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11 tháng 7 2023
`@` `\text {Ans}`
`\downarrow`
\(\dfrac{2^8-2^3}{2^5-1}=\dfrac{2^3\left(2^5-1\right)}{2^5-1}=\dfrac{2^3}{1}=2^3=8\)
_____
\(\dfrac{4^8\cdot9^4}{6^6\cdot8^3}\)
`=`\(\dfrac{\left(2^2\right)^8\cdot\left(3^2\right)^4}{2^6\cdot3^6\cdot\left(2^3\right)^3}\)
`=`\(\dfrac{2^{16}\cdot3^8}{2^6\cdot3^6\cdot2^9}\)
`=`\(\dfrac{2^{16}\cdot3^8}{2^{15}\cdot3^6}\)
`=`\(\dfrac{3^2}{2}\) `=`\(\dfrac{9}{2}\)
______
\(\dfrac{27^4\cdot2^3-3^{10}\cdot4^3}{6^4\cdot9^3}\)
`=`\(\dfrac{\left(3^3\right)^4\cdot2^3-3^{10}\cdot\left(2^2\right)^3}{2^4\cdot3^4\cdot\left(3^2\right)^3}\)
`=`\(\dfrac{3^{12}\cdot2^3-3^{10}\cdot2^6}{2^4\cdot3^4\cdot3^6}\)
`=`\(\dfrac{3^{10}\cdot\left(3^2\cdot2^3-2^6\right)}{3^{10}\cdot2^4}\)
`=`\(\dfrac{72-2^6}{2^4}=\dfrac{8}{16}=\dfrac{1}{2}\)
11 tháng 7 2023
\(\dfrac{2^8-2^3}{2^5-1}=\dfrac{2^3\left(2^5-1\right)}{2^5-1}=2^3=8\)
\(\dfrac{4^8.9^4}{6^6.8^3}=\dfrac{2^{16}.3^8}{2^6.3^6.2^9}=2.3^2=18\)
\(\dfrac{27^4.2^3-3^{10}.4^3}{6^4.9^3}=\dfrac{3^{12}.2^3-3^{10}.2^6}{2^4.3^4.3^6}=\dfrac{2^3.3^{10}.\left(3^2-2^3\right)}{2^4.3^{10}}=\dfrac{9-8}{2}=\dfrac{1}{2}\)
HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023
a)
\(\frac{{{4^3}{{.9}^7}}}{{{{27}^5}{{.8}^2}}} = \frac{{{{\left( {{2^2}} \right)}^3}.{{\left( {{3^2}} \right)}^7}}}{{{{\left( {{3^3}} \right)}^5}.{{\left( {{2^3}} \right)}^2}}} =\frac{2^{2.3}.3^{2.7}}{3^{3.5}.2^{2.3}}= \frac{{{2^6}{{.3}^{14}}}}{{{3^{15}}{{.2}^6}}} = \frac{1}{3}\)
b)
\(\frac{{{{\left( { - 2} \right)}^3}.{{\left( { - 2} \right)}^7}}}{{{{3.4}^6}}} =\frac{(-2)^{3+7}}{3.(2^2)^6}= \frac{{{{\left( { - 2} \right)}^{10}}}}{{3.{{\left( {{2^{2.6}}} \right)}}}} = \frac{{{2^{10}}}}{{{{3.2}^{12}}}} = \frac{1}{{{{3.2}^2}}} = \frac{1}{{12}}\)
c)
\(\begin{array}{l}\frac{{{{\left( {0,2} \right)}^5}.{{\left( {0,09} \right)}^3}}}{{{{\left( {0,2} \right)}^7}.{{\left( {0,3} \right)}^4}}} = \frac{{{{\left( {0,2} \right)}^5}.{{\left[ {{{\left( {0,3} \right)}^2}} \right]}^3}}}{{{{\left( {0,2} \right)}^7}.{{\left( {0,3} \right)}^4}}} = \frac{{{{\left( {0,2} \right)}^5}.{{\left( {0,3} \right)}^6}}}{{{{\left( {0,2} \right)}^7}.{{\left( {0,3} \right)}^4}}}\\ = \frac{{{{\left( {0,3} \right)}^2}}}{{{{\left( {0,2} \right)}^2}}} = \frac{{0,9}}{{0,4}} = \frac{9}{4}\end{array}\)
d)
Cách 1: \(\frac{{{2^3} + {2^4} + {2^5}}}{{{7^2}}} = \frac{{8 + 16 + 32}}{{49}} = \frac{{56}}{{49}} = \frac{8}{7}\)
Cách 2: \(\frac{{{2^3} + {2^4} + {2^5}}}{{{7^2}}} = \frac{{2^3.(1+2+2^2)}}{{7^2}} = \frac{{2^3.7}}{{7^2}} = \frac{8}{7}\)
21 tháng 7 2023
a: \(=\dfrac{2^9\cdot5^9\cdot3^{40}}{2^{12}\cdot5^{10}\cdot3^{20}}=\dfrac{3^{20}}{5\cdot2^3}\)
b: \(=\dfrac{-3^8\cdot2^{10}\cdot5^6}{2^9\cdot\left(-1\right)\cdot3^6\cdot5^7}=\dfrac{-2}{5}\cdot3^2=-\dfrac{18}{5}\)
c: \(=\dfrac{3^{186}\cdot5^{100}}{5^{100}\cdot3^{187}}=\dfrac{1}{3}\)
26 tháng 7 2017
a, \(\dfrac{1}{2!}+\dfrac{2}{3!}+...+\dfrac{99}{100!}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}< 1\)
\(\Rightarrowđpcm\)
d, \(D=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow3D=1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\)
\(\Rightarrow3D-D=\left(1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\right)\)
\(\Rightarrow2D=1-\dfrac{1}{3^{99}}\)
\(\Rightarrow D=\dfrac{1}{2}-\dfrac{1}{3^{99}.2}< \dfrac{1}{2}\)
\(\Rightarrowđpcm\)
26 tháng 7 2017
\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=\left(1+\dfrac{1}{3}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{49}+\dfrac{1}{50}-1-\dfrac{1}{2}-...-\dfrac{1}{25}\)
\(=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\)
\(\Rightarrowđpcm\)
4 tháng 1 2022
\(\dfrac{45^{10}\cdot5^{20}}{75^{15}}=\dfrac{\left(3^2\cdot5\right)^{10}\cdot5^{20}}{\left(3\cdot5^2\right)^{15}}=\dfrac{3^{20}\cdot5^{10}\cdot5^{20}}{3^{15}\cdot5^{30}}=3^5=243\\ \dfrac{6^6+6^3+3^3+3^6}{-73}=\dfrac{46656+216+27+729}{-73}=-\dfrac{47628}{73}\\ \dfrac{27^7+3^{15}}{9^9-27}=\dfrac{\left(3^3\right)^7+3^{15}}{\left(3^2\right)^9-3^3}=\dfrac{3^{21}+3^{15}}{3^{18}-3^3}=\dfrac{3^{15}\left(3^6+1\right)}{3^3\left(3^{15}-1\right)}=\dfrac{3^5\cdot730}{3^{15}-1}\\ \dfrac{8^{20}+4^{20}}{4^{25}+64^5}=\dfrac{\left(2^3\right)^{20}+\left(2^2\right)^{20}}{\left(2^2\right)^{25}+\left(2^6\right)^5}=\dfrac{2^{60}+2^{40}}{2^{50}+2^{30}}=\dfrac{2^{40}\left(2^{20}+1\right)}{2^{30}\left(2^{20}+1\right)}=2^{10}=1024\)
H9
HT.Phong (9A5)
CTVHS
22 tháng 7 2023
a) \(\dfrac{27^3\cdot11+9^5\cdot5}{3^9\cdot2^4}\)
\(=\dfrac{3^9\cdot11+3^{10}\cdot5}{3^9\cdot2^4}\)
\(=\dfrac{3^9\cdot\left(11+3\cdot5\right)}{3^9\cdot2^4}\)
\(=\dfrac{11+15}{16}\)
\(=\dfrac{26}{16}\)
\(=\dfrac{13}{8}\)
b) \(\dfrac{5^8+2^2\cdot25^4+2^3\cdot125^3-15^4\cdot5^4}{4^2\cdot625^2}\)
\(=\dfrac{5^8+2^2\cdot5^8+2^3\cdot5^9-3^4\cdot5^4\cdot5^4}{2^4\cdot5^8}\)
\(=\dfrac{5^8\cdot\left(1+2^2+2^3\cdot5-3^4\right)}{5^8\cdot2^4}\)
\(=\dfrac{1+4+40-81}{16}\)
\(=\dfrac{-36}{16}\)
\(=\dfrac{-9}{4}\)
c) \(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)
\(=\dfrac{2^{10}\cdot3^8-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8\cdot\left(1-3\right)}{2^{10}\cdot3^8\cdot\left(1+5\right)}\)
\(=\dfrac{-2}{6}\)
\(=-\dfrac{1}{3}\)
8 tháng 10 2023
a: \(A=\dfrac{3^3\cdot2^3+3^3\cdot2^2+3^3\cdot1}{-13}=\dfrac{27\left(2^3+2^2+1\right)}{-13}=-27\)
b: \(B=\dfrac{2\cdot2^{12}\cdot3^6+2^{11}\cdot3^9}{2^3\cdot2^7\cdot3^7+2^7\cdot2^3\cdot5\cdot3^8}\)
\(=\dfrac{2^{13}\cdot3^6+2^{11}\cdot3^9}{2^{10}\cdot3^7+2^{10}\cdot5\cdot3^8}\)
\(=\dfrac{2^{11}\cdot3^6\left(2^2+3^3\right)}{2^{10}\cdot3^7\left(1+5\cdot3\right)}=\dfrac{2}{3}\cdot\dfrac{4+27}{1+15}=\dfrac{2}{3}\cdot\dfrac{31}{16}=\dfrac{31}{24}\)
c: \(C=\dfrac{5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}}{5\cdot2^{35}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\)
\(=\dfrac{2^{29}\cdot3^{18}\left(5\cdot2-3^2\right)}{2^{29}\cdot3^{18}\left(5\cdot2^6-7\right)}=\dfrac{10-9}{5\cdot64-7}=\dfrac{1}{313}\)
YP
1
13 tháng 12 2021
\(a,=\dfrac{1}{6}-\dfrac{5}{6}=-\dfrac{4}{6}=-\dfrac{2}{3}\\ b,=\dfrac{3^2}{3^{12}\cdot3^{18}}=\dfrac{1}{3^{28}}\)
18 tháng 9 2021
\(\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{49\cdot50}\\ =1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{49}-\dfrac{1}{50}\\ =\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{50}\right)\\ =\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{50}\right)\\ =\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{25}\right)\)
\(=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{49}+\dfrac{1}{50}\)
\(\dfrac{27^3 . 4^3}{6^8}\)
-> \(\dfrac{3}{4}\)
\(\dfrac{27^3.4^3}{6^8}\)=\(\dfrac{\left(3^3\right)^3.\left(2^2\right)^3}{\left(3.2\right)^8}\)=\(\dfrac{3^9.2^6}{3^8.2^8}\)=\(\dfrac{3}{4}\)