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\(\frac{5^4.18^4}{125.9^5.2^4}=\frac{\left(5.18\right)^4}{5^3.9^3.9^2.2^3.2}\)

\(=\frac{90^4}{\left(5.9.2\right)^3.9^2.2}\)\(=\frac{90^4}{90^3.81.2}\)

\(=\frac{90}{162}=\frac{5}{9}\)

20 tháng 11 2016

Câu 1 :

\(\frac{\left(-5\right)^{32}.20^{43}}{\left(-8\right)^{29}.125^5}\)

= \(\frac{5^{32}.2^{86}.5^{43}}{\left(-2\right)^{87}.5^{15}}\)

= \(\frac{5^{72}.\left(-2\right)^{86}}{\left(-2\right)^{87}.5^{75}}\)

= \(\frac{1}{-2}\)

Câu 2 :

\(\frac{5^4.18^4}{125.9^5.16}\)

= \(\frac{5^4.2^4.3^8}{5^3.3^{10}.2^4}\)

= \(\frac{5}{3^2}\)

= \(\frac{5}{9}\)

Câu 3 :

\(\frac{9^{18}.2^{29}}{8^9.27^{12}}\)

= \(\frac{3^{36}.2^{29}}{2^{27}.3^{36}}\)

= \(2^2\)

= 4

20 tháng 11 2016

mk k bít hihi

13 tháng 11 2015

bài này không khó. Nhưng đánh máy để giải cho bạn thì thực sự khó

10 tháng 8 2020

Ta có :\(\frac{6^8.2^4-4^5.18^4}{27^3.8^4-3^9.2^{13}}=\frac{\left(2.3\right)^8.2^4-\left(2^2\right)^5.\left(3^2.2\right)^4}{\left(3^3\right)^3.\left(2^3\right)^4-3^9.2^{13}}=\frac{2^{12}.3^8-2^{14}.3^8}{3^9.2^{12}-3^9.2^{13}}=\frac{3^8.2^{12}.\left(2^2-1\right)}{3^9.2^{12}.\left(1-2\right)}\)

\(=\frac{3^9.2^{12}}{-3^9.2^{12}}=-1\)

10 tháng 8 2020

\(\frac{6^8\cdot2^2-4^5\cdot18^4}{27^3\cdot8^4-3^9\cdot2^{13}}\)

\(=\frac{\left(2.3\right)^8.2^4-\left(2^2\right)^5.\left(3^2.2\right)^4}{\left(3^3\right)^3.\left(2^3\right)^4-3^9.2^{13}}\)

\(=\frac{2^{12}.3^8-2^{14}.3^8}{3^9.2^{12}-3^9.2^{14}}\)

\(=\frac{3^8.2^{12}.\left(2^2-1\right)}{3^9.2^{12}.\left(1-2\right)}\)

\(=\frac{3^9.2^{12}}{-3^9.2^{12}}=-1\)

27 tháng 4 2017

A=1.5.(3.2)+2.10.(6.2)+3.15.(9.2)+4.20.(12.2)+5.25.(15.2)

1.3.5+2.6.10+3.9.15+4.12.20+5.15.25

A=1.5.3+2.10.6+3.15.9+4.20.12+5.25.15(2.2.2.2.2)

1.3.5+2.6.10+3.9.15+4.12.20+5.15.25

A=2.2.2.2.2

A=32

27 tháng 4 2017

\(\frac{1\cdot3\cdot5\cdot2+2\cdot10\cdot6\cdot2+3\cdot15\cdot9\cdot2+4\cdot20\cdot12\cdot2+5\cdot25\cdot15\cdot2}{1\cdot3\cdot5+2\cdot10\cdot6+3\cdot15\cdot9+4\cdot20\cdot12+5\cdot25\cdot15 }\)

\(2\cdot2\cdot2\cdot2\cdot2=2^5\)

\(=32\)

\(\Rightarrow A=4.\left[\frac{6}{2.\left(2.4\right)}+\frac{5}{\left(2.4\right).13}+\frac{3}{13.\left(4.4\right)}+\frac{2}{\left(4.4\right).18}+\frac{10}{18.\left(7.4\right)}\right]\)

\(=4.\left(\frac{6}{2.8}+\frac{5}{8.13}+\frac{3}{13.16}+\frac{2}{16.18}+\frac{10}{18.28}\right)=4.\left(\frac{1}{2}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{18}+\frac{1}{18}-\frac{1}{28}\right)\)

\(=4.\left(\frac{1}{2}-\frac{1}{28}\right)=4.\frac{13}{28}=\frac{13}{7}\)

4 tháng 8 2020

Thank you <3

14 tháng 4 2017

=\(\frac{30\left(1+8+27+64+125\right)}{5\left(3+24+81+64+375\right)}\)

\(\frac{30.225}{5.574}\)

=\(\frac{6750}{2870}\)

=\(\frac{675}{287}\)

K mình với!!!!

11 tháng 12 2019

Bài 1:

a) \(\left(\frac{9}{25}-2.18\right):\left(3\frac{4}{5}+0,2\right)\)

\(=\left(\frac{9}{25}-36\right):\left(\frac{19}{5}+\frac{1}{5}\right)\)

\(=\left(\frac{9}{25}-\frac{900}{25}\right):4\)

\(=-\frac{891}{25}.\frac{1}{4}\)

\(=-\frac{891}{100}\)

b) \(\frac{3}{8}.19\frac{1}{3}-\frac{3}{8}.33\frac{1}{3}\)

\(=\frac{3}{8}.\frac{58}{3}-\frac{3}{8}.\frac{100}{3}\)

\(=\frac{3}{8}\left(\frac{58}{3}-\frac{100}{3}\right)\)

\(=\frac{3}{8}\left(-\frac{42}{3}\right)\)

\(=\frac{3}{8}.\left(-14\right)\)

\(=-\frac{21}{4}\)

c) \(1\frac{4}{23}+\frac{5}{21}-\frac{4}{23}+0,5+\frac{16}{21}\)

\(=\frac{27}{23}+\frac{5}{21}-\frac{4}{23}+\frac{1}{2}+\frac{16}{21}\)

\(=\frac{27}{23}+\frac{5}{21}+\left(-\frac{4}{23}\right)+\frac{1}{2}+\frac{16}{21}\)

\(=\left[\frac{27}{23}+\left(-\frac{4}{23}\right)\right]+\left(\frac{5}{21}+\frac{16}{21}\right)+\frac{1}{2}\)

\(=1+1=2\)

d) \(\frac{21}{47}+\frac{9}{45}+\frac{26}{47}+\frac{4}{5}\)

\(=\frac{21}{47}+\frac{9}{45}+\frac{26}{47}+\frac{36}{45}\)

\(=\left(\frac{21}{47}+\frac{26}{47}\right)+\left(\frac{9}{45}+\frac{36}{45}\right)\)

\(=1+1=2\)

15 tháng 7 2019

b)=1/5.(1/4-1/9+1/9-1/14+1/14-1/19+...+1/44-1/49).2-1-3-5-7-...-49/89

=1/5.(1/4-1/49).2-(1+3+5+7...+49)/89

=1/5.45/196.2-625/89

=9/196.-623/89

=9/196.-7

=9/28

h cho mình nha ! Chúc bạn học tốt

15 tháng 7 2019

\(a,\frac{27^4\cdot2^3-3^{10}\cdot4^3}{6^4\cdot9^3}=\frac{3^{12}\cdot2^3-3^{10}\cdot2^6}{2^3\cdot3^4\cdot3^6}=\frac{3^{10}\cdot2^3\cdot\left(3^2-2^3\right)}{2^3\cdot3^{10}}=3^2-2^3=1\)

\(b,\left(\frac{1}{4\cdot9}+\frac{1}{9\cdot14}+\frac{1}{14\cdot19}+...+\frac{1}{44\cdot49}\right)\cdot\frac{1-3-5-7-...-49}{89}\)

\(=\frac{1}{5}\left(\frac{5}{4\cdot9}+\frac{5}{9\cdot14}+\frac{5}{14\cdot19}+...+\frac{1}{44\cdot49}\right)\cdot\frac{1-\left(3+5+7+...+49\right)}{89}\)

\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{44}-\frac{1}{49}\right)\cdot\frac{1-\left(3+49\right)\cdot24\div2}{89}\)

\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{49}\right)\cdot\frac{505}{89}\)

\(=\frac{1}{5}\cdot\frac{45}{196}\cdot\frac{505}{89}\)