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a) \(2^3+3.\left(\frac{1}{2}\right)^0+\left[\left(-2\right)^2:\frac{1}{2}\right]\)
\(=8+3.1+4:\frac{1}{2}\)
\(=8+3+8=19\)
b)\(\frac{2^{15}.9^4}{6^6.8^3}=\frac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^6.\left(2^3\right)^3}=\frac{2^{15}.3^8}{2^6.3^6.2^9}\)\(=\frac{2^{15}.3^8}{2^{15}.3^6}=3^2=9\)
c) \(\left(1+\frac{2}{3}-\frac{1}{4}\right).\left(\frac{4}{5}-\frac{3}{4}\right)^2\)
\(=\frac{17}{12}.\frac{1}{400}=\frac{17}{4800}\)
d) \(\left(-\frac{10}{3}\right)^3.\left(\frac{-6}{5}\right)^4=-\frac{100}{27}.\frac{1296}{625}\)\(=\frac{-4.48}{1.25}=-\frac{192}{25}\)
(22+21+22+23).20.21.22.23
=(4+2+4+8).1.2.4.8
=18.1.2.4.8
=1152
1 3/8+1/8:(0,75-1/2)-25%.1/2
=11/8+1/8:(3/4-1/2)-1/4.1/2
=12/8:1/4-1/8
=6/1-1/8
=47/8
12 1/3-5/6:(24-23 5/7)
=37/3-5/6:(24-166/7)
=37/3-5/6:2/7
=37/3-35/2
=31/6
(-1/2)2-(-2)2-50
=1/4-4-1
=-19/4
\(\left(2^2+2^1+2^2+2^3\right)×2^0×2^1×2^2×2^3\)
\(=\left(4+2+4+8\right)×1×2×4×8\)
\(=18×1×2×4×8\)
\(=1152\)
\(1\frac{3}{8}+\frac{1}{8}:\left(0,75-\frac{1}{2}\right)-25\%×\frac{1}{2}\)
\(=\frac{11}{8}+\frac{1}{8}:\left(\frac{3}{4}-\frac{1}{2}\right)-\frac{1}{4}×\frac{1}{2}\)
\(=\frac{11}{8}+\frac{1}{8}:\frac{1}{4}-\frac{1}{8}\)
\(=\frac{11}{8}+\frac{1}{2}-\frac{1}{8}\)
\(=\frac{7}{4}\)
\(12\frac{1}{3}-\frac{5}{6}:\left(24-23\frac{5}{7}\right)\)
\(=\frac{37}{3}-\frac{5}{6}:\left(24-\frac{166}{7}\right)\)
\(=\frac{37}{3}-\frac{5}{6}:\frac{2}{7}\)
\(=\frac{37}{3}-\frac{35}{12}\)
\(=\frac{113}{12}\)
\(\left(\frac{-1}{2}\right)^2-\left(-2\right)^2-5^0\)
\(=\frac{1}{4}-4-1\)
\(=\frac{-19}{4}\)
\(\left(\frac{3}{8}+-\frac{3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
\(=\left(\frac{9}{24}+-\frac{18}{24}+\frac{14}{24}\right):\frac{5}{6}+\frac{1}{2}\)
\(=\frac{5}{24}:\frac{5}{6}+\frac{1}{2}\)
\(=\frac{5}{24}.\frac{6}{5}+\frac{1}{2}\)
\(=\frac{1}{4}+\frac{1}{2}\)
\(=\frac{1}{4}+\frac{2}{4}\)
\(=\frac{3}{4}\)
\(\frac{1}{2}+\frac{3}{4}-\left(\frac{3}{4}-\frac{4}{5}\right)\)
\(=\frac{1}{2}+\frac{3}{4}-\left(\frac{15}{20}-\frac{16}{20}\right)\)
\(=\frac{1}{2}+\frac{3}{4}-\frac{-1}{20}\)
\(=\frac{10}{20}+\frac{15}{20}-\frac{-1}{20}\)
\(=\frac{25}{20}-\frac{-1}{20}\)
\(=\frac{26}{20}\)
\(=\frac{13}{10}\)
a) \(2\frac{3}{4}\cdot\left(-0,4\right)-1\frac{3}{5}\cdot2,75+1,2:\frac{4}{11}\)
\(=2\frac{3}{4}\cdot\left(-\frac{2}{5}\right)-1\frac{3}{5}\cdot\frac{11}{4}+\frac{6}{5}:\frac{4}{11}\)
\(=\frac{11}{4}\cdot\left(-\frac{2}{5}\right)-1\frac{3}{5}\cdot\frac{11}{4}+\frac{6}{5}\cdot\frac{11}{4}\)
\(=\frac{11}{4}\left(-\frac{2}{5}-1\frac{3}{5}+\frac{6}{5}\right)\)
\(=\frac{11}{4}\left(-\frac{2}{5}-\frac{8}{5}+\frac{6}{5}\right)\)
\(=\frac{11}{4}\cdot\left(-\frac{4}{5}\right)=\frac{11}{1}\cdot\left(-\frac{1}{5}\right)=-\frac{11}{5}\)
b) \(\left(\frac{1}{2}+1\right)\cdot\left(\frac{1}{3}+1\right)\cdot\left(\frac{1}{4}+1\right)....\left(\frac{1}{31}+1\right)\)
\(=\left(\frac{1}{2}+\frac{2}{2}\right)\left(\frac{1}{3}+\frac{3}{3}\right)\left(\frac{1}{4}+\frac{4}{4}\right)...\left(\frac{1}{31}+\frac{31}{31}\right)\)
\(=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot...\cdot\frac{32}{31}\)
\(=\frac{3\cdot4\cdot5\cdot...\cdot32}{2\cdot3\cdot4\cdot...\cdot31}=\frac{32}{2}=16\)
c) Đặt \(C=1+2+3+...+30\)
Số số hạng là : \(\left(30-1\right):1+1=30\)(số)
Tổng của dãy số là : \(\frac{\left(1+30\right)\cdot30}{2}=465\)
Do đó : \(\frac{930}{C}=\frac{930}{465}=2\)
a) \(\left(-\frac{1}{4}\right)^0=1\)
b) \(\left(-2\frac{1}{3}\right)^2=\left(-\frac{7}{3}\right)^2=\frac{49}{9}\)
c) \(\left(\frac{4}{5}\right)^{-2}=\frac{25}{16}\)
d) \(\left(0,5\right)^{-3}=8\)
e) \(\left(-1\frac{1}{3}\right)^4=\left(-\frac{4}{3}\right)^4=\frac{256}{81}\)
a, \(\left(\frac{-1}{4}\right)^0\) = 1
Bất kỳ số nguyên nào nếu có mũ bằng 0 đều bằng 1
b, \(\left(-2\frac{1}{3}\right)^2=\left(-\frac{7}{3}\right)^2=\frac{49}{9}\)
Bài 1:
1) \(\frac{11}{3}\): 3\(\frac{1}{3}\)- 3
= \(\frac{11}{3}\): \(\frac{10}{3}\)- 3
= \(\frac{11}{3}\). \(\frac{3}{10}\)- 3
= \(\frac{11}{10}\)- 3
= \(\frac{-19}{10}\)
2) \(\frac{5}{6}\): \(\frac{3}{52}\) - \(\frac{5}{6}\). 47\(\frac{1}{3}\)
= \(\frac{5}{6}\) . \(\frac{52}{3}\)- \(\frac{5}{6}\). 47\(\frac{1}{3}\)
= \(\frac{5}{6}\).(\(\frac{52}{3}\)- 47\(\frac{1}{3}\))
= \(\frac{5}{6}\).( -30)
= -25
a)\(5-\left(-\frac{5}{11}\right)^0+\left(\frac{1}{3}\right)^2:3=5-1+\frac{1}{9}\cdot\frac{1}{3}=4+\frac{1}{27}=\frac{108}{27}+\frac{1}{27}=\frac{109}{27}\)
b)\(2^3+3.\left(\frac{1}{2}\right)^0+\left[\left(-2\right)^3:\frac{1}{2}\right]=8+3.1+\left[\left(-8\right)\cdot2\right]=8+3-16=-5\)
a/ \(5-\left(-\frac{5}{11}\right)^0+\left(\frac{1}{3}\right)^2:3=5-1+\frac{1}{9}:3=5-1+\frac{1}{27}=4+\frac{1}{27}=\frac{109}{27}\)
b/ \(2^3+3.\left(\frac{1}{2}\right)^0+\left[\left(-2\right)^3:\frac{1}{2}\right]=8+3.1+\left[-8:\frac{1}{2}\right]=11+-16=-5\)