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Cũng khuya rồi , mình làm câu 1 thôi nhé !
\(\frac{2.5^{22}-9.5^{21}}{25^{10}}=\frac{2.5^{22}-9.5^{21}}{\left(5^2\right)^{10}}\)
\(\frac{5^{21}.\left(2.5-9\right)}{5^{20}}=5.\left(10-9\right)=5\)
a) \(\left(-\frac{5}{2}\right)^2:\left(-15\right)-\left(-0,45+\frac{3}{4}\right).\left(-1\frac{5}{9}\right)\)
= \(-\frac{25}{4}:\left(-15\right)-\left(\frac{9}{20}+\frac{15}{20}\right).\left(-\frac{14}{9}\right)\)
=\(-\frac{25}{4}.\frac{1}{-15}-\frac{6}{5}.\left(-\frac{14}{9}\right)\)
= \(\frac{-5}{12}-\frac{8}{5}\)
= \(\frac{\left(-25\right)-96}{60}\)
= \(\frac{\left(-25\right)+\left(-96\right)}{60}\)
=\(\frac{121}{60}\)
b) \(\left(\frac{-1}{3}\right)-\left(\frac{-3}{5}\right)^0+\left(1-\frac{1}{2}\right)^2:2\)
= \(\left(\frac{-1}{3}\right)-1+\left(\frac{1}{2}\right)^2.\frac{1}{2}\)
=\(\left(\frac{-1}{3}\right)-\frac{3}{3}+\frac{1}{4}.\frac{1}{2}\)
= \(\frac{-4}{3}+\frac{1}{8}\)=\(\frac{-32+3}{24}\)
=\(\frac{-29}{24}\)
c) E=\(\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)
=\(\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.6^9}{2^{10}.3^8+6^8.20}\)
=\(\frac{2^{10}.3^8-2.6^9}{2^{10}.3^8+6^8.20}\)
=\(\frac{3}{5}\)
d)\(\frac{5^4.20^4}{25^5.4^5}\)
=\(\frac{\left(5.20\right)^4}{\left(25.4\right)^5}\)
=\(\frac{100^4}{100^5}\)
=\(\frac{1}{100}\)
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: \(\left(\frac{1}{16}\right)^{10}=\left(\frac{1}{2^4}\right)^{10}=\frac{1}{2^{40}}\)
\(\left(\frac{1}{2}\right)^{50}=\frac{1}{2^{50}}\)
\(\Rightarrow\frac{1}{2^{40}}>\frac{1}{2^{50}}\)
\(\Rightarrow\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)
\(3\frac{1}{2}-\frac{1}{2}.\left(-4,25-\frac{3}{4}\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-4,25-0,75\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-5\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.5.\frac{4}{5}\)
\(=\frac{7}{2}-2\)
\(=\frac{7}{2}-\frac{4}{2}\)
\(=\frac{3}{2}\)
\(\frac{3}{7}.1\frac{1}{2}+\frac{3}{7}.0,5-\frac{3}{7}.9\)
\(=\frac{3}{7}.\left(\frac{3}{2}+\frac{1}{2}-9\right)\)
\(=\frac{3}{7}.\left(2-9\right)\)
\(=\frac{3}{7}.\left(-7\right)\)
\(=-3\)
\(\frac{125^{2016}.8^{2017}}{50^{2017}.20^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^2\right)^{2017}.2^{2017}.\left(2^2\right)^{2018}.5^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^3\right)^{2017}.\left(2^3\right)^{2017}.2.5}=\frac{1}{5^4.2}=\frac{1}{1250}\)( 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) =