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1
5 tháng 4 2016
a) -7/25 . 11/13 + -7/25 . 2/13 - 18/25=-1
b) 5/7 . 1/3 - 5/7 . 1/4 - 5/7 . 1/12=0
c) 5 + 2/5 . 4+ 2/7 + 5 + 5/7 . 5+ 2/5=18
d) 75% - 3/2 + 0,5 - [ -1/2]^2=-1/2
NH
1
ML
11 tháng 5 2016
\(\frac{5}{7}\).(\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{4}{7}\)) +(\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{4}{7}\)):\(\frac{7}{5}\)
= \(\frac{5}{7}\).(\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{4}{7}\))+(\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{4}{7}\)).\(\frac{5}{7}\)
=2.\(\frac{5}{7}\).(\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{4}{7}\))
=\(\frac{10}{7}\).(\(\frac{21}{42}\)-\(\frac{8}{42}\)+\(\frac{24}{42}\))
=\(\frac{10}{7}\).\(\frac{37}{42}\)=\(\frac{185}{147}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
27 tháng 3
\(\left(\dfrac{5}{7}-\dfrac{7}{7}\right)-\left[0,2-\left(-\dfrac{2}{7}-\dfrac{1}{10}\right)\right]\)
=\(-\dfrac{2}{7}-\left[\dfrac{1}{5}+\dfrac{2}{7}+\dfrac{1}{10}\right]\)
=\(-\dfrac{2}{7}-\dfrac{1}{5}-\dfrac{2}{7}-\dfrac{1}{10}\)
=\(\left(-\dfrac{2}{7}-\dfrac{2}{7}\right)-\left(\dfrac{1}{5}+\dfrac{1}{10}\right)\)
=\(-\dfrac{4}{7}-\left(\dfrac{2}{10}+\dfrac{1}{10}\right)\)
=\(-\dfrac{4}{7}-\dfrac{3}{10}\)
=\(-\dfrac{40}{70}-\dfrac{21}{70}\)
=\(-\dfrac{61}{70}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
27 tháng 3
(3 - \(\dfrac{1}{4}\) + \(\dfrac{2}{3}\)) - (5 - \(\dfrac{1}{3}\) - \(\dfrac{5}{6}\)) - (6 - \(\dfrac{7}{4}\) - \(\dfrac{3}{2}\))
= 3 - \(\dfrac{1}{4}\) + \(\dfrac{2}{3}\) - 5 + \(\dfrac{1}{3}\) + \(\dfrac{5}{6}\) - 6 + \(\dfrac{7}{4}\) + \(\dfrac{3}{2}\)
= (3 - 5 - 6) + ( \(\dfrac{7}{4}\) - \(\dfrac{1}{4}\)) + (\(\dfrac{2}{3}\) + \(\dfrac{1}{3}\)) + \(\dfrac{3}{2}\) + \(\dfrac{5}{6}\)
= - 8 + \(\dfrac{3}{2}\) + 1 + \(\dfrac{3}{2}\) + \(\dfrac{5}{6}\)
= (- 8 + 1) + (\(\dfrac{3}{2}\) + \(\dfrac{3}{2}\)) + \(\dfrac{5}{6}\)
= -7 + 3 + \(\dfrac{5}{6}\)
= - 4 + \(\dfrac{5}{6}\)
= \(\dfrac{-19}{6}\)
2 tháng 5 2019
a) A=\(\frac{178}{179}+\frac{179}{180}+\frac{183}{181}\)
ta có :
\(A=\left(1-\frac{1}{179}\right)+\left(1-\frac{1}{180}\right)+\left(1+\frac{2}{181}\right)\)
\(\Rightarrow A=\left(1+1+1\right)-\left(\frac{1}{179}-\frac{1}{180}+\frac{2}{181}\right)\)
\(\Rightarrow A=3-\left(\frac{1}{179}-\frac{1}{180}+\frac{2}{181}\right)< 3\)
Vậy \(A< 3\)
KN
2 tháng 5 2019
a. Ta có :
\(\frac{178}{179}< 1\left(\frac{1}{179}\right)\)
\(\frac{179}{180}< 1\left(\frac{1}{180}\right)\)
\(\frac{183}{181}>1\left(\frac{3}{181}\right)\left(1\right)\)
Mà \(\frac{3}{181}>\frac{1}{179}+\frac{1}{180}\left(=\frac{359}{32220}< \frac{3}{181}\right)\left(2\right)\)
Từ \(\left(1\right)\&\left(2\right)\Rightarrow\frac{178}{179}+\frac{179}{180}+\frac{183}{181}< 1+1+1\)
Vậy \(A< 3\)
28 tháng 7 2023
a: =7+5/11-2-3/7-3-5/11
=2-3/7=11/7
b: =-3/5(5/7+3/7+6/7)
=-3/5*2=-6/5
c: =1/3(4/5+6/5)-4/3
=2/3-4/3=-2/3
d: =5/9(7/13+13-3/13)
=5/9*165/13=275/39
HY
1
29 tháng 1
a: \(\dfrac{6}{7}:\left(\dfrac{2}{5}\cdot\dfrac{6}{7}\right)\)
\(=\dfrac{6}{7}:\dfrac{12}{35}\)
\(=\dfrac{6}{7}\cdot\dfrac{35}{12}=\dfrac{6}{12}\cdot\dfrac{35}{7}=\dfrac{5}{2}\)
b: \(\dfrac{6}{7}+\dfrac{5}{7}:5-\dfrac{8}{9}\)
\(=\dfrac{6}{7}+\dfrac{1}{7}-\dfrac{8}{9}\)
\(=1-\dfrac{8}{9}=\dfrac{1}{9}\)
c: \(\dfrac{6}{7}+\dfrac{5}{8}\cdot\dfrac{1}{5}-\dfrac{3}{16}\cdot4\)
\(=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}\)
\(=\dfrac{48+7-42}{56}=\dfrac{13}{56}\)
d: \(\dfrac{-1}{6}+\dfrac{2}{3}\cdot\dfrac{-3}{4}+\dfrac{4}{5}\)
\(=-\dfrac{1}{6}-\dfrac{1}{2}+\dfrac{4}{5}\)
\(=\dfrac{-5-15+24}{30}=\dfrac{4}{30}=\dfrac{2}{15}\)
11 tháng 8 2016
a)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{3}{4}+...+\frac{1}{9}-\frac{1}{10}\)
= \(1+\left(\frac{-1}{2}+\frac{1}{2}\right)+\left(\frac{-1}{3}+\frac{1}{3}\right)+...+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{10}\)
= \(1-\frac{1}{10}\)
=\(\frac{9}{10}\)
b)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
=\(1+\left(\frac{-1}{3}+\frac{1}{3}\right)+\left(\frac{-1}{5}+\frac{1}{5}\right)+\left(\frac{-1}{7}+\frac{1}{7}\right)+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{11}\)
=\(1-\frac{1}{11}\)
= \(\frac{10}{11}\)
c) đặt A=\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}\)
\(\frac{1}{3}A\)=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(\frac{2}{3}A\)=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
\(\frac{2}{3}A\)=\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
\(\frac{2}{3}A\)=\(1+\left(\frac{-1}{3}+\frac{1}{3}\right)+\left(\frac{-1}{5}+\frac{1}{5}\right)+\left(\frac{-1}{7}+\frac{1}{7}\right)+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{11}\)
\(\frac{2}{3}A\)=\(\frac{10}{11}\)
A= \(\frac{10}{11}:\frac{2}{3}\)
A= \(\frac{10}{11}.\frac{3}{2}\)=\(\frac{15}{11}\)
d) giả tương tự câu c kết quả \(\frac{25}{11}\)
11 tháng 8 2016
tổng đặc biệt đó bạn
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{9\times10}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(1-\frac{1}{10}=\frac{9}{10}\)
những câu sau cũng áp dụng như vậy nhé
(-5)7 . (1/5)7= -78125. 1/78125= -1
chúc bn hoctot!
\(\left(-5\right)^7.\left(\frac{1}{5}\right)^7=\left(-\frac{5.1}{5}\right)^7=\left(-\frac{1}{5}\right)^7\)\(^7\)