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\(1\frac{13}{15}\cdot3\cdot(0,5)^2\cdot3+\left[\frac{8}{15}-1\frac{19}{60}:1\frac{23}{24}\right]\)
\(=\frac{28}{15}\cdot3\cdot0,5\cdot0,5\cdot3+\left[\frac{8}{15}-\frac{79}{60}:\frac{47}{24}\right]\)
\(=\frac{28}{5}\cdot0,25\cdot3+\left[\frac{32}{60}-\frac{79}{60}\cdot\frac{24}{47}\right]\)
\(=\frac{28}{5}\cdot\frac{25}{100}\cdot3+\left[\frac{32}{60}-\frac{158}{235}\right]\)
\(=\frac{28}{5}\cdot\frac{1}{4}\cdot3+\frac{-98}{705}=\frac{7}{5}\cdot1\cdot3+\frac{-98}{705}\)
Đến đây là tính dễ rồi :v
\((-3,2)\cdot\frac{-15}{64}+\left[0,8-2\frac{4}{15}\right]:1\frac{23}{24}\)
\(=\frac{-32}{10}\cdot\frac{-15}{64}+\left[\frac{8}{10}-\frac{34}{15}\right]:\frac{47}{24}\)
\(=\frac{-32\cdot(-15)}{10\cdot64}+\left[\frac{4}{5}-\frac{34}{15}\right]:\frac{47}{24}\)
\(=\frac{-1\cdot(-3)}{2\cdot2}+\frac{4\cdot3-34}{15}:\frac{47}{24}\)
\(=\frac{3}{4}+\frac{-22}{15}:\frac{47}{24}\)
\(=\frac{3}{4}+\frac{-517}{180}=\frac{-191}{90}\)
Bài 2 : \(\frac{2\cdot(-13)\cdot9\cdot10}{(-3)\cdot4\cdot(-5)\cdot26}=\frac{1\cdot(-1)\cdot3\cdot2}{(-1)\cdot2\cdot(-1)\cdot2}=\frac{1\cdot3}{-1\cdot2}=\frac{3}{-2}=\frac{-3}{2}\)
\(\frac{15\cdot8+15\cdot4}{12\cdot3}=\frac{15\cdot(8+4)}{12\cdot3}=\frac{15\cdot12}{12\cdot3}=\frac{15}{3}=5\)
\(A=49\frac{8}{23}-\left(5\frac{7}{32}+14\frac{8}{23}\right)\)
\(A=49\frac{8}{23}-5\frac{7}{32}+14\frac{8}{23}\)
\(A= \left(49\frac{8}{23}-14\frac{8}{23}\right)-5\frac{7}{32}\)
\(A=\left[\left(49-14\right)-\left(\frac{8}{23}-\frac{8}{23}\right)\right]-5\frac{7}{32}\)
\(A=\left[35-0\right]-5\frac{7}{32}\)
\(A=35-5\frac{7}{32}\)
\(A=\frac{953}{32}\)
\(B=71\frac{38}{45}-\left(43\frac{38}{45}-1\frac{17}{57}\right)\)
\(B=71\frac{38}{45}-\frac{36377}{855}\)
\(B=\frac{1670}{57}\)
\(C=\left(19\frac{5}{8}:\frac{7}{12}-13\frac{1}{4}:\frac{7}{12}\right):\frac{4}{5}\)
\(C=\left[\left(19\frac{5}{8}-13\frac{1}{4}\right):\frac{7}{12}\right]:\frac{4}{5}\)
\(C=\left[\frac{51}{8}:\frac{7}{12}\right]:\frac{4}{5}\)
\(C=\frac{153}{14}:\frac{4}{5}\)
\(C=\frac{765}{56}\)
\(D=\left[\left(\frac{10}{15}-\frac{2}{3}\right):\frac{1}{7}\right]\cdot0,15-\frac{1}{4}\)
\(D=\left[0:\frac{1}{7}\right]\cdot\frac{3}{20}-\frac{1}{4}\)
\(D=0\cdot\frac{3}{20}-\frac{1}{4}\)
\(D=0-\frac{1}{4}\)
\(D=-\frac{1}{4}\)
\(E=\frac{13}{30}+\frac{28}{45}\cdot2\frac{1}{2}-\left[\left(\frac{1}{2}+\frac{1}{3}\right):\frac{53}{90}\right]:\frac{50}{53}\)
\(E=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\left[\frac{5}{6}:\frac{53}{90}\right]:\frac{50}{53}\)
\(E=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\frac{75}{53}:\frac{50}{53}\)
\(E=\frac{13}{30}+\frac{14}{9}-\frac{3}{2}\)
\(\)\(E=\frac{22}{45}\)
CHUC BAN HOC TOT >.<
a/ (-3,2).\(\frac{-15}{64}\)+(0,8-2\(\frac{4}{5}\)):1\(\frac{23}{24}\)
=(\(\frac{-16}{5}\)).\(\frac{-15}{64}\)+(\(\frac{4}{5}\)-\(\frac{14}{5}\)):\(\frac{47}{24}\)
=(\(\frac{-16}{5}\)).\(\frac{-15}{64}\)+(-2):\(\frac{47}{24}\)
= \(\frac{3}{4}\)+\(\frac{-48}{47}\)
=\(\frac{-51}{188}\)
b/ 1\(\frac{13}{15}\).3.(0,5)\(^2\).3+(\(\frac{8}{15}\)-1\(\frac{19}{60}\)):1\(\frac{23}{24}\)
= \(\frac{28}{15}\).3.\(\frac{1}{4}\).3+(\(\frac{8}{15}\)-\(\frac{79}{60}\)):\(\frac{47}{24}\)
= \(\frac{28}{15}\).3.\(\frac{1}{4}\).3+(\(\frac{-47}{60}\)):\(\frac{47}{24}\)
= \(\frac{28}{5}\).\(\frac{1}{4}\).3+(\(\frac{-47}{60}\)):\(\frac{47}{24}\)
= \(\frac{7}{5}\).3+(\(\frac{-47}{60}\)):\(\frac{47}{24}\)
= \(\frac{21}{5}\)+(\(\frac{-47}{60}\)):\(\frac{47}{24}\)
= \(\frac{21}{5}\)+(\(\frac{-2}{5}\))
= \(\frac{19}{5}\)
mk làm hơi dài dòng chút
CHÚC BẠN HỌC TỐT
\(1\frac{13}{15}.\left(0,5\right)^2.3+\left(\frac{8}{15}-1\frac{19}{60}\right):1\frac{23}{24}\)
\(=1\frac{13}{15}.0,25.3+\left(-\frac{47}{60}\right):1\frac{23}{24}\)
\(=1\frac{13}{14}.0,75+\left(-\frac{2}{5}\right)\)
\(=1,4-0,4\)
\(=1\)
\(1\frac{13}{15}.\left(0,5\right)^2.3+\left(\frac{8}{15}-1\frac{19}{60}\right);1\frac{23}{24}=\frac{7}{5}+\left(-\frac{2}{5}\right)=1\)