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a) \(\sqrt{0,01}+\sqrt{0,16}-12^0=0,1+0,4-1=-0,5\)
b) \(\left(\frac{1}{16}\right)^7:\left(\frac{1}{8}\right)^9=\left[\left(\frac{1}{2}\right)^4\right]^7:\left[\left(\frac{1}{2}\right)^3\right]^9=\left(\frac{1}{2}\right)^{28}:\left(\frac{1}{2}\right)^{27}=\frac{1}{2}\)
a)
\(=\sqrt{0,1^2}+\sqrt{0,4^2}-1\)
\(=|0,1|+|0,4|-1\)
\(=0,1+0,4-1=-0,5\)
b)
= \(\left(\left(\frac{1}{2}\right)^4\right)^7:\left(\left(\frac{1}{2}\right)^3\right)^9\)
=\(\left(\frac{1}{2}\right)^{4\cdot7}:\left(\frac{1}{2}\right)^{3\cdot9}\)
= \(\left(\frac{1}{2}\right)^{28}:\left(\frac{1}{2}\right)^{27}\)
\(=\left(\frac{1}{2}\right)^{28-27}\)
\(=\left(\frac{1}{2}\right)^1=\frac{1}{2}\)
\(a,\frac{-5}{9}.\left(\frac{3}{10}-\frac{2}{5}\right)\)
\(=\frac{-5}{9}.\frac{-1}{10}\)
\(=\frac{1}{18}\)
\(b,2^8:2^5+3^3.2-12\)
\(=2^3+9.2-12\)
\(=8+18-12\)
\(=26-12\)
\(=14\)
Câu c,d em chưa học nên không biết làm ạ, mong mọi người thông cảm!!!
Sửa lại câu b
\(=2^3+27.2-12\)
\(=8+54-12\)
\(=62-12\)
\(=50\)
a) \(\frac{3}{5}:\left(\frac{-1}{15}-\frac{1}{6}\right)+\frac{3}{5}:\left(\frac{-1}{3}-1\frac{1}{15}\right)\)
\(=\frac{3}{5}:\frac{-7}{30}+\frac{3}{5}:\frac{-7}{5}\)
\(=\frac{3}{5}\cdot\frac{30}{-7}+\frac{3}{5}\cdot\frac{5}{-7}\)
\(=\frac{3}{5}\left(\frac{-30}{7}+\frac{-5}{7}\right)=\frac{3}{5}\cdot-5=-3\)
b) \(10\cdot\sqrt{0,01}\cdot\sqrt{\frac{16}{9}}+3\sqrt{49}-\frac{1}{6}\sqrt{4}\)
\(=10\cdot\frac{1}{10}\cdot\frac{4}{3}+3\cdot7-\frac{1}{6}\cdot2\)
\(=\frac{4}{3}+21-\frac{2}{6}=22\)
\(3\frac{1}{2}-\frac{1}{2}.\left(-4,25-\frac{3}{4}\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-4,25-0,75\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-5\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.5.\frac{4}{5}\)
\(=\frac{7}{2}-2\)
\(=\frac{7}{2}-\frac{4}{2}\)
\(=\frac{3}{2}\)
\(\frac{3}{7}.1\frac{1}{2}+\frac{3}{7}.0,5-\frac{3}{7}.9\)
\(=\frac{3}{7}.\left(\frac{3}{2}+\frac{1}{2}-9\right)\)
\(=\frac{3}{7}.\left(2-9\right)\)
\(=\frac{3}{7}.\left(-7\right)\)
\(=-3\)
\(\frac{125^{2016}.8^{2017}}{50^{2017}.20^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^2\right)^{2017}.2^{2017}.\left(2^2\right)^{2018}.5^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^3\right)^{2017}.\left(2^3\right)^{2017}.2.5}=\frac{1}{5^4.2}=\frac{1}{1250}\)( 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) =
1. a) 3+2=5
b) 0,5-0,1=0,4
c) 4/5-1/9=31/45
d) 2-0,6=1,4
2. a) 8-4+3=7
b) 11+5-3=13
c) 3/2-4/6-7-37/6
d) 4+5-6=3
\(a,\left(\frac{1}{\sqrt{625}}+\frac{1}{5}+1\right):\left(\frac{1}{25}-\frac{1}{\sqrt{25}}-1\right)\)
\(=\left(\frac{1}{25}+\frac{1}{5}+1\right):\left(\frac{1}{25}-\frac{1}{5}-1\right)\)
\(=\frac{31}{25}:\left(-\frac{29}{25}\right)\)
\(=\frac{31}{25}.\frac{-25}{29}\)
\(=-\frac{31}{29}\)
\(b,\left[18\frac{1}{6}-\left(0,06:7\frac{1}{2}+3\frac{2}{5}.0,38\right)\right]:\left(19-2\frac{2}{3}.4\frac{3}{4}\right)\)
\(=\left[\frac{109}{6}-\left(\frac{3}{50}:\frac{15}{2}+\frac{17}{5}.\frac{19}{50}\right)\right]:\left(19-\frac{8}{3}.\frac{19}{4}\right)\)
\(=\left(\frac{109}{6}-\frac{13}{10}\right):\frac{19}{3}\)
\(=\frac{253}{15}.\frac{3}{19}\)
\(=\frac{253}{95}\)
Số to :v
\(a)=\frac{7}{25}+\frac{4}{13}-\frac{5}{2}+\frac{18}{25}-\frac{17}{13}\)
\(=1-1-\frac{5}{2}\)
\(=-\frac{5}{2}\)