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\(C=\dfrac{5\times2^{12}\times3^8-3^9\times2^{12}}{2^2\times2^{13}\times3^8+2\times2^{12}\times\left(-3^9\right)}=\dfrac{3^8\times2^{12}\times\left(5-3\right)}{2^{15}\times3^8+2^{13}\times\left(-3\right)^9}\)
\(=\dfrac{3^8\times2^{12}\times2}{2^{13}\times3^8\times\left(4-3\right)}=\dfrac{1}{1}=1\)
\(#PaooNqoccc\)
Ta có : D = (1 + 2 + 22 + 23 + ....... + 22004) - 22005
Đặt A = 1 + 2 + 22 + 23 + ....... + 22004
=> 2A = 2 + 22 + 23 + ....... + 22005
=> 2A - A = 22005 - 1
=> A = 22005 - 1
Thay vào ta có : D = (1 + 2 + 22 + 23 + ....... + 22004) - 22005
=> D = 22005 - 1 - 22005
=> D = -1
a,\(A=1993^{1^{2\times3\times4\times...\times1994}}=1993^1=1993\)
b,\(B=1994^{\left(225-1^2\right)\times\left(225-2^2\right).....\left(225-50^2\right)}\)
\(=1994^{\left(225-1^2\right)\times\left(225-2^2\right)...\left(225-15^2\right)...\left(225-50^2\right)}\)
\(=1994^{\left(225-1^2\right)\times\left(225-2^2\right)...\left(225-225\right)...\left(225-50^2\right)}\)
\(=1994^{\left(225-1^2\right)\times\left(225-2^2\right)...\times0\times...\left(225-50^2\right)}\)
\(=1994^0=1\)
c, \(C=\frac{2^{10}\times3^{31}+2^{40}\times3^6}{2^{11}\times3^{31}+2^{41}\times3^6}\)
\(=\frac{2^{10}\times3^6\times\left(1\times3^{25}+2^{30}\times1\right)}{2^{11}\times3^6\times\left(1\times3^{25}+2^{30}\times1\right)}\)
\(=\frac{2^{10}}{2^{11}}=\frac{1}{2}\)
\(1\frac{13}{15}\cdot3\cdot(0,5)^2\cdot3+\left[\frac{8}{15}-1\frac{19}{60}:1\frac{23}{24}\right]\)
\(=\frac{28}{15}\cdot3\cdot0,5\cdot0,5\cdot3+\left[\frac{8}{15}-\frac{79}{60}:\frac{47}{24}\right]\)
\(=\frac{28}{5}\cdot0,25\cdot3+\left[\frac{32}{60}-\frac{79}{60}\cdot\frac{24}{47}\right]\)
\(=\frac{28}{5}\cdot\frac{25}{100}\cdot3+\left[\frac{32}{60}-\frac{158}{235}\right]\)
\(=\frac{28}{5}\cdot\frac{1}{4}\cdot3+\frac{-98}{705}=\frac{7}{5}\cdot1\cdot3+\frac{-98}{705}\)
Đến đây là tính dễ rồi :v
\((-3,2)\cdot\frac{-15}{64}+\left[0,8-2\frac{4}{15}\right]:1\frac{23}{24}\)
\(=\frac{-32}{10}\cdot\frac{-15}{64}+\left[\frac{8}{10}-\frac{34}{15}\right]:\frac{47}{24}\)
\(=\frac{-32\cdot(-15)}{10\cdot64}+\left[\frac{4}{5}-\frac{34}{15}\right]:\frac{47}{24}\)
\(=\frac{-1\cdot(-3)}{2\cdot2}+\frac{4\cdot3-34}{15}:\frac{47}{24}\)
\(=\frac{3}{4}+\frac{-22}{15}:\frac{47}{24}\)
\(=\frac{3}{4}+\frac{-517}{180}=\frac{-191}{90}\)
Bài 2 : \(\frac{2\cdot(-13)\cdot9\cdot10}{(-3)\cdot4\cdot(-5)\cdot26}=\frac{1\cdot(-1)\cdot3\cdot2}{(-1)\cdot2\cdot(-1)\cdot2}=\frac{1\cdot3}{-1\cdot2}=\frac{3}{-2}=\frac{-3}{2}\)
\(\frac{15\cdot8+15\cdot4}{12\cdot3}=\frac{15\cdot(8+4)}{12\cdot3}=\frac{15\cdot12}{12\cdot3}=\frac{15}{3}=5\)
\(\frac{4^5\times3\div0,8\%-\left(-150\right)\times\left(-20\right)\times\left(-5\right)+1000\times5^4\div0,25}{\left(-64\right)\times\left(-1,25\right)\times\left(-2,5\right)\times\left(-5\right)}\)=\(\frac{4^5\times3\times1000:2^3-\left(-150\right)\times100+1000\times5^4\times100:5^2}{64\times125\times25\times5:1000}\)
= \(\frac{\left(2^2\right)^5\times3\times10^3:2^3+15\times10^3+10^5\times5^2}{64\times5^3\times5^2\times5:10^3}\) = \(\frac{2^7\times3\times10^3+3\times5\times10^3+10^5\times5^2}{\left(2^6\times5^6\right):10^3}=\frac{10^3\times\left(2^7\times3+3\times5+10^2\times5^2\right)}{10^3}\)
= \(\left(2^7\times3+3\times5+10^2\times5^2\right)\)
=> \(A=-2^{10}+2^7\times3+3\times5+10^2\times5^2=2^7\left(3-2^3\right)+3\times5+10^2\times5^2\)
= - 27 . 5 + 3.5 + 100 .25 = 5. (-27 + 3 + 500) = 5. 375 = 1875
A=\(\frac{5x\left(2^2x3^2\right)^9-2x\left(2^2x3\right)^{14}x3^4}{5x2^{28}x3^{18}-7x2^{29}x3^{18}}\)=\(\frac{5x2^{18}x3^{18}-2x2^{28}x3^{14}x3^4}{2^{28}x3^{18}x\left(5-7x2\right)}\)=\(\frac{5x2^{18}x3^{18}-2^{29}x3^{18}}{2^{28}x3^{18}x\left(-9\right)}\)=
= \(\frac{2^{18}x3^{18}\left(5-2^{11}\right)}{-9x2^{28}x3^{18}}=\frac{5-2^{11}}{-9x2^{10}}=\frac{2043}{9216}=\frac{227}{1024}\)
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