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\(\frac{5}{7}\)<\(\frac{15}{12}\)
\(\frac{17}{21}\)<\(\frac{17}{19}\)
a) \(\frac{13}{19}+\frac{18}{19}+\frac{19}{19}=\left(\frac{13}{19}+\frac{17}{19}\right)+\frac{18}{19}=\frac{20}{19}+\frac{18}{19}=\frac{48}{19}\)
b) \(\frac{3}{5}+\frac{3}{16}+\frac{13}{16}=\left(\frac{3}{16}+\frac{13}{16}\right)+\frac{3}{5}=1+\frac{3}{5}=\frac{5}{5}+\frac{3}{5}=\frac{8}{5}\)
a) \(\frac{13}{19}+\frac{18}{19}+\frac{17}{19}\)
= \(\frac{31}{19}+\frac{17}{19}\)
= \(\frac{48}{19}.\)
b) \(\frac{3}{5}+\frac{3}{16}+\frac{13}{16}\)
= \(\frac{3}{5}+\left(\frac{3}{16}+\frac{13}{16}\right)\)
= \(\frac{3}{5}+\frac{16}{16}\)
= \(\frac{3}{5}+1\)
= \(\frac{3}{5}.\)
1.
a) \(\frac{16}{24}-\frac{1}{3}=\frac{16}{24}-\frac{8}{24}=\)\(\frac{8}{24}=\frac{1}{3}\)
b) \(\frac{4}{5}-\frac{12}{60}=\frac{48}{60}-\frac{12}{60}=\frac{36}{60}=\frac{9}{15}\)
3.
a)\(\frac{17}{6}-\frac{2}{6}=\frac{17-2}{6}=\frac{15}{6}\)
b) \(\frac{16}{15}-\frac{11}{15}=\frac{16-11}{15}=\frac{5}{15}=\frac{1}{3}\)
c) \(\frac{19}{12}-\frac{13}{12}=\frac{19-13}{12}=\frac{6}{12}=\frac{1}{2}\)
a) 16 24 − 1 3 = 16 24 − 8 24 = 24 16 − 3 1 = 24 16 − 24 8 = 8 24 = 1 3 24 8 = 3 1 b) 4 5 − 12 60 = 48 60 − 12 60 = 36 60 = 9 15 5 4 − 60 12 = 60 48 − 60 12 = 60 36 = 15 9 3. a) 17 6 − 2 6 = 17 − 2 6 = 15 6 6 17 − 6 2 = 6 17−2 = 6 15 b) 16 15 − 11 15 = 16 − 11 15 = 5 15 = 1 3 15 16 − 15 11 = 15 16−11 = 15 5 = 3 1 c) 19 12 − 13 12 = 19 − 13 12 = 6 12 = 1 2 12 19 − 12 13 = 12 19−13 = 12 6 = 2 1
Ta có:
\(A=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{3999.4000}}\)
\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{3999}-\frac{1}{4000}}\)
\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\left(1+\frac{1}{3}+...+\frac{1}{3999}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{4000}\right)}\)
\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{3999}+\frac{1}{4000}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{4000}\right)}\)
\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{3999}+\frac{1}{4000}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2000}\right)}\)
\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}=1\)
Ta lại có:
\(B=\frac{\left(17+1\right)\left(\frac{17}{2}+1\right)...\left(\frac{17}{19}+1\right)}{\left(1+\frac{19}{17}\right)\left(1+\frac{19}{16}\right)...\left(1+19\right)}\)
\(=\frac{\frac{18}{1}.\frac{19}{2}.\frac{20}{3}...\frac{36}{19}}{\frac{36}{17}.\frac{35}{16}.\frac{34}{15}...\frac{20}{1}}\)
\(=\frac{1.2.3...36}{1.2.3...36}=1\)
Từ đây ta suy ra được
\(A-B=1-1=0\)
a,\(\frac{26}{7}\)
b,\(\frac{78}{19}\)
c,\(\frac{13}{11}\)
d,\(1\)
a, \(4-\frac{2}{7}\)=\(\frac{26}{7}\)
b,5-\(\frac{17}{19}=\frac{78}{19}\)
c,\(\frac{35}{11}-2=\frac{13}{11}\)
d,\(\frac{68}{17}-3=1\)
k mình nha
a) \(\frac{55}{32}\)
b) \(\frac{90}{48}\)
c) \(\frac{76}{15}\)
d) \(\frac{552}{119}\)
\(\frac{9}{16}+\frac{37}{32}=\frac{18}{32}+\frac{37}{32}=\frac{55}{32}\)
\(\frac{126}{48}-\frac{12}{16}=\frac{21}{8}-\frac{6}{8}=\frac{15}{8}\)
\(\frac{19}{15}\times\frac{32}{8}=\frac{19}{15}\times4=\frac{76}{15}\)
\(\frac{23}{17}:\frac{7}{24}=\frac{23}{17}.\frac{24}{7}=4\frac{76}{119}\)
a/ \(\left[\frac{4}{6}+\frac{14}{6}\right]+\left[\frac{7}{13}+\frac{19}{13}\right]+\left[\frac{17}{9}+\frac{1}{9}\right]=\frac{18}{6}+\frac{26}{13}+\frac{18}{9}=3+2+2=7\)
\(\frac{10}{17}< \frac{19}{ }< \frac{10}{16}\)
\(\frac{190}{323}< \frac{190}{...}< \frac{190}{304}\)
... phải chia hết cho 10 => ... = 310 hoặc 320
\(\Rightarrow\frac{10}{17}< \frac{19}{31}< \frac{10}{16}\)và \(\frac{10}{17}< \frac{19}{32}< \frac{10}{16}\)
12/17+19/17=31/17
31/17 bạn nhé