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\(a,\left(-37-17\right).\left(-9\right)+35.\left(-9-11\right)\)
\(=\left(-54\right).\left(-9\right)+35.\left(-20\right)\)
\(=486+\left(-700\right)\)
\(=-214\)
\(b,\left(-25\right).\left(75-25\right)-75.\left(45-25\right)\)
\(=\left(-25\right).50-75.20\)
\(=\left(-25\right).50+\left(-25\right).3.20\)
\(=\left(-25\right).50+\left(-25\right).60\)
\(=\left(-25\right).\left(50+60\right)\)
\(=\left(-25\right).110\)
\(=-2750\)
\(c,136.\left(-47\right)+36.47\)
\(=47.\left(-136+36\right)\)
\(=47.\left(-100\right)\)
\(=-4700\)
\(a.\frac{27.45+27.55}{2+4+6+...+14+16+18}=\frac{27.100}{\frac{\left(2+18\right).9}{2}}=30\)
\(b.\frac{26.108-26.12}{32-28+24-20+16-12+8-4}=\frac{26\left(108-12\right)}{\left(32-28\right).4}=\frac{26.96}{4.4}=156\)
\(c.\frac{27.4500+135.550.2}{2+4+6+...+14+16+18}=\frac{270.450+270.550}{\frac{\left(2+8\right).9}{2}}\)
\(d.\frac{48.700-24.45.20}{45-40+35-30+25-20+15-10+5}=\frac{48.700-48.450}{5.5}\)\(=\frac{48\left(700-450\right)}{25}=\frac{48.250}{25}=480\)
#ĐinhBa
\(a,\left(-25\right).\left(75-45\right)-75.\left(45-25\right)\)
\(=\left(-25\right).20-75.20\)
\(=20.\left(-25-75\right)\)
\(=20.\left(-100\right)\)
\(=-2000\)
\(b,\frac{18.12-48.15}{-3.270-3.30}\)
\(=\frac{18.12-12.4.15}{-3.270-3.30}\)
\(=\frac{12.\left(18-60\right)}{-3.\left(270+30\right)}\)
\(=\frac{12.\left(-42\right)}{-3.300}\)
\(=\frac{14}{25}\)
\(c,\frac{2^5.7+2^5}{2^5.5^2-2^5.3}\)
\(=\frac{2^5.\left(7+1\right)}{2^5\left(25-3\right)}\)
\(=\frac{2^5.8}{2^5.22}\)
\(=\frac{8}{22}\)
\(=\frac{4}{11}\)
3.(-4)^3-2.(-5)+(-7)^2.(-6)=3.-64-(-10)+49.(-6)=(-192)-(-10)+(-294)=(-182)+(-294)=(-476)
(5).(-26-25)+(-4).(-51)=5.(-51)+204=(-255)+204=-51
34.(47-75)+47.(75-34)=34.(-28)+47.41=(-952)+1927=975
(-25).(75-45)-75.(45-25)=(-25).30-75.20=(-750)-1500=-2250
\(a,\frac{3}{27}+\frac{1}{4}-\frac{1}{9}=\frac{3}{27}+\frac{1}{4}-\frac{3}{27}=\frac{3}{27}-\frac{3}{27}+\frac{1}{4}=\frac{1}{4}\)
\(b,\frac{72}{25}\cdot\frac{75}{36}=\frac{72\cdot75}{25\cdot36}=\frac{2\cdot36\cdot25\cdot3}{25\cdot36}=\frac{2\cdot3}{1\cdot1}=6\)
\(c,\frac{36}{49}:\frac{48}{14}=\frac{36}{49}\cdot\frac{14}{48}=\frac{36\cdot14}{49\cdot48}=\frac{3\cdot12\cdot2\cdot7}{7\cdot7\cdot4\cdot12}=\frac{3\cdot2}{7\cdot4}=\frac{6}{28}=\frac{3}{14}\)
\(d,\frac{39}{23}:26=\frac{39}{23}\cdot\frac{1}{26}=\frac{39}{23\cdot26}=\frac{3\cdot13}{23\cdot2\cdot13}=\frac{3}{23\cdot2}=\frac{3}{46}\)
\(e,45:\frac{15}{2}=\frac{45}{1}\cdot\frac{2}{15}=\frac{45\cdot2}{15}=\frac{3\cdot15\cdot2}{15}=\frac{6}{1}=6\)
\(g,\frac{57}{8}:\frac{3}{4}-\frac{15}{7}\cdot\frac{14}{25}=\frac{57\cdot4}{8\cdot3}-\frac{15\cdot14}{7\cdot25}=\frac{3\cdot19\cdot4}{2\cdot4\cdot3}-\frac{3\cdot5\cdot2\cdot7}{7\cdot5\cdot5}\)
\(=\frac{19}{2}-\frac{3\cdot2}{5}=\frac{95}{10}-\frac{12}{10}=\frac{83}{10}\)
Câu a:
(-37 + 72).(-10) + 35.(-9-11)
= 35.(-10) + 35.(-20)
= -35.(10 + 20)
= -35.30
= - 1050
Câu b:
(-25).(75 - 45) - 75.(45 - 25)
= -25.75 + 25.45 - 75.45 + 75.25
= -(25.75 - 25.75) - 45.(75 - 25)
= 0 - 45.50
= - 2250
Câu c:
(36 - 16).(-5) + 6.(-14 - 6)
= 20.(-5) + 6.(-20)
= -20.(5 + 6)
= - 20.11
= -220
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