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\(A=\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}-\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}-\frac{3}{293}}\)
\(A=\frac{2.\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}-\frac{1}{293}\right)}{3.\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}-\frac{1}{293}\right)}\)
\(A=\frac{2}{3}\)
\(A=\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}\)
\(=\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}\)
\(=\frac{2}{3}\)
1. \(\frac{9}{30}=\frac{3}{10};\frac{98}{80}=\frac{49}{40};\frac{15}{1000}=\frac{3}{200}\)
Vì \(200⋮10;200⋮40\)
=> BCNN(10; 40; 200) = 200
200 : 10 = 20
200 : 40 = 5
=> \(\frac{3}{10}=\frac{3\cdot20}{10\cdot20}=\frac{60}{200}\), \(\frac{49}{40}=\frac{49\cdot5}{40\cdot5}=\frac{245}{200}\)
b: \(27D=3^{14}+3^{17}+...+3^{2024}\)
\(\Leftrightarrow26D=3^{2024}-3^{11}\)
hay \(D=\dfrac{3^{2024}-3^{11}}{26}\)
c: \(25E=-5^4-5^6-...-5^{1002}\)
\(\Leftrightarrow24E=-5^{1002}+5^2\)
hay \(E=\dfrac{-5^{1002}+5^2}{24}\)
a) \(2^5\cdot2^7\)
\(=2^{5+7}\)
\(=2^{12}\)
b) \(2^3\cdot2^2\)
\(=2^{3+2}\)
\(=2^5\)
c) \(2^4\cdot2^3\cdot2^5\)
\(=2^{4+3+5}\)
\(=2^{12}\)
d) \(2^2\cdot2^4\cdot2^6\cdot2\)
\(=2^{2+4+6+1}\)
\(=2^{13}\)
e) \(2\cdot2^3\cdot2^7\cdot2^4\)
\(=2^{1+3+7+4}\)
\(=2^{15}\)
f) \(3^8\cdot3^7\)
\(=3^{8+7}\)
\(=3^{15}\)
g) \(3^2\cdot3\)
\(=3^{2+1}\)
\(=3^3\)
h) \(3^4\cdot3^2\cdot3\)
\(=3^{4+2+1}\)
\(=3^7\)
I) \(3\cdot3^5\cdot3^4\cdot3^2\)
\(=3^{1+5+4+2}\)
\(=3^{12}\)
Đáp án C
A = 2 5 + 2 7 + 2 17 - 2 395 3 5 + 3 7 + 3 17 - 3 395 = 2 . 1 5 + 2 . 1 7 + 2 . 17 - 2 1 395 3 . 1 5 + 3 . 1 7 + 3 . 1 17 - 3 . 1 195 = 2 1 5 + 1 7 + 1 17 - 1 395 3 1 5 + 1 7 + 1 17 - 1 395 = 2 3