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a. \(5^{10}.125^2.625^3=5^{10}.\left(5^3\right)^2.\left(5^4\right)^3=5^{10}.5^6.5^{12}=5^{10+6+12}=5^{28}\)
a: \(=5^{10}\cdot5^6\cdot5^{12}=5^{28}\)
b: \(=10^3\cdot10^8\cdot10^{15}=10^{26}\)
c: \(=2^{20}\cdot2^{20}=2^{40}\)
d: \(=2^{16}\cdot2^{16}\cdot3^8=2^{32}\cdot3^8\)
e: \(=\dfrac{3^{24}}{3^8}=3^{16}\)
f: \(=2^{12}\cdot2^{20}\cdot2^5=2^{37}\)
e) \(E=0,7.2\frac{2}{3}.20.0,375.\frac{5}{28}\)
\(=\left(0,7.20\right)\left(2\frac{2}{3}.0,375\right)\frac{5}{28}\)
\(=14.1.\frac{5}{28}\)
\(=14.\frac{5}{28}\)
\(=\frac{5}{2}\)
f) \(F=\left(9,75.21\frac{3}{7}+\frac{39}{4}.18\frac{4}{7}\right)\frac{15}{78}\)
\(=\left(\frac{39}{4}.21\frac{3}{7}+\frac{39}{4}.18\frac{4}{7}\right)\frac{15}{78}\)
\(=\frac{39}{4}\left(21\frac{3}{7}+18\frac{4}{7}\right)\frac{15}{78}\)
\(=\frac{39}{4}.40.\frac{15}{78}\)
\(=390.\frac{15}{78}\)
\(=78\)
Chúc bạn học tốt môn Toán!!!
Mình ko ghi đề đâu
\(A=49\frac{8}{23}-5\frac{7}{32}-14\frac{8}{23}=\left(49\frac{8}{23}-14\frac{8}{23}\right)-5\frac{7}{32}=35-5\frac{7}{32}=29\frac{25}{32}\)
\(B=71\frac{38}{45}-43\frac{8}{45}+1\frac{17}{57}=\left(71\frac{38}{45}-43\frac{8}{45}\right)+1\frac{17}{57}=28\frac{2}{3}+1\frac{17}{57}=29\frac{55}{57}\)
\(C=-\frac{3}{7}.\left(\frac{5}{9}+\frac{4}{9}\right)+2\frac{3}{7}=-\frac{3}{7}.1+2\frac{3}{7}=-\frac{3}{7}+2\frac{3}{7}=2\)
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 >.<
a) \(\left(56.27+56.35\right):62\)
\(=56.\left(27+35\right):62\)
\(=56.62:62\)
\(=56\)
b) \(3\frac{2}{7}.12\frac{1}{2}-3\frac{2}{7}.5\frac{1}{2}+1\frac{1}{2}:\frac{3}{4}\)
\(=3\frac{2}{7}.\left(12\frac{1}{2}-5\frac{1}{2}\right)+\frac{3}{2}.\frac{4}{3}\)
\(=\frac{23}{7}.7+2\)
\(=23+2\)
\(=25\)
c) \(\frac{16.17-5}{16.16+11}\)
\(=\frac{16.\left(16+1\right)-5}{16.16+11}\)
\(=\frac{16.16+16-5}{16.16+11}\)
\(=\frac{16.16+11}{16.16+11}\)
\(=1\)
a) \(5^6.4^3.19=5^6.\left(2^2\right)^3.19=\left(5.2\right)^6.19=1000000.19=19000000\)
b)\(123.25^4.16^2=123.5^8.2^8=123.\left(5.2\right)^8=123.10^8=123.100000000=12300000000\)
c) \(73.5^{12}.8^4=73.5^{12}.\left(2^3\right)^4=73.\left(5.2\right)^{12}=73.10^{12}=73.1000000000000=73000000000000\)
d)\(1111.8^4.25^7=1111.\left(2^3\right)^4.\left(5^2\right)^7=1111.2^{12}.5^{14}=\left(1111.25\right).\left(2.5\right)^{12}=27775.10^{12}\)
\(=27775.1000000000000=27775000000000000\)
a)56x43=1000000
1000000x19=19000000
b) 12300000000
c) 73000000000000
d) Ko ra đc