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\(\frac{\frac{2}{45}-\frac{4}{13}-\frac{1}{3}}{\frac{3}{13}-\frac{4}{15}+\frac{2}{3}}=\frac{-\frac{349}{585}}{\frac{41}{65}}=-\frac{349}{369}\)
Speed Ninja\(A=\frac{\frac{5}{22}+\frac{3}{13}-\frac{1}{2}}{\frac{4}{13}-\frac{2}{11}+\frac{3}{2}}\)
A=\(\left(\frac{\frac{5}{22}+\frac{3}{13}-\frac{1}{2}}{\frac{4}{13}-\frac{2}{11}+\frac{3}{2}}\right)\)x\(\frac{\left(2.11.13\right)}{\left(2.11.13\right)}\)
A=\(\frac{65+66-143}{88-52+429}\)
A=\(\frac{-12}{465}=\frac{-4}{155}\)
\(B=\frac{5}{2}+\left(\frac{4}{1.11}+\frac{3}{11.2}\right)+\left(\frac{1}{2.15}+\frac{13}{15.4}\right)\)
\(B=\frac{5}{2}+\frac{1}{11}.\left(4+\frac{3}{2}\right)+\frac{1}{15}\left(\frac{1}{2}+\frac{13}{4}\right)=\frac{5}{2}+\frac{1}{11}.\frac{11}{2}+\frac{1}{15}.\frac{15}{4}\)
=> \(B=\frac{5}{2}+\frac{1}{2}+\frac{1}{4}=\frac{10}{4}+\frac{2}{4}+\frac{1}{4}=\frac{13}{4}\)
Ta có
1/7.B = 5/2.7 + 4/7.11 + 3/11.14 + 1/14.15 + 13/15.28
1/7.B = 1/2 - 1/7 + 1/7 - 1/11 + 1/11 - 1/14 + 1/14 - 1/15 + 1/15 - 1/28
1/7.B = 1/2 - 1/28
1/7.B = 14/28 - 1/28
1/7.B = 13/28
B = 13/28 : 1/7
B = 13/28 . 7
B = 13/4
a)\(\frac{2}{3}+\frac{3}{4}+\frac{5}{6}\)
\(=\frac{8+9+10}{12}\)
\(=\frac{27}{12}=\frac{9}{4}\)
b)\(\frac{15}{8}-\frac{7}{12}+\frac{5}{6}\)
\(=\frac{45-14+20}{24}\)
\(=\frac{51}{24}=\frac{17}{8}\)
2)
a)\(\frac{2}{5}+\frac{7}{13}+\frac{3}{5}+\frac{1}{7}\)
\(=\frac{2}{5}+\frac{3}{5}+\frac{7}{13}+\frac{1}{7}\)
\(=1+\frac{7}{13}+\frac{1}{7}\)
\(=\frac{20}{13}+\frac{1}{7}\)
\(=\frac{153}{91}\)
Tí tớ trả lời tiếp
\(A=49\frac{8}{23}-\left(5\frac{7}{32}+14\frac{8}{23}\right)\)
\(A=49\frac{8}{23}-5\frac{7}{32}+14\frac{8}{23}\)
\(A= \left(49\frac{8}{23}-14\frac{8}{23}\right)-5\frac{7}{32}\)
\(A=\left[\left(49-14\right)-\left(\frac{8}{23}-\frac{8}{23}\right)\right]-5\frac{7}{32}\)
\(A=\left[35-0\right]-5\frac{7}{32}\)
\(A=35-5\frac{7}{32}\)
\(A=\frac{953}{32}\)
\(B=71\frac{38}{45}-\left(43\frac{38}{45}-1\frac{17}{57}\right)\)
\(B=71\frac{38}{45}-\frac{36377}{855}\)
\(B=\frac{1670}{57}\)
\(C=\left(19\frac{5}{8}:\frac{7}{12}-13\frac{1}{4}:\frac{7}{12}\right):\frac{4}{5}\)
\(C=\left[\left(19\frac{5}{8}-13\frac{1}{4}\right):\frac{7}{12}\right]:\frac{4}{5}\)
\(C=\left[\frac{51}{8}:\frac{7}{12}\right]:\frac{4}{5}\)
\(C=\frac{153}{14}:\frac{4}{5}\)
\(C=\frac{765}{56}\)
\(D=\left[\left(\frac{10}{15}-\frac{2}{3}\right):\frac{1}{7}\right]\cdot0,15-\frac{1}{4}\)
\(D=\left[0:\frac{1}{7}\right]\cdot\frac{3}{20}-\frac{1}{4}\)
\(D=0\cdot\frac{3}{20}-\frac{1}{4}\)
\(D=0-\frac{1}{4}\)
\(D=-\frac{1}{4}\)
\(E=\frac{13}{30}+\frac{28}{45}\cdot2\frac{1}{2}-\left[\left(\frac{1}{2}+\frac{1}{3}\right):\frac{53}{90}\right]:\frac{50}{53}\)
\(E=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\left[\frac{5}{6}:\frac{53}{90}\right]:\frac{50}{53}\)
\(E=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\frac{75}{53}:\frac{50}{53}\)
\(E=\frac{13}{30}+\frac{14}{9}-\frac{3}{2}\)
\(\)\(E=\frac{22}{45}\)
CHUC BAN HOC TOT >.<
B = \(\left(\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\right)\))
B = 7 . \(\left(\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\right)\)
B = 7 . \(\left(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\right)\)
B = 7 . ( 1/2 - 1/28 )
B = 7 . 13/28
B = 13/4
1/7B= 5/2.7 + 4/7.11 + 3/11.14 + 1/14.15 + 13/15.28
1/7B= 1/2 - 1/7 + 1/7 - 1/11 + 1/11 - 1/14 + 1/14 - 1/15 + 1/15 - 1/28
1/7B= 1/2 - 1/28= 13/28
B= 13/28 : 1/7= 13/28.7= 13/4
Vậy B= 13/4
a) \(\frac{1}{3}.\frac{-6}{13}.\frac{-9}{10}.\frac{-13}{36}\)
\(=\left(\frac{1}{3}.\frac{-9}{10}\right)\left(\frac{-6}{13}.\frac{-13}{36}\right)\)
\(=\frac{-3}{10}.\frac{1}{6}\)
\(=\frac{-1}{20}\)
b) \(\frac{-1}{3}.\frac{-15}{17}.\frac{34}{45}\)
\(=\frac{-1}{3}.\frac{-2}{3}\)
\(=\frac{2}{9}\)
c) \(\left(1-\frac{1}{5}\right)\left(\frac{-3}{10}+\frac{1}{5}\right)\)
\(=\frac{4}{5}.\frac{-1}{10}\)
\(=\frac{-2}{25}\)
d) \(A=\frac{1}{3}.\frac{4}{5}+\frac{1}{3}.\frac{6}{5}+\frac{2}{3}\)
\(=\frac{1}{3}\left(\frac{4}{5}+\frac{6}{5}\right)+\frac{2}{3}\)
\(=\frac{1}{3}.2+\frac{2}{3}\)
\(=\frac{2}{3}+\frac{2}{3}\)
\(=\frac{4}{3}\)
e) \(11\frac{1}{4}-\left(2\frac{5}{7}+5\frac{1}{4}\right)\)
\(=\left(11\frac{1}{4}-5\frac{1}{4}\right)-2\frac{5}{7}\)
\(=6-2\frac{5}{7}\)
\(=5\frac{7}{7}-2\frac{5}{7}\)
\(=3\frac{2}{7}\)
\(\frac{\frac{2}{45}-\frac{4}{13}-\frac{1}{3}}{\frac{3}{13}-\frac{4}{15}+\frac{2}{3}}=\frac{\frac{-349}{585}}{\frac{41}{65}}=\frac{-349}{585}:\frac{41}{65}=\frac{-349}{585}\cdot\frac{65}{41}=-3\frac{52}{99}\)