Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\left(\frac{11}{12}:\frac{44}{16}\right).\left(\frac{-1}{3}+\frac{1}{2}\right)\) \(=\left(\frac{11}{12}.\frac{16}{44}\right).\left(\frac{-2}{6}+\frac{3}{6}\right)\) \(=\frac{1}{3}.\frac{1}{6}\) \(=\frac{1}{18}\)
b) \(\frac{\left(-5\right)^2.\left(-5\right)^3.16}{5^4.\left(-2\right)^4}\) \(=\frac{\left(-5\right)^5.2^4}{5^4.\left(-2\right)^4}\) \(=5\) (Có sửa đề lại, nếu có sai thì ib mình sửa lại nhé!)
c) \(7,5:\left(\frac{-5}{3}\right)+2\frac{1}{2}:\left(\frac{-5}{3}\right)\) \(=\frac{15}{2}.\left(\frac{-3}{5}\right)+\frac{5}{2}.\left(\frac{-3}{5}\right)\) \(=\frac{-3}{5}.\left(\frac{15}{2}+\frac{5}{2}\right)\)
\(=\frac{-3}{5}.10\) \(=-6\)
d) \(\left(\frac{-1}{2}+\frac{1}{3}\right).\frac{4}{5}+\left(\frac{2}{3}+\frac{1}{2}\right):\frac{5}{4}\) \(=\left(\frac{-1}{2}+\frac{1}{3}\right).\frac{4}{5}+\left(\frac{2}{3}+\frac{1}{2}\right).\frac{4}{5}\)
\(=\frac{4}{5}.\left(\frac{-1}{2}+\frac{1}{3}+\frac{2}{3}+\frac{1}{2}\right)\) \(=\frac{4}{5}.\left(\frac{0}{2}+1\right)\) \(=\frac{4}{5}.1=\frac{4}{5}\)
a) (1112:4416).(−13+12)(1112:4416).(−13+12) =(1112.1644).(−26+36)=(1112.1644).(−26+36) =13.16=13.16 =118=118
b) (−5)2.(−5)3.1654.(−2)4(−5)2.(−5)3.1654.(−2)4 =(−5)5.2454.(−2)4=(−5)5.2454.(−2)4 =5=
c) 7,5:(−53)+212:(−53)7,5:(−53)+212:(−53) =152.(−35)+52.(−35)=152.(−35)+52.(−35) =−35.(152+52)=−35.(152+52)
=−35.10=−35.10 =−6=−6
d) (−12+13).45+(23+12):54(−12+13).45+(23+12):54 =(−12+13).45+(23+12).45=(−12+13).45+(23+12).45
=45.(−12+13+23+12)=45.(−12+13+23+12) =45.(02+1)=45.(02+1) =45.1=45
\(D=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2017}{2018}\)
\(D=\frac{1}{2018}\)
Vậy \(D=\frac{1}{2018}\)
\(E=\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+...+\frac{1}{64.69}+\frac{1}{69.74}\)
\(E=\frac{1}{5}\cdot\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{69}-\frac{1}{74}\right)\)
\(E=\frac{1}{5}\cdot\left(\frac{1}{4}-\frac{1}{74}\right)\)
\(E=\frac{1}{5}\cdot\frac{35}{148}=\frac{7}{148}\)
Vậy E = ...
a) \(\frac{-6}{21}.\frac{3}{2}=-\frac{3}{7}\) b) \(\left(-3\right).\left(\frac{-7}{12}\right)=\frac{21}{12}=\frac{7}{4}\)
c) \(\left(\frac{11}{12}:\frac{33}{16}\right).\frac{3}{5}=\frac{11}{12}.\frac{16}{33}.\frac{3}{5}=\frac{4}{15}\)
d) \(\sqrt{\left(-7\right)^2}+\sqrt{\frac{2}{16}}=7+\sqrt{\frac{1}{8}}\)
c) \(\frac{1}{2}.\sqrt{100}-\sqrt{\frac{1}{16}}+\left(\frac{1}{3}\right)^0=\frac{1}{2}.10-\frac{1}{4}+1=5\frac{3}{4}\)
b)
\(x-2.\left(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)=\frac{16}{9}\)
\(x-2\cdot\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)
\(x-2=\frac{16}{9}:\left(\frac{1}{3}-\frac{1}{9}\right)\)
\(x-2=8\)
=> x = 10
a)
\(A=\frac{1}{2}.\frac{2}{3}\cdot\frac{3}{4}\cdot\cdot\cdot\frac{2013}{2014}\cdot\frac{2014}{2015}\cdot\frac{2015}{2016}\)
\(A=\frac{1}{2016}\)
\(\frac{2^{76}-2^{74}}{2^{78}-2^{76}}=\frac{2^{74}\left(2^2-1\right)}{2^{74}\left(2^4-2^2\right)}=\frac{4-1}{16-4}=\frac{1}{4}\)
\(1+\frac{1}{11}=\frac{12}{11}\)
\(1+\frac{1}{12}=\frac{13}{12}\)
\(1+\frac{1}{13}=\frac{14}{13}\)
\(1+\frac{1}{14}=\frac{15}{14}\)
\(1+\frac{1}{15}=\frac{16}{15}\)
\(1+\frac{1}{16}=\frac{17}{16}\)
\(\left(1+\frac{1}{11}\right)\left(1+\frac{1}{12}\right)...\left(1+\frac{1}{16}\right)=\frac{12.13.14.15.16.17}{11.12.13.14.15.16}=\frac{17}{11}\)