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a) \(74\frac{19}{35}.\frac{7}{90}+15\frac{16}{35}.\frac{7}{90}+2\frac{14}{90}\)
= \(\left(74\frac{19}{35}+15\frac{16}{35}\right).\frac{7}{90}+2\frac{14}{90}\)
= 90 . 7/90 + 194/90
= 630/90 + 194/90
= 824/90 = 412/45
b) (-2/5 + 3/7) - (4/9 + 12/20 - 13/35) + 7/35
= -2/5 + 3/7 - 4/9 - 3/5 + 13/35 + 7/35
= (-2/5 - 3/5) + 3/7 - 4/9 + (13/35 + 7/35)
= -1 + 3/7 - 4/9 + 4/7
= -1 + (3/7 + 4/7) - 4/9
= -1 + 1 - 4/9 = -4/9
a, 3 \(\frac{14}{19}\)+ \(\frac{13}{17}\)+ \(\frac{35}{43}\)+ 6\(\frac{5}{19}\)+ \(\frac{8}{43}\)= \(\left(3\frac{14}{19}+6\frac{5}{19}\right)+\left(\frac{35}{43}+\frac{8}{43}\right)+\frac{13}{17}=\)\(9+1+\frac{13}{17}=8+\frac{13}{17}=8\frac{13}{17}\)
b, \(\frac{-5}{7}.\frac{2}{11}+\frac{-5}{7}.\frac{9}{11}+1\frac{5}{7}\)\(=\frac{-5}{7}\left(\frac{2}{11}+\frac{9}{11}\right)+1\frac{5}{7}\)\(=\frac{-5}{7}.1+1\frac{5}{7}\)\(=\frac{-5}{7}+\frac{12}{7}=\frac{7}{7}=1\)
Chúc bn học tốt
\(3\frac{14}{19}+\frac{13}{17}+\frac{35}{43}+6\frac{5}{19}+\frac{8}{43}\)
\(=\left(3\frac{14}{19}+6\frac{5}{19}\right)+\left(\frac{35}{43}+\frac{8}{43}\right)+\frac{13}{17}\)
\(=10+1+\frac{13}{17}=11+\frac{13}{17}=11\frac{13}{17}\)
a) \(\frac{4}{11}-\frac{7}{15}+\frac{7}{11}-\frac{5}{15}\)
\(=\left(\frac{4}{11}+\frac{7}{11}\right)-\left(\frac{7}{15}+\frac{5}{15}\right)\)
\(=1-\frac{4}{5}\)
\(=\frac{1}{5}\)
b) \(\frac{7}{3}-\frac{4}{9}-\frac{1}{3}-\frac{5}{9}\)
\(=\left(\frac{7}{3}-\frac{1}{3}\right)-\left(\frac{4}{9}+\frac{5}{9}\right)\)
\(=2-1\)
\(=1\)
c) \(\frac{1}{4}+\frac{7}{33}-\frac{5}{3}\)
\(=\frac{-1}{4}+\frac{-16}{11}\)
\(=\frac{-75}{44}\)
d) \(\frac{-3}{4}\times\frac{8}{11}-\frac{3}{11}\times\frac{1}{2}\)
\(=\frac{-6}{11}-\frac{3}{22}\)
\(=\frac{15}{22}\)
e) \(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)
\(=\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+\frac{1}{9\times11}+\frac{1}{11\times13}+\frac{1}{13\times15}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)
\(=\frac{1}{3}-\frac{1}{15}\)
\(=\frac{4}{15}\)
Câu 1:
\(S=\frac{10}{7}+\frac{10}{7^2}+\frac{10}{7^3}+...+\frac{10}{7^{10}}\)
\(\frac{1}{7}S=\frac{10}{7^2}+\frac{10}{7^3}+....+\frac{10}{7^{11}}\)
\(\rightarrow\)\(\left(1-\frac{1}{7}\right).S=\frac{10}{7}-\frac{10}{7^{11}}\)
=> \(S=\frac{10.7^{10}-10}{7^{10}.6}\)
\(3\frac{14}{19}+\frac{13}{17}+\frac{35}{43}+6\frac{5}{19}+\frac{8}{43}\)
\(=\left(3\frac{14}{19}+6\frac{5}{19}\right)+\left(\frac{35}{43}+\frac{8}{43}\right)+\frac{13}{17}\)
\(=10+1+\frac{13}{17}=11+\frac{13}{17}=11\frac{13}{17}\)
\(\frac{-5}{7}.\frac{2}{11}+\frac{-5}{7}.\frac{9}{11}+1\frac{5}{7}\)
\(=\frac{-5}{7}.\frac{2}{11}+\frac{-5}{7}.\frac{9}{11}+1+\frac{-5}{7}.\left(-1\right)\)
\(=\frac{-5}{7}\left(\frac{2}{11}+\frac{9}{11}-1\right)+1\)
\(=\frac{-5}{7}.0+1==0+1=1\)
\(3\frac{14}{19}+\frac{13}{17}+\frac{35}{43}+6\frac{5}{19}+\frac{8}{43}\)
\(\left(3\frac{14}{19}+6\frac{5}{19}\right)+\left(\frac{35}{43}+\frac{8}{43}\right)+\frac{13}{17}\)
\(10+1+\frac{13}{17}=11+\frac{13}{17}=11\frac{13}{17}\)
\(11\frac{3}{13}-\left(2\frac{4}{7}+5\frac{3}{13}\right)\)
\(=11\frac{3}{13}-2\frac{4}{7}-5\frac{3}{13}\)
\(=\left(11\frac{3}{13}-5\frac{3}{13}\right)-2\frac{4}{7}\)
\(=6-2-\frac{4}{7}\)
\(=4-\frac{4}{7}=3\frac{3}{7}\)
Ta có:
\(74\frac{19}{35}.\frac{7}{90}+15\frac{16}{35}:\frac{90}{7}\)
\(=74\frac{19}{35}.\frac{7}{90}+15\frac{16}{35}.\frac{7}{90}\)
\(=\frac{7}{90}.\left(74\frac{19}{35}+15\frac{16}{35}\right)\)
\(=\frac{7}{90}.90\)
\(=7\)
\(74\frac{19}{35}.\frac{7}{90}+15\frac{16}{35}.\frac{7}{90}\)
\(\frac{7}{90}.\left(74\frac{19}{35}+15\frac{16}{35}\right)\)
\(\frac{7}{90}.90\)
\(7\)