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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) \(\left(-\frac{10}{3}\right)^3.\left(\frac{-6}{5}\right)^4=-\frac{100}{27}.\frac{1296}{625}\)\(=\frac{-4.48}{1.25}=-\frac{192}{25}\)
B
từ 1 đến 2012 có tất cả:
2012-1:1+1 = 2012 (số)
=>có: 2012:2 = 1006 (cặp)
Mà mỗi cặp bằng (-1)nên
tổng dãy số trên là: 1006 . (-1) = -1006
(1-2)+(2-3)+(3-4)+(5-6)+...+(2011-2012)
=-1+(-1)+(-1)+(-1)+...+(-1)
có tất cả các số -1 trên dãy số trên là
(2012-2);2+1=1006
vậy suy ra ; -1x1006=(-1006)
chac chan la dung
a) \(\left(2\frac{5}{6}+1\frac{4}{9}\right):\left(10\frac{1}{12}-9\frac{1}{2}\right)\)
\(=\left(\frac{17}{6}+\frac{13}{9}\right):\left(\frac{121}{12}-\frac{19}{2}\right)\)
\(=\frac{77}{18}:\frac{7}{12}\)
\(=\frac{22}{3}\)
b) \(1\frac{5}{18}-\frac{5}{18}.\left(\frac{1}{15}+1\frac{1}{12}\right)\)
\(=\frac{23}{18}-\frac{5}{18}.\left(\frac{1}{15}+\frac{13}{12}\right)\)
\(=\frac{23}{18}-\frac{5}{18}.\frac{23}{20}\)
\(=\frac{23}{18}-\frac{23}{72}\)
\(=\frac{23}{24}\)
c) \(-1\frac{1}{7}.\left(9\frac{1}{2}-8,75\right):\frac{2}{7}+0,625:1\frac{2}{3}\)
\(=\frac{-8}{7}.\left(\frac{19}{2}-\frac{35}{4}\right):\frac{2}{7}+\frac{5}{8}:\frac{5}{3}\)
\(=\frac{-8}{7}.\frac{3}{4}:\frac{2}{7}+\frac{3}{8}\)
\(=\frac{-8}{7}.\frac{3}{4}.\frac{7}{2}+\frac{3}{8}\)
\(=-3+\frac{3}{8}\)
\(=\frac{-21}{8}\)
Chúc bn học tốt !!
\(\left(\frac{3}{8}+-\frac{3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
\(=\left(\frac{9}{24}+-\frac{18}{24}+\frac{14}{24}\right):\frac{5}{6}+\frac{1}{2}\)
\(=\frac{5}{24}:\frac{5}{6}+\frac{1}{2}\)
\(=\frac{5}{24}.\frac{6}{5}+\frac{1}{2}\)
\(=\frac{1}{4}+\frac{1}{2}\)
\(=\frac{1}{4}+\frac{2}{4}\)
\(=\frac{3}{4}\)
\(\frac{1}{2}+\frac{3}{4}-\left(\frac{3}{4}-\frac{4}{5}\right)\)
\(=\frac{1}{2}+\frac{3}{4}-\left(\frac{15}{20}-\frac{16}{20}\right)\)
\(=\frac{1}{2}+\frac{3}{4}-\frac{-1}{20}\)
\(=\frac{10}{20}+\frac{15}{20}-\frac{-1}{20}\)
\(=\frac{25}{20}-\frac{-1}{20}\)
\(=\frac{26}{20}\)
\(=\frac{13}{10}\)
a/ \(\frac{1}{-2}:\frac{-1}{3}-\frac{1}{6}\)
\(=\frac{-1}{2}.\frac{3}{-1}-\frac{1}{6}\)
= \(\frac{-3}{-2}-\frac{1}{6}\)
\(=\frac{3}{2}-\frac{1}{6}\)
\(=\frac{9}{6}-\frac{1}{6}\)
\(=\frac{8}{6}\)
\(=\frac{4}{3}\)
b/ \(\left(2\frac{4}{15}-80\%\right):\left(-3\frac{2}{3}\right)+\frac{-45}{-75}\)
\(=\left(\frac{34}{15}-\frac{80}{100}\right):\left(\frac{-11}{3}\right)+\frac{45}{75}\)
\(=\left(\frac{34}{15}-\frac{4}{5}\right).\frac{3}{-11}+\frac{3}{5}\)
\(=\left(\frac{34}{15}-\frac{12}{15}\right).\frac{-3}{11}+\frac{3}{5}\)
\(=\frac{22}{15}.\frac{-3}{11}+\frac{3}{5}\)
\(=\frac{-66}{165}+\frac{3}{5}\)
\(=\frac{-66}{165}+\frac{99}{165}\)
= \(\frac{33}{165}\)
= \(\frac{11}{53}\)
Nhầm:
b/ \(=\frac{11}{55}\)