Bài 4:

a; \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) = \(\dfrac{5}{20}\) - \(\dfrac{4}{20}\) = \(\dfrac{1}{20}\)

b; \(\dfrac{3}{5}\) - \(\dfrac{-1}{2}\) = \(\dfrac{6}{10}\) + \(\dfrac{5}{10}\) = \(\dfrac{11}{10}\)

c; \(\dfrac{3}{5}\) - \(\dfrac{-1}{3}\) = \(\dfrac{9}{15}\) + \(\dfrac{5}{15}\) = \(\dfrac{14}{15}\)

d; \(\dfrac{-5}{7}\) - \(\dfrac{1}{3}\)= \(\dfrac{-15}{21}\) - \(\dfrac{7}{21}\)= \(\dfrac{-22}{21}\)

Bài 5

a; 1 + \(\dfrac{3}{4}\) = \(\dfrac{4}{4}\) + \(\dfrac{3}{4}\) = \(\dfrac{7}{4}\)       b; 1 - \(\dfrac{1}{2}\) = \(\dfrac{2}{2}\) - \(\dfrac{1}{2}\) = \(\dfrac{1}{2}\)

c; \(\dfrac{1}{5}\) - 2 = \(\dfrac{1}{5}\) - \(\dfrac{10}{5}\) = \(\dfrac{-9}{5}\)     d; -5 - \(\dfrac{1}{6}\) = \(\dfrac{-30}{6}\) - \(\dfrac{1}{6}\) = \(\dfrac{-31}{6}\)

e; - 3 - \(\dfrac{2}{7}\)= \(\dfrac{-21}{7}\) - \(\dfrac{2}{7}\)= \(\dfrac{-23}{7}\)     f; - 3 + \(\dfrac{2}{5}\) = \(\dfrac{-15}{5}\) + \(\dfrac{2}{5}\)= - \(\dfrac{13}{5}\)

g; - 3 - \(\dfrac{2}{3}\) = \(\dfrac{-9}{3}\) - \(\dfrac{2}{3}\) = \(\dfrac{-11}{3}\)     h; - 4 - \(\dfrac{-5}{7}\) = \(\dfrac{-28}{7}\)+ \(\dfrac{5}{7}\) = - \(\dfrac{23}{7}\)

 Do C là trung điểm của OB

⇒ OC = OB : 2 = 6 : 2 = 3 (cm)

⇒ OC > OA

⇒ O không là trung điểm của AC

Lời giải:

\(E=\frac{\frac{2013}{1}.\frac{2014}{2}.\frac{2015}{3}....\frac{3012}{1000}}{\frac{1001}{1}.\frac{1002}{2}.\frac{1003}{3}....\frac{3012}{2012}}\\ =\frac{2013.2014.2015....3012}{1001.1002.1003....3012}.\frac{1.2.3...2012}{1.2.3..1000}\\ =\frac{1}{1001.1002...2012}.(1001.1002....2012)=1\)

giúp với ahhh

Có quá nhiều bài, thứ nhất em đăng tách ra, thứ hai chụp gần cận cho rõ, thứ ba em chỉ đăng bài cần giúp

Bài 3:

a; \(\frac56-\frac89\) = \(\frac{15}{18}-\) \(\frac{16}{18}\) = - \(\frac{1}{18}\)

d; \(\frac{5}{16}-\frac{5}{24}\)

= \(\frac{15}{48}\) - \(\frac{10}{48}\)

= \(\frac{5}{48}\)

e; - \(\frac{7}{30}\) + \(\frac{8}{45}\)

= \(-\frac{21}{90}\) + \(\frac{16}{90}\)

= - \(\frac{5}{90}\)

= - \(\frac{1}{18}\)

f; \(\frac{7}{12}-\frac{-9}{20}\)

= \(\frac{35}{60}\) + \(\frac{27}{60}\)

= \(\frac{62}{60}\)

= 31/30

h; - \(\frac48\) + \(-\frac{3}{10}\)

= -\(\frac{20}{40}-\frac{12}{40}\)

= - \(\frac{32}{40}\)

= - \(\frac45\)

\(a,MSC:180\\ Có:-5=\dfrac{-5.180}{180}=\dfrac{-900}{180};\dfrac{17}{-20}=\dfrac{17.\left(-9\right)}{\left(-9\right).\left(-20\right)}=\dfrac{-153}{180};\dfrac{-16}{9}=\dfrac{-16.20}{9.20}=\dfrac{-320}{180}\\ ---\\ b.MSC:75\\ Có:\dfrac{13}{-15}=\dfrac{13.\left(-5\right)}{\left(-15\right).\left(-5\right)}=\dfrac{-65}{75};\dfrac{-18}{25}=\dfrac{-18.3}{25.3}=\dfrac{-54}{75};-3=\dfrac{-3.75}{75}=\dfrac{-225}{75}\)

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