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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) \(2\frac{3}{4}\cdot\left(-0,4\right)-1\frac{3}{5}\cdot2,75+1,2:\frac{4}{11}\)
\(=2\frac{3}{4}\cdot\left(-\frac{2}{5}\right)-1\frac{3}{5}\cdot\frac{11}{4}+\frac{6}{5}:\frac{4}{11}\)
\(=\frac{11}{4}\cdot\left(-\frac{2}{5}\right)-1\frac{3}{5}\cdot\frac{11}{4}+\frac{6}{5}\cdot\frac{11}{4}\)
\(=\frac{11}{4}\left(-\frac{2}{5}-1\frac{3}{5}+\frac{6}{5}\right)\)
\(=\frac{11}{4}\left(-\frac{2}{5}-\frac{8}{5}+\frac{6}{5}\right)\)
\(=\frac{11}{4}\cdot\left(-\frac{4}{5}\right)=\frac{11}{1}\cdot\left(-\frac{1}{5}\right)=-\frac{11}{5}\)
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 : \(\frac{930}{C}=\frac{930}{465}=2\)
\(=\frac{12}{7}\cdot\frac{3}{4}-\frac{6}{7}\cdot\frac{4}{3}+\frac{6}{7}\)
\(=\frac{6}{7}\left(\frac{3}{2}-\frac{4}{3}+1\right)\)
\(=\frac{6}{7}\left(\frac{1}{6}+1\right)=\frac{6}{7}\cdot\frac{7}{6}=1\)
2.
\(=2017\cdot2018\cdot\left[\left(2016\cdot2018\right)-\left(2016\cdot2017\right)\right]\)
\(=2017\cdot2018\cdot2016\left(2018-2017\right)=2016\cdot2017\cdot2018\)
3.
\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)....\left(\frac{1}{100}-1\right)=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot....\cdot\frac{99}{100}\)
\(=\frac{1}{100}\)
4.
\(=\frac{1+2+2^2+2^4+...+2^9}{2\left(1+2+2^2+2^3+2^4+...+2^9\right)}\)
\(=\frac{1}{2}\)
mình chỉ làm được câu 3 thôi
có \(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)....\left(\frac{1}{100}-1\right)\)
\(=\frac{-1}{2}\times\frac{-2}{3}\times....\times\frac{-99}{100}\)
\(=\frac{\left(-1\right)\left(-2\right)....\left(-99\right)}{2\times3\times....\times100}\)
\(=\frac{-\left(1\times2\times....\times99\right)}{2\times3\times....\times100}\)
\(=\frac{-1}{100}\)
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 !!
Ta có :
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{10}\right)\)
\(=\)\(\frac{2-1}{2}.\frac{3-1}{3}.....\frac{10-1}{10}\)
\(=\)\(\frac{1}{2}.\frac{2}{3}.....\frac{9}{10}\)
\(=\)\(\frac{1.2.....9}{2.3.....10}\)
\(=\)\(\frac{2.3.....9}{2.3.....9}.\frac{1}{10}\)
\(=\)\(\frac{1}{10}\)
Vậy \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{10}\right)\)
Chúc bạn học tốt ~
=1/2 . 2/3 ....... 9/10
=1.2.3......9/2.3.4........10
=1/10
Chúc bạn học tốt