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NB
2
24 tháng 9 2017
\(A=\frac{\left(2^2\right)^5\cdot\left(3^2\right)^4-2\cdot69}{2^{10}\cdot3^8+6^8\cdot20}\)
\(A=\frac{2^{10}\cdot3^8-2\cdot3\cdot23}{2^{10}\cdot3^8+\left(2\cdot3\right)^8\cdot20}\)
\(A=\frac{6718464-1\cdot23}{6718464-6^7\cdot20}\)
\(A=\frac{6718464-23}{6718464-5598720}\)
\(A=\frac{6718441}{1119744}\)
31 tháng 5 2015
\(A=\frac{11\cdot3^{22}\cdot3^7-9^{15}}{\left(2\cdot3^{14}\right)^2}=6\)
\(B=\frac{\left(3\cdot4\cdot2^{16}\right)^2}{11\cdot2^{13}\cdot4^{11}-16^9}=2\)
\(C=\frac{4^5\cdot9^{4-2\cdot6^9}}{2^{10}\cdot3^8+6^8\cdot20}=0\)
15 tháng 8 2016
A=\(\frac{11\cdot3^{22}\cdot3^7-9^{15}}{\left(2\cdot3^{14}\right)^2}=6\)
TT
0
MT
12 tháng 8 2015
\(\frac{x+1}{5}=\frac{-10}{16}\Rightarrow x+1=\frac{5.\left(-10\right)}{16}=\frac{-25}{8}\Rightarrow x=\frac{-25}{8}-1=-\frac{33}{8}\)
\(x+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}=\frac{-37}{45}\)
\(x+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}=\frac{-37}{45}\)
\(x+\frac{1}{5}-\frac{1}{45}=\frac{-37}{45}\)
\(x+\frac{8}{45}=\frac{-37}{45}\)
\(x=\frac{-37}{45}-\frac{8}{45}\)
\(x=-1\)
13 tháng 2 2018
Ta có :
\(M=-\frac{4}{1.5}-\frac{4}{5.9}-\frac{4}{9.13}-...-\frac{4}{\left(n+4\right)n}\)
\(\Leftrightarrow\)\(M=-\left(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{n+4}-\frac{1}{n}\right)\)
\(\Leftrightarrow\)\(M=-\left(1-\frac{1}{n}\right)\)
\(\Leftrightarrow\)\(M=-\frac{n}{n}+\frac{1}{n}\)
\(\Leftrightarrow\)\(M=\frac{-n+1}{n}\)
Vậy \(M=\frac{-n+1}{n}\)
VM
4 tháng 12 2019
a) \(\frac{5^4.20^4}{25^5.4^5}=\frac{5^4.\left(5.2^2\right)^4}{\left(5^2\right)^5.\left(2^2\right)^5}=\frac{5^4.5^4.2^8}{5^{10}.2^{10}}=\frac{5^8.2^8}{5^{10}.2^{10}}=\frac{1}{5^2.2^2}=\frac{1}{25.4}=\frac{1}{100}.\)
Chúc bạn học tốt!
17 tháng 9 2020
a) \(\frac{x-1}{21}=\frac{3}{x+1}\)( ĐKXĐ : x khác -1 )
<=> ( x - 1 )( x + 1 ) = 21.3
<=> x2 - 1 = 63
<=> x2 = 64
<=> x2 = ( ±8 )2
<=> x = ±8 ( tmđk )
b) \(\frac{7}{x}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+...+\frac{4}{41\cdot45}=\frac{29}{45}\)( ĐKXĐ : x khác 0 )
<=> \(\frac{7}{x}+\left(\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+...+\frac{4}{41\cdot45}\right)=\frac{29}{45}\)
<=> \(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)=\frac{29}{45}\)
<=> \(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\)
<=> \(\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)
<=> \(\frac{7}{x}=\frac{7}{15}\)
<=> x = 15 ( tmđk )
VC
Vũ Cao Minh⁀ᶦᵈᵒᶫ ( ✎﹏IDΣΛ亗 )
CTV
VIP
17 tháng 9 2020
a) \(\frac{x-1}{21}=\frac{3}{x+1}\Leftrightarrow\left(x-1\right)\left(x+1\right)=3.21\)
\(\Leftrightarrow x^2-1=63\Rightarrow x^2=63+1=64\Rightarrow x=\pm8\)
b) \(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\Leftrightarrow\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}=\frac{29}{45}-\frac{8}{45}=\frac{21}{45}=\frac{7}{15}\Rightarrow x=15\)
\(=\frac{4^5.\left(3^2\right)^4}{\left(4^2\right)^2.\left(3^3\right)^3}=\frac{4^5.3^8}{4^4.3^9}=\frac{4}{3}\)
\(=\frac{\left(2^2\right)^5\cdot\left(3^2\right)^4}{\left(2^4\right)^2\cdot\left(3^3\right)^3}\)
\(=\frac{2^{10}\cdot3^8}{2^8\cdot3^9}\)
\(=\frac{4}{3}\)