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NT
2
S
21 tháng 6 2016
a) \(\left(-\frac{2}{3}+\frac{3}{7}\right):\frac{4}{5}+\left(-\frac{1}{3}+\frac{4}{7}\right):\frac{4}{5}\)
\(=\left[\left(-\frac{2}{3}+\frac{3}{7}\right)+\left(-\frac{1}{3}+\frac{4}{7}\right)\right]:\frac{4}{5}\)
\(=\left[\left(-\frac{2}{3}-\frac{1}{3}\right)+\left(\frac{3}{7}+\frac{4}{7}\right)\right]:\frac{4}{5}\)
\(=0:\frac{4}{5}\)
\(=0\)
b) \(\frac{5}{9}\left(\frac{1}{11}-\frac{5}{22}\right)+\frac{5}{9}\left(\frac{1}{15}-\frac{2}{3}\right)\)
\(=\frac{5}{9}\left[\left(\frac{2}{22}-\frac{5}{22}\right)+\left(\frac{1}{15}-\frac{10}{15}\right)\right]\)
\(=\frac{5}{9}\left[-\frac{3}{22}+\left(-\frac{3}{5}\right)\right]\)
\(=\frac{5}{9}\left(-\frac{81}{110}\right)\)
\(=-\frac{9}{22}\)
T
23 tháng 7 2018
\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
\(B=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{92.95}\)
\(=\frac{5-2}{2.5}+\frac{8-5}{5.8}+\frac{11-8}{8.11}+...+\frac{92-92}{92.95}\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{92}-\frac{1}{95}\)
\(=\frac{1}{2}-\frac{1}{95}=\frac{93}{190}\)
\(C=\frac{5}{6}+\frac{5}{66}+\frac{5}{176}+\frac{5}{336}\)
\(=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+\frac{5}{16.21}\)
\(=\frac{6-1}{1.6}+\frac{11-6}{6.11}+\frac{16-11}{11.16}+\frac{21-16}{16.21}\)
\(=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}\)
\(=1-\frac{1}{21}=\frac{20}{21}\)
[ HỌC TỐT]
23 tháng 7 2018
\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{2}-\frac{1}{100}\)
\(A=\frac{100}{200}-\frac{2}{200}\)
\(A=\frac{98}{200}=\frac{49}{100}\)
26 tháng 7 2018
a) \(A=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{31.34}\)
\(A=\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{31}-\frac{1}{34}\right)\)
\(A=\frac{2}{3}.\left(1-\frac{1}{34}\right)\)
\(A=\frac{2}{3}\cdot\frac{33}{34}=\frac{11}{17}\)
b) \(B=\frac{3}{1}+\frac{3}{3}+\frac{3}{6}+...+\frac{3}{210}\)
\(B=\frac{6}{2}+\frac{6}{6}+\frac{6}{12}+...+\frac{6}{420}\) ( 3/1 = 6/2; 6/6=3/3;..)
\(B=\frac{6}{1.2}+\frac{6}{2.3}+\frac{6}{3.4}+...+\frac{6}{20.21}\)
\(B=6.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{20}-\frac{1}{21}\right)\)
\(B=6.\left(1-\frac{1}{21}\right)=6\cdot\frac{20}{21}=\frac{40}{7}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
9 tháng 10 2024
a; - \(\dfrac{2}{3}\) + \(\dfrac{3}{4}\) - (- \(\dfrac{1}{6}\)) + (- \(\dfrac{2}{5}\))
= - \(\dfrac{2}{3}\) + \(\dfrac{3}{4}\) + \(\dfrac{1}{6}\) - \(\dfrac{2}{5}\)
= \(-\dfrac{40}{60}\) + \(\dfrac{45}{60}\) + \(\dfrac{10}{60}\) - \(\dfrac{24}{60}\)
= \(\dfrac{5}{60}\) + \(\dfrac{10}{60}\) - \(\dfrac{24}{60}\)
= \(\dfrac{15}{60}\) - \(\dfrac{24}{60}\)
= - \(\dfrac{3}{20}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
9 tháng 10 2024
b; (- \(\dfrac{2}{3}\)) + (- \(\dfrac{1}{5}\)) + \(\dfrac{3}{4}\) - \(\dfrac{5}{6}\) - \(\dfrac{-7}{10}\)
= - \(\dfrac{2}{3}\) - \(\dfrac{1}{5}\) + \(\dfrac{3}{4}\) - \(\dfrac{5}{6}\) + \(\dfrac{7}{10}\)
= - \(\dfrac{40}{60}\) - \(\dfrac{12}{60}\) + \(\dfrac{45}{60}\) - \(\dfrac{50}{60}\) + \(\dfrac{42}{60}\)
= - \(\dfrac{52}{60}\) + \(\dfrac{45}{60}\) - \(\dfrac{50}{60}\) + \(\dfrac{42}{60}\)
= - \(\dfrac{7}{60}\) - \(\dfrac{50}{60}\) + \(\dfrac{42}{60}\)
= - \(\dfrac{57}{60}\) + \(\dfrac{42}{60}\)
= - \(\dfrac{1}{4}\)
YN
3
Q
3 tháng 7 2016
a, [ (- 1) . 3 ] + (- 4) - (7.0) + 1
= [- 3 + ( - 4 )] - 0 + 1
= - 7 - 0 + 1
= - 6
b, (-1).(-2)+(-3).(-4)-(-2).(-3)
= [(-1).(-2)] + [(-3).(-4)] - [(-2).(-3)]
= 2 + 12 - 6
= 8
c, (-1).(-2).(-3).(-4).(-5):[(-3)-(-5)]
= (-1).(-2).(-3).(-4).(-5) : 2
= - 120 : 2
= - 60
Chắc chắn đấy!
3 tháng 7 2016
a, (-1).3+(-4)-7.0+1=-3-4-0+1=-6
b, (-1).(-2)+(-3).(-4)-(-2).(-3)=2+12-6=8
c, (-1).(-2).(-3).(-4).(-5):[(-3)-(-5)]=(-120):2=-60