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a) \(\frac{5}{6}\)= \(\frac{15}{18}\); b) \(\frac{99}{100}\)< \(\frac{100}{99}\); c ) \(\frac{15}{17}\)> \(\frac{13}{18}\)vì \(\frac{15}{17}\)> \(\frac{15}{18}\)> \(\frac{13}{18}\);
d) \(\frac{222}{333}\)= \(\frac{2}{3}\)\(=1-\frac{1}{3}\); \(\frac{3333}{4444}\)= \(\frac{3}{4}\)= \(1-\frac{1}{4}\); vì \(\frac{1}{3}\)> \(\frac{1}{4}\)nên \(\frac{222}{333}\)< \(\frac{3333}{4444}\)
e) \(\frac{292929}{272727}\)= \(\frac{29}{27}\)= \(1+\frac{2}{17}\); \(\frac{347347}{345345}\)= \(\frac{347}{345}\)= \(1+\frac{2}{345}\)nên \(\frac{292929}{272727}\)> \(\frac{347347}{345345}\)
\(\left(a\right)\frac{9}{8}+\frac{15}{32}=\frac{36}{32}+\frac{15}{32}=\frac{36+15}{32}=\frac{51}{32}\)
\(\left(b\right)4+\frac{35}{45}=\frac{4}{1}+\frac{35}{45}=\frac{180}{45}+\frac{35}{45}=\frac{180+35}{45}=\frac{43}{9}\)
\(\left(c\right)\frac{11}{4}-\frac{15}{16}=\frac{44}{16}-\frac{15}{16}=\frac{44-15}{16}=\frac{29}{16}\)
\(\left(d\right)3-\frac{13}{9}=\frac{3}{1}-\frac{13}{9}=\frac{27}{9}-\frac{13}{9}=\frac{27-13}{9}=\frac{14}{9}\)
\(\left(e\right)\frac{5}{6}-\frac{5}{8}=\frac{20}{24}-\frac{15}{24}=\frac{20-15}{24}=\frac{5}{24}\)
\(\left(g\right)\frac{196}{64}-2=\frac{196}{64}-\frac{2}{1}=\frac{196}{64}-\frac{128}{64}=\frac{196-128}{64}=\frac{17}{16}\)
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a) \(\frac{9}{8}+\frac{15}{32}=\frac{36}{32}+\frac{15}{32}=\frac{51}{32}\)
b) \(4+\frac{35}{45}=4+\frac{7}{9}=\frac{36}{9}+\frac{7}{9}=\frac{43}{9}\)
c)\(\frac{11}{4}+\frac{15}{16}=\frac{44}{16}+\frac{15}{16}=\frac{59}{16}\)
d )\(3-\frac{13}{9}=\frac{27}{9}-\frac{13}{9}=\frac{14}{9}\)
e ) \(\frac{5}{6}+\frac{5}{8}=\frac{40}{48}+\frac{30}{48}\)\(=\frac{70}{48}\)
A)\(-\frac{3}{29}+\frac{16}{58}=-\frac{3}{29}+\frac{8}{29}=\frac{5}{29}\)
B)\(-\frac{1}{5}+\frac{5}{18}+-\frac{7}{9}=-\frac{1}{5}+\frac{5}{18}+-\frac{14}{18}=-\frac{1}{5}+-\frac{1}{2}\)\(=-\frac{2}{10}+-\frac{5}{10}=-\frac{7}{10}\)
C)\(-\frac{8}{18}+-\frac{15}{27}=-\frac{4}{9}+-\frac{5}{9}=-1\)
Đăt S=1/15+1/35+1/63+1/99+...+1/2915+1/3135
=1/3.5+1/5.7+1/7.9+1/9.11+...+1/53.55+1/55.57
=1/2(2/3.5+2/5.7+2/7.9+...+2/53.55+2/55.57)
=1/2(1/3-1/5+1/5-1/7+1/7-1/9+...+1/53-1/55+1/55-1/57)
=1/2(1/3-1/57)
=1/2(19/57-1/57)
=1/2.18/57
=3/19
Vậy 1/15+1/35+1/63+1/99+...+1/2915+1/3135=3/19
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Đặt \(A=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{2915}+\frac{1}{3135}\)
\(\Leftrightarrow A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+....+\frac{1}{53\cdot55}+\frac{1}{55\cdot57}\)
\(\Leftrightarrow2A=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+...+\frac{2}{53\cdot55}+\frac{2}{55\cdot57}\)
\(\Leftrightarrow2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-....+\frac{1}{53}-\frac{1}{55}+\frac{1}{55}-\frac{1}{57}\)
\(\Leftrightarrow2A=\frac{1}{3}-\frac{1}{57}=\frac{6}{19}\)
\(\Leftrightarrow A=\frac{6}{19}:2=\frac{3}{19}\)
\(\frac{4}{27}\cdot\frac{9}{2}+\frac{7}{15}=\frac{2}{3}+\frac{7}{15}=\frac{17}{15}\)
( \(\cdot\)là dấu nhân đó)
a) 67 + 135 + 33
= ( 67 + 33 ) + 135
= 100 + 135
= 235
b) 277 + 113 + 323 + 87
= ( 277 + 323 ) + ( 113 + 87 )
= 600 + 200
= 800