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a) \(\left(-7\right)-\left[\left(-19\right)+\left(-21\right)\right].\left(-3\right)-\left[\left(+32\right)+\left(-7\right)\right]\)
\(=\left(-7\right)-\left(-40\right).\left(-3\right)-25\)
\(=\left(-7\right)-120-25\)
\(=-152\)
b) \(\left(-2\right)^3.3-\left(1^{10}+8\right):\left(-3\right)^2\)
\(=\left(-8\right).3-\left(1+8\right):9\)
\(=\left(-24\right)-9:9\)
\(=\left(-24\right)-1\)
\(=-25\)
Bài giải
a, \(\left(-7\right)-\left[\left(-19\right)+\left(-21\right)\right]\cdot\left(-3\right)-\left[\left(+32\right)+\left(-7\right)\right]\)
\(=\left(-7\right)-\left(-40\right)\cdot\left(-3\right)-25\)
\(=-7-120-25\)
\(=-127-25\)
\(=-152\)
b, \(\left(-2\right)^3\cdot3-\left(1^{10}+8\right)\text{ : }\left(-3\right)^2\)
\(=-8\cdot3-\left(1+8\right)\text{ : }9\)
\(=-24-9\text{ : }9\)
\(=-24-1\)
\(=-25\)
\(a,-7-\left[\left(-19\right)+\left(21\right)\right].\left(-3\right)-\left[\left(32\right)+\left(-7\right)\right]\)
\(=-7-\left[21-19\right].\left(-3\right)-\left[32-7\right]\)
\(=-7-2.\left(-3\right)-25\)
\(=-7+6-25=-26\)
\(b,\left(-2\right)^2.3-\left(1^{10}+8\right):\left(-3\right)^2\)
\(=4.3-9:9\)
\(=12-1=11\)
\(a,=\left(-\frac{3}{7}+\frac{3}{7}\right)+\frac{5}{13}=0+\frac{5}{13}=\frac{5}{13}\)
\(b,=-\frac{5}{21}+\left(-\frac{2}{21}\right)+\frac{8}{24}=-\frac{1}{3}+\frac{1}{3}=0\)
\(c,=\left[-\frac{6}{11}+\frac{\left(-5\right)}{11}\right]+2=-1+2=1\)
\(d,=\frac{-1+\left(-15\right)}{32}+\frac{1}{2}=-\frac{1}{2}+\frac{1}{2}=0\)
a) \(\frac{-3}{7}+\frac{5}{13}+\frac{3}{7}\)
\(=\left(\frac{-3}{7}+\frac{3}{7}\right)+\frac{5}{13}\)
\(=0+\frac{5}{13}=\frac{5}{13}\)
b) \(\frac{-5}{21}+\frac{-2}{21}+\frac{8}{24}\)
\(=\frac{-7}{21}+\frac{1}{3}\)
\(=\frac{-1}{3}+\frac{1}{3}=0\)
c) \(\frac{-5}{11}+\left(\frac{-6}{11}+2\right)\)
\(=\frac{-5}{11}+\frac{-6}{11}+2\)
\(=\left(-1\right)+2=1\)
d) \(\left(\frac{-1}{32}+\frac{1}{2}\right)+\frac{-15}{32}\)
\(=\frac{-1}{32}+\frac{1}{2}+\frac{-15}{32}\)
\(=\left(\frac{-1}{32}+\frac{-15}{32}\right)+\frac{1}{2}\)
\(=\frac{-16}{32}+\frac{1}{2}\)
\(=\frac{-1}{2}+\frac{1}{2}=0\)
\(A=49\frac{8}{23}-\left(5\frac{7}{32}+14\frac{8}{23}\right)\)
\(A=49\frac{8}{23}-5\frac{7}{32}+14\frac{8}{23}\)
\(A= \left(49\frac{8}{23}-14\frac{8}{23}\right)-5\frac{7}{32}\)
\(A=\left[\left(49-14\right)-\left(\frac{8}{23}-\frac{8}{23}\right)\right]-5\frac{7}{32}\)
\(A=\left[35-0\right]-5\frac{7}{32}\)
\(A=35-5\frac{7}{32}\)
\(A=\frac{953}{32}\)
\(B=71\frac{38}{45}-\left(43\frac{38}{45}-1\frac{17}{57}\right)\)
\(B=71\frac{38}{45}-\frac{36377}{855}\)
\(B=\frac{1670}{57}\)
\(C=\left(19\frac{5}{8}:\frac{7}{12}-13\frac{1}{4}:\frac{7}{12}\right):\frac{4}{5}\)
\(C=\left[\left(19\frac{5}{8}-13\frac{1}{4}\right):\frac{7}{12}\right]:\frac{4}{5}\)
\(C=\left[\frac{51}{8}:\frac{7}{12}\right]:\frac{4}{5}\)
\(C=\frac{153}{14}:\frac{4}{5}\)
\(C=\frac{765}{56}\)
\(D=\left[\left(\frac{10}{15}-\frac{2}{3}\right):\frac{1}{7}\right]\cdot0,15-\frac{1}{4}\)
\(D=\left[0:\frac{1}{7}\right]\cdot\frac{3}{20}-\frac{1}{4}\)
\(D=0\cdot\frac{3}{20}-\frac{1}{4}\)
\(D=0-\frac{1}{4}\)
\(D=-\frac{1}{4}\)
\(E=\frac{13}{30}+\frac{28}{45}\cdot2\frac{1}{2}-\left[\left(\frac{1}{2}+\frac{1}{3}\right):\frac{53}{90}\right]:\frac{50}{53}\)
\(E=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\left[\frac{5}{6}:\frac{53}{90}\right]:\frac{50}{53}\)
\(E=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\frac{75}{53}:\frac{50}{53}\)
\(E=\frac{13}{30}+\frac{14}{9}-\frac{3}{2}\)
\(\)\(E=\frac{22}{45}\)
CHUC BAN HOC TOT >.<
a) 2011 + 5[300 - (17 - 7)2 ]
= 2011 + 5[300 - 100]
= 2011 + 5.200
= 2011 + 1000
= 3011
b) 695 - [200 + (11 - 1)2 ]
= 695 - [200 + 100]
= 695 - 300
= 395
c) 129 - 5[29 - (6-1)2 ]
= 129 - 5[29 - 25]
= 129 - 5 . 4
= 129 - 20
= 109
d) 2345 - 1000 : [19 - 2(21-18)2 ]
= 2345 - 1000 : [19 - 2 . 9]
= 2345 - 1000 : 1
= 2345 - 1000
= 1345
a, 2011+5 [300-(17-7)2 ] =2011+5[300-102]
=2011+5[300-100]=2011+5.200
=2011+1000=3011
b, 695-[200+(11-1)2]=695-[200+102]
=695-[200+100]=695-300=395