\times\dfrac{11}{8}=\dfrac{11}{14}\)
e) \(\dfrac{3}{5}+\dfrac{4}{5}:\dfrac{2}{5}=\dfrac{3}{5}+\dfrac{4}{5}\times\dfrac{5}{2}=\dfrac{3}{5}+2=\dfrac{3}{5}+\dfrac{10}{5}=\dfrac{13}{5}\)
\(D=\dfrac{4}{8\cdot13}+\dfrac{4}{13\cdot18}+...+\dfrac{4}{253\cdot258}\)
\(=\dfrac{4}{5}\left(\dfrac{1}{8}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{18}+...+\dfrac{1}{253}-\dfrac{1}{258}\right)\)
\(=\dfrac{4}{5}\cdot\dfrac{125}{1032}=\dfrac{25}{258}\)