Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(\frac{9}{16}\right)^{2016}.\left(\frac{16}{9}\right)^{2015}.\frac{4}{3}=\left(\frac{9}{16}\right)^{2016}:\left(\frac{9}{16}\right)^{2015}.\frac{4}{3}\)
\(=\frac{9}{16}.\frac{4}{3}\)
\(=\frac{3}{4}\)
\(\left(\frac{9}{16}\right)^{2016}\cdot\left(\frac{16}{9}\right)^{2015}\cdot\frac{4}{3}=\left(\frac{9}{16}\right)^{2016}\cdot\left(\frac{9}{16}\right)^{2015}\cdot\frac{4}{3}\)
\(=\frac{9}{16}\cdot\frac{4}{3}\)
\(=\frac{3}{4}\)
\(\left(\frac{9}{16}\right)^{2016}.\left(\frac{16}{9}\right)^{2015}.\frac{4}{3}\)
=\(\left(\frac{3}{4}\right)^{4032}.\left(\frac{4}{3}\right)^{4030}.\frac{4}{3}\)
=
=\(\frac{9^{2016}}{16^{2016}}.\frac{16^{2015}}{9^{2015}}.\frac{4}{3}\)
=\(\frac{9}{16}.\frac{4}{3}\)
=\(\frac{3}{4}\)
k cho mk nhoa
\(\left(\frac{9}{16}\right)^{2016}.\left(\frac{16}{9}\right)^{2015}.\frac{4}{3}\)
\(=\left[\frac{9}{16}\left(\frac{9}{16}\right)^{2015}\right].\left(\frac{16}{9}\right)^{2015}.\frac{4}{3}\)
\(=\frac{9}{16}\left[\left(\frac{9}{16}\right)^{2015}.\left(\frac{16}{9}\right)^{2015}\right].\frac{4}{3}\)
\(=\frac{9}{16}\left[\left(\frac{9}{16}.\frac{16}{9}\right)^{2015}\right].\frac{4}{3}\)
\(=\frac{9}{16}.1^{2015}.\frac{4}{3}\)
\(=\frac{9}{16}.\frac{4}{3}\)
\(=\frac{3}{4}\)
=(16/10)2015.(16/9)(4/3)
=(8/5)2015.(4/3)3
(bn xem lại đề, có thể là 9/16 chứ k phải là 9/10)?
\(b,\left(\sqrt{1\frac{9}{16}-\sqrt{\frac{9}{16}}}\right):5\)
\(=\left(\sqrt{\frac{25}{16}-\frac{3}{4}}\right):5\)
\(=\sqrt{\frac{13}{16}}:5\)
\(=\frac{\sqrt{13}}{4}:5\)
\(=\frac{\sqrt{13}}{20}\)
\(=\frac{16}{5}.\frac{15}{16}-\left(\frac{3}{4}+\frac{2}{7}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{21}{28}+\frac{8}{28}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{29}{28}\right).\left(\frac{-28}{29}\right)\)
\(=3-\left(-1\right)\)
\(=4\)
b) \(=\left(\frac{1}{4}+\frac{25}{2}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3}{8}-\frac{1}{12}\right)\right)\)
\(=\left(\frac{4}{16}+\frac{200}{16}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3.3}{2.3.4}-\frac{2}{2.3.4}\right)\right)\)
\(=\left(\frac{199}{16}\right):\left(12-\frac{7}{12}:\left(\frac{9}{24}-\frac{2}{24}\right)\right)\)
\(=\frac{199}{16}:\left(12-\frac{7}{12}.\frac{24}{7}\right)\)
\(=\frac{199}{16}:\left(12-2\right)\)
\(=\frac{199}{16}:10\)
\(=\frac{199}{160}\)
c) \(\left(\frac{-3}{5}+\frac{5}{11}\right):\frac{-3}{7}+\left(\frac{-2}{5}+\frac{6}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-33}{55}+\frac{25}{55}\right):\frac{-3}{7}+\left(\frac{4}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-8}{55}\right).\frac{-7}{3}+\frac{4}{5}.\frac{-7}{3}\)
\(\frac{-7}{3}\left(\frac{-8}{55}+\frac{4}{5}\right)\)
\(\frac{-7}{3}.\frac{36}{55}=\frac{-84}{55}\)
\(3\frac{1}{2}-\frac{1}{2}.\left(-4,25-\frac{3}{4}\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-4,25-0,75\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-5\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.5.\frac{4}{5}\)
\(=\frac{7}{2}-2\)
\(=\frac{7}{2}-\frac{4}{2}\)
\(=\frac{3}{2}\)
\(\frac{3}{7}.1\frac{1}{2}+\frac{3}{7}.0,5-\frac{3}{7}.9\)
\(=\frac{3}{7}.\left(\frac{3}{2}+\frac{1}{2}-9\right)\)
\(=\frac{3}{7}.\left(2-9\right)\)
\(=\frac{3}{7}.\left(-7\right)\)
\(=-3\)
\(\frac{125^{2016}.8^{2017}}{50^{2017}.20^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^2\right)^{2017}.2^{2017}.\left(2^2\right)^{2018}.5^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^3\right)^{2017}.\left(2^3\right)^{2017}.2.5}=\frac{1}{5^4.2}=\frac{1}{1250}\)( tính nhẩm, ko chắc đúng )
1
a) \(3\frac{1}{2}-\frac{1}{2}\cdot\left(-4,25-\frac{3}{4}\right)^2\) : \(\frac{5}{4}\)
= \(3\cdot25:\frac{5}{4}\)
= \(3\cdot\left(25:\frac{5}{4}\right)\)
=\(3\cdot20\)
=60
b)=\(\frac{3}{7}\cdot\left(1\frac{1}{2}+0,5-9\right)\)
=\(\frac{3}{7}\cdot\left(-7\right)\)
=\(-3\)
c) =
Ta có : \(\frac{\left(\frac{2}{5}\right)^9.10^9-\left(\frac{-9}{4}\right)^5:\left(\frac{-3}{16}\right)^{10}}{4^{12}+16^9}\)
=\(\frac{\left(\frac{2}{5}.10\right)^9-\left[\left(\frac{-3}{2}\right)^2\right]^5:\left(\frac{-3}{16}\right)^{10}}{\left(2^2\right)^{12}+\left(2^4\right)^9}\)=\(\frac{4^9-\left(\frac{-3}{2}\right)^{10}:\left(\frac{-3}{16}\right)^{10}}{2^{24}+2^{36}}\)=\(\frac{4^9-\left[\left(\frac{-3}{2}\right):\left(\frac{-3}{16}\right)\right]^{10}}{2^{24}\left(1+2^{12}\right)}\)
=\(\frac{\left(2^2\right)^9-8^{10}}{2^{24}\left(1+2^{12}\right)}\)=\(\frac{\left(2^2\right)^9-\left(2^3\right)^{10}}{2^{24}\left(1+2^{12}\right)}\)=\(\frac{2^{18}-2^{30}}{2^{24}\left(1+2^{12}\right)}\)=\(\frac{2^{18}\left(1-2^{12}\right)}{2^{24}\left(1+2^{12}\right)}\)=\(\frac{1-2^{12}}{2^6\left(1+2^{12}\right)}\)
Không biết đúng hong nha!
=\(\frac{3}{4}\)
= 3/8
(calculator đúng)