B
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B
0
14 tháng 2 2020
\(C=...\)
\(=\frac{1}{3^2}+\frac{1}{4^2}-\frac{1}{4^2}+\frac{1}{5^2}-\frac{1}{5^2}+...-\frac{1}{14^2}+\frac{1}{14^2}-\frac{1}{15^2}\)
\(=\frac{1}{3^2}-\frac{1}{15^2}\)
\(=\frac{1}{9}-\frac{1}{225}\)
Do \(\frac{1}{9}-\frac{1}{225}\)<\(\frac{1}{5}\)
\(=>C< \frac{1}{5}\)( ĐPCM )
14 tháng 2 2020
C = ...
=> C = \(\frac{1}{3^2}-\frac{1}{4^2}+\frac{1}{4^2}-\frac{1}{5^2}+...+\frac{1}{14^2}-\frac{1}{15^2}\)
C = \(\frac{1}{3^2}-\frac{1}{15^2}\)
Ta thấy : \(\frac{1}{3^2}< \frac{1}{5}\Leftrightarrow\frac{1}{3^2}-\frac{1}{15^2}< \frac{1}{5}\)
=> C < \(\frac{1}{5}\)
VM
23 tháng 7 2019
a) \(\frac{2}{3}-4.\left(\frac{1}{2}+\frac{3}{4}\right)\)
= \(\frac{2}{3}-4.\frac{5}{4}\)
= \(\frac{2}{3}-5\)
= \(-\frac{13}{3}\).
b) Câu này là 117 hay \(\frac{11}{7}\) vậy bạn?
Chúc bạn học tốt!
25 tháng 7 2017
1.
a) \(\frac{-7}{9}.2\frac{3}{4}=\frac{-7}{9}.\frac{11}{4}=\frac{-77}{36}\)
b) \(\frac{2}{3}+\frac{1}{3}.\frac{-2}{5}=\frac{2}{3}+\frac{-2}{15}=\frac{8}{15}\)
c) \(\frac{3}{4}.15\frac{1}{3}-\frac{3}{4}.43\frac{1}{3}=\frac{3}{4}.\frac{46}{3}-\frac{3}{4}.\frac{130}{3}=\frac{23}{2}-\frac{65}{2}=-21\)
d) \(\left(-49,1\right).\frac{13}{27}-58,9.\frac{13}{27}=\frac{13}{27}.\left(-49,1-58,9\right)=\frac{13}{27}.\left(-108\right)=-52\)
e) \(0,375:\left(-4,5\right)=\frac{-1}{12}\)
f) \(3\frac{1}{7}:\left(-1\frac{3}{7}\right)=\frac{22}{7}:\frac{-10}{7}=\frac{-11}{5}\)
g) \(9\frac{1}{3}:4\frac{2}{3}-2=\frac{28}{3}:\frac{14}{3}-2=2-2=0\)
h) \(\left(7\frac{3}{4}:0,3125+4,5.2\frac{2}{45}\right):\left(-8,5\right)=\left(\frac{31}{4}:\frac{5}{16}+\frac{9}{2}.\frac{92}{45}\right):\frac{-17}{2}=\left(\frac{124}{5}+\frac{46}{5}\right):\frac{-17}{2}=34:\frac{-17}{2}=-4\)
25 tháng 7 2017
Bài 1 : Tính:
a)
\(\frac{-7}{9}.2\frac{3}{4}=\frac{-7}{9}.\frac{11}{4}=\frac{-77}{36}\)
b)
\(\frac{2}{3}+\frac{1}{3}.\frac{-2}{5}=\frac{2}{3}+\frac{-2}{15}=\frac{10}{15}+\frac{-2}{15}=\frac{8}{15}\)
c)
\(\frac{3}{4}.15\frac{1}{3}-\frac{3}{4}.43\frac{1}{3}=\frac{3}{4}.\frac{46}{3}-\frac{3}{4}.\frac{130}{3}\)\(=\frac{23}{2}-\frac{65}{2}=\frac{-42}{2}=-21\)
....
Tự lm tiếp dạng như v
Bài 2 :
\(A=\frac{-6}{11}.\frac{7}{10}.\frac{11}{-6}.-20=\left(\frac{-6}{11}.\frac{11}{-6}\right).\left(\frac{7}{10}.-20\right)\)\(=1.\left(-14\right)=-14\)
.....
Bài 3 :
\(\frac{3}{7}.x-\frac{2}{5}.x=\frac{-17}{35}\)
\(\Leftrightarrow\frac{3}{7}-\frac{2}{5}.x=\frac{-17}{35}\)
\(\Leftrightarrow\frac{1}{35}x=\frac{-17}{35}\)
\(\Leftrightarrow x=\frac{-17}{35}:\frac{1}{35}\)
\(\Leftrightarrow x=\frac{-17}{35}.35=-17\)
25 tháng 7 2019
Bài 2:
a) \(\frac{2}{3}\) - 4 .(\(\frac{1}{2}\) + \(\frac{3}{4}\)) = \(\frac{2}{3}\) - 4 . \(\frac{5}{4}\)
= \(\frac{2}{3}\) - 5 = \(\frac{-13}{3}\)
b) (\(\frac{-1}{3}\) + \(\frac{5}{6}\)) . 11 - 7 = \(\frac{1}{2}\). 11 - 7
= \(\frac{11}{2}\)- 7 = \(\frac{-3}{2}\)
c) \(\frac{-5}{9}\). \(\frac{3}{11}\)+ (\(\frac{-13}{18}\)) . \(\frac{3}{11}\)= \(\frac{3}{11}\). (\(\frac{-5}{9}\)- \(\frac{13}{18}\))
= \(\frac{3}{11}\). (\(\frac{-23}{18}\))= \(\frac{-23}{66}\)
d) \(\frac{-2}{3}\) . \(\frac{3}{11}\)+ (\(\frac{-16}{9}\)) . \(\frac{3}{11}\)= \(\frac{3}{11}\). (\(\frac{-2}{3}\)- \(\frac{16}{9}\))
= \(\frac{3}{11}\). (\(\frac{-22}{9}\)) = \(\frac{-2}{3}\)
Bài 1:
a) =\(\frac{-1}{2}\)+ \(\frac{3}{5}\)- \(\frac{1}{9}\)+ \(\frac{1}{71}\) + \(\frac{2}{7}\) + \(\frac{4}{35}\)- \(\frac{7}{18}\)
= (\(\frac{-1}{2}\)- \(\frac{1}{9}\)- \(\frac{7}{18}\)) + ( \(\frac{3}{5}\)+ \(\frac{2}{7}\)+ \(\frac{4}{35}\)) + \(\frac{1}{71}\)
= -1 + 1 + \(\frac{1}{71}\)= \(\frac{1}{71}\)
b) = 3 - \(\frac{1}{4}\)+ \(\frac{2}{3}\)- 5 + \(\frac{1}{3}\)+ \(\frac{6}{5}\)- 6 + \(\frac{7}{4}\)- \(\frac{3}{2}\)
= (3 - 5 - 6) + (\(\frac{-1}{4}\)+ \(\frac{7}{4}\)- \(\frac{3}{2}\)) + (\(\frac{2}{3}\)+ \(\frac{1}{3}\)) + \(\frac{6}{5}\)
= -8 + 0 + 1 + \(\frac{6}{5}\)
= \(\frac{-29}{5}\)