\(\dfrac{9^5.5^7}{45^7}\)
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a,(0,8)5:(0,4)6 = (\(\dfrac{0,8}{0,4}\))5 : 0,4 = 25:0,4 = 80
b, (-25)7: 323 - \(\dfrac{6103515625}{3276}\) = - 186264,5149
c, \(\dfrac{4^2.4^3}{2^{10}}\) = \(\dfrac{4^5}{2^{10}}\) = \(\dfrac{2^{10}}{2^{10}}\) = 1
d, \(\dfrac{9^5.5^7}{45^7}\) = \(\dfrac{9^5.5^7}{9^7.5^7}\) = \(\dfrac{1}{81}\)
\(\dfrac{\left(0.8\right)^5}{\left(0.4\right)^4}\)=\(\dfrac{\left(2.0,4\right)^5}{\left(0.4^4\right)}\)=\(\dfrac{2^5.\left(0.4\right)^5}{\left(0,4\right)^4}\)=\(2^5\).\(\left(0.4\right)^1\)=12,8
b)câu b không biết có sai đề không nhưng đáp án câu b là -186264,5149
c) \(\dfrac{4^2.4^3}{2^{10}}\)=\(\dfrac{4^5}{\left(2^2\right)^5}\)=\(\dfrac{4^5}{4^5}\)=1
d)\(\dfrac{9^5.5^7}{45^7}\)=\(\dfrac{9^5.5^5.5^2}{45^7}\)=\(\dfrac{45^5.5^2}{45^7}\)=\(\dfrac{5^2}{45^2}\)=\(\left(\dfrac{5}{45}\right)^2\)=\(\left(\dfrac{1}{9}\right)^2\)=\(\dfrac{1}{81}\)
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\(9^5.5^6:45^7=\frac{9^5.5^6}{\left(9.5\right)^7}\)
\(=\frac{9^5.5^6}{9^7.5^7}\)
\(=\frac{1}{9^2.5}=\frac{1}{405}\)
95 . 56 : 457
= \(\frac{9^5.5^6}{9^7.5^7}\)
= \(\left(9\right)^{-2}.\left(5\right)^{-1}\)
= \(\frac{1}{81}.\frac{1}{5}\)
= \(\frac{1}{405}\)
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\(\dfrac{14}{7}>\dfrac{9}{7}\)
\(\dfrac{6}{13}< 1\)
\(\dfrac{7}{9}=\dfrac{14}{18}\)
\(\dfrac{4}{9}=\dfrac{20}{45}< \dfrac{26}{45}\)
\(\dfrac{11}{24}>\dfrac{9}{24}=\dfrac{3}{8}\)
\(\dfrac{42}{91}=\dfrac{42:7}{91:7}=\dfrac{6}{13}< \dfrac{7}{13}\)
\(\dfrac{34}{64}< \dfrac{52}{64}=\dfrac{13}{16}\)
\(\dfrac{17}{30}>\dfrac{16}{30}=\dfrac{8}{15}\)
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\(\dfrac{9}{5}+\dfrac{5}{7}+\dfrac{7}{5}+\dfrac{3}{7}=\left(\dfrac{9}{5}+\dfrac{7}{5}\right)+\left(\dfrac{5}{7}+\dfrac{3}{7}\right)=\dfrac{16}{5}+\dfrac{8}{7}=\dfrac{112}{35}+\dfrac{40}{35}=\dfrac{152}{35}\)
\(\dfrac{1}{2}\times\dfrac{45}{33}\times\dfrac{1}{9}\times\dfrac{11}{6}=\dfrac{1}{2}\times\dfrac{15}{11}\times\dfrac{1}{9}\times\dfrac{11}{6}=\left(\dfrac{1}{2}\times\dfrac{1}{9}\right)\times\left(\dfrac{15}{11}\times\dfrac{11}{6}\right)=\dfrac{1}{18}\times\dfrac{15}{6}=\dfrac{5}{36}\)
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a)\(\dfrac{2}{5}-\dfrac{3}{7}+\dfrac{9}{45}\\ =\dfrac{2}{5}-\dfrac{3}{7}+\dfrac{1}{5}\\ =\dfrac{14-15+1}{35}\\ =\dfrac{0}{35}=0\)
b)\(\dfrac{5}{15}+\dfrac{14}{25}-\dfrac{12}{9}+\dfrac{2}{7}+\dfrac{11}{25}\\ =\left(\dfrac{1}{3}-\dfrac{4}{3}\right)+\left(\dfrac{14}{25}+\dfrac{11}{25}\right)+\dfrac{2}{7}\\ =-1+1+\dfrac{2}{7}\\ =0+\dfrac{2}{7}=\dfrac{2}{7}\)
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13)\(\dfrac{45}{8}\left(\dfrac{4}{15}-\dfrac{7}{8}\right)+\dfrac{45}{8}\left(\dfrac{11}{15}+\dfrac{9}{8}\right)\)
=\(\dfrac{45}{8}\left(\dfrac{4}{15}-\dfrac{7}{8}+\dfrac{11}{15}+\dfrac{9}{8}\right)\)
=\(\dfrac{45}{8}\left[\left(\dfrac{4}{15}+\dfrac{11}{15}\right)-\left(\dfrac{7}{8}-\dfrac{9}{8}\right)\right]\)
=\(\dfrac{45}{8}.\dfrac{5}{4}\)=\(\dfrac{225}{32}\)
14)\(\dfrac{15}{7}\left(\dfrac{49}{35}-\dfrac{27}{8}\right)-\left(\dfrac{2}{5}-\dfrac{27}{4}\right):\dfrac{7}{15}\)
=\(\dfrac{15}{7}\left(\dfrac{49}{35}-\dfrac{27}{8}\right)-\left(\dfrac{2}{5}-\dfrac{27}{4}\right).\dfrac{15}{7}\)
=\(\dfrac{15}{7}\left[\left(\dfrac{49}{35}-\dfrac{27}{8}\right)-\left(\dfrac{2}{5}-\dfrac{27}{4}\right)\right]\)
=\(\dfrac{15}{7}\left(\dfrac{49}{35}-\dfrac{27}{8}-\dfrac{2}{5}+\dfrac{27}{4}\right)\)
=\(\dfrac{15}{7}.\dfrac{35}{8}\)=\(\dfrac{75}{8}\)
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\(A=\left(\dfrac{5.5^3}{125^2.49}\right)^{15}:\left(\dfrac{5^6}{434}\right)^7\)
\(=\left(\dfrac{5^4}{5^6.7^2}\right)^{15}.\dfrac{434^7}{5^{42}}\)
\(=\left(\dfrac{1}{5^2.7^2}\right)^{15}.\dfrac{434^7}{5^{42}}\)
\(=\dfrac{62^7.7^7}{5^{30}.7^{30}.5^{42}}\)
\(=\dfrac{62^7}{5^{72}.7^{23}}\)
\(\dfrac{9^5.5^7}{45^7}=\dfrac{9^5.5^7}{9^7.5^7}=\dfrac{1.1}{9^2.1}\)
=\(\dfrac{1}{81}\)
\(\dfrac{9^5.5^7}{45^7}=\dfrac{9^5.5^7}{9^7.5^7}=\dfrac{1.1}{9^2.1}=\dfrac{1}{81}\)
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