Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
( -1,25) . 14,7. ( -8 )
= \(\frac{-5}{4}\). \(\frac{147}{10}\)\(.\frac{\left(-8\right)}{1}\)
= \(\frac{\left(-5\right).147.\left(-8\right)}{4.10.1}.\)
= \(\frac{\left[\left(-5\right).\left(-8\right)\right].147}{\left(4.10\right).1}\)
= \(\left[\frac{-5}{4}.\frac{-8}{10}\right].147\)
=147
\(3\frac{1}{2}-\frac{1}{2}.\left(-4,25-\frac{3}{4}\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-4,25-0,75\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-5\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.5.\frac{4}{5}\)
\(=\frac{7}{2}-2\)
\(=\frac{7}{2}-\frac{4}{2}\)
\(=\frac{3}{2}\)
\(\frac{3}{7}.1\frac{1}{2}+\frac{3}{7}.0,5-\frac{3}{7}.9\)
\(=\frac{3}{7}.\left(\frac{3}{2}+\frac{1}{2}-9\right)\)
\(=\frac{3}{7}.\left(2-9\right)\)
\(=\frac{3}{7}.\left(-7\right)\)
\(=-3\)
\(\frac{125^{2016}.8^{2017}}{50^{2017}.20^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^2\right)^{2017}.2^{2017}.\left(2^2\right)^{2018}.5^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^3\right)^{2017}.\left(2^3\right)^{2017}.2.5}=\frac{1}{5^4.2}=\frac{1}{1250}\)( 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) =
\(a)=\frac{7}{25}+\frac{4}{13}-\frac{5}{2}+\frac{18}{25}-\frac{17}{13}\)
\(=1-1-\frac{5}{2}\)
\(=-\frac{5}{2}\)
Bài 1 và Bài 2 dễ, bn có thể tự làm được!
Bài 3:
a) ta có: 1020 = (102)10 = 10010
=> 10010>910
=> 1020>910
b) ta có: (-5)30 = 530 =( 53)10 = 12510 ( vì là lũy thừa bậc chẵn)
(-3)50 = 350 = (35)10= 24310
=> 12510 < 24310
=> (-5)30 < (-3)50
c) ta có: 648 = (26)8= 248
1612 = ( 24)12 = 248
=> 648 = 1612
d) ta có: \(\left(\frac{1}{16}\right)^{10}=\left(\frac{1}{2^4}\right)^{10}=\frac{1}{2^{40}}\)
\(\left(\frac{1}{2}\right)^{50}=\frac{1}{2^{50}}\)
\(\Rightarrow\frac{1}{2^{40}}>\frac{1}{2^{50}}\)
\(\Rightarrow\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)
Bài 1:
a) \(0,5-\frac{5}{41}+\frac{1}{2}-\frac{36}{41}\)
\(=\frac{1}{2}-\frac{5}{41}+\frac{1}{2}-\frac{36}{41}\)
\(=\left(\frac{1}{2}+\frac{1}{2}\right)-\left(\frac{5}{41}+\frac{36}{41}\right)\)
\(=1-1\)
\(=0.\)
b) \(\left(-\frac{2}{3}+\frac{3}{7}\right):\frac{4}{5}+\left(-\frac{1}{3}+\frac{4}{7}\right):\frac{4}{5}\)
\(=-\frac{2}{3}+\frac{3}{7}:\frac{4}{5}-\frac{1}{3}+\frac{4}{7}:\frac{4}{5}\)
\(=\left[\left(-\frac{2}{3}\right)-\frac{1}{3}\right]+\left(\frac{3}{7}+\frac{4}{7}\right):\frac{4}{5}\)
\(=\left(-1\right)+1:\frac{4}{5}\)
\(=\left(-1\right)+\frac{5}{4}\)
\(=\frac{1}{4}.\)
c) \(\left(-\frac{3}{4}\right).\sqrt{\frac{16}{9}+3.\sqrt{49}}\)
\(=\left(-\frac{3}{4}\right).\sqrt{\frac{16}{9}+3.7}\)
\(=\left(-\frac{3}{4}\right).\sqrt{\frac{16}{9}+21}\)
\(=\left(-\frac{3}{4}\right).\sqrt{\frac{205}{9}}\)
\(=\left(-\frac{3}{4}\right).\frac{\sqrt{205}}{3}\)
\(=-\frac{\sqrt{205}}{4}.\)
d) \(\left(-\frac{1}{3}\right)^2.\frac{4}{11}+1\frac{5}{11}.\left(\frac{1}{3}\right)^2\)
\(=\frac{1}{9}.\frac{4}{11}+\frac{16}{11}.\frac{1}{9}\)
\(=\frac{1}{9}.\left(\frac{4}{11}+\frac{16}{11}\right)\)
\(=\frac{1}{9}.\frac{20}{11}\)
\(=\frac{20}{99}.\)
Chúc bạn học tốt!
\(\text{a,}\frac{2}{13}.\frac{-5}{3}+\frac{11}{13}.\frac{-5}{3}=-\frac{5}{3}\left(\frac{2}{13}+\frac{11}{13}\right)\)
\(=\frac{-5}{3}.\frac{13}{13}\)
\(=-\frac{5}{3}\)
\(\text{b,}\left(-\frac{1}{3}\right)^2+\left(-\frac{1}{3}\right)^3.27+\left(\frac{-2017}{2018}\right)^0=\frac{1}{9}-\frac{1}{27}.27+1\)
\(=\frac{1}{9}-1+1\)
\(=\frac{1}{9}\)
\(\text{c,}1,2-\sqrt{\frac{1}{4}}:1\frac{1}{20}+\left|\frac{3}{4}-1,25\right|-\left(\frac{-3}{2}\right)^2=\frac{6}{5}-\frac{1}{2}:\frac{21}{20}+\left|\frac{3}{4}-\frac{5}{4}\right|-\frac{9}{4}\)
\(=\frac{6}{5}-\frac{10}{21}+\frac{1}{2}-\frac{9}{4}\)
\(=\frac{-431}{420}\)
Thanks you !