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\(=\) \(9^{100}+3^{200}\)
\(=\left(3^2\right)^{100}+3^{200}\)
\(=3^{200}+3^{200}\)
\(=2.3^{200}\)
\(=9^{100}+3^{200}\)
\(=\left(3^2\right)^{100}+3^{200}\)
\(=3^{200}+3^{200}\)
\(=\left(2\times3\right)^{100}\)
\(=6^{100}\)
A = \(\dfrac{4}{3}\) + \(\dfrac{17}{3}\) + \(\dfrac{17}{9}\) + \(\dfrac{19}{13}\) + \(\dfrac{1}{9}\) + \(\dfrac{14}{6}\)
A = (\(\dfrac{4}{3}\) + \(\dfrac{17}{3}\)) + (\(\dfrac{17}{9}\) + \(\dfrac{1}{9}\)) + (\(\dfrac{19}{13}\) + \(\dfrac{14}{6}\))
A = \(\dfrac{21}{3}\) + \(\dfrac{18}{9}\) + \(\dfrac{148}{39}\)
A = 7 + 2 + \(\dfrac{148}{39}\)
A = 9 + \(\dfrac{148}{39}\)
A = \(\dfrac{39\times9}{39}\) + \(\dfrac{148}{39}\)
A = \(\dfrac{499}{39}\)
\((\)\(\dfrac{11}{9}\)+\(\dfrac{7}{9}\)\()\)-\(\dfrac{4}{3}\)
= 2-\(\dfrac{4}{3}\)
= \(\dfrac{2}{3}\)
\(\dfrac{11}{9}\) - \(\dfrac{4}{3}\) + \(\dfrac{7}{9}\)
= (\(\dfrac{11}{9}\) + \(\dfrac{7}{9}\)) - \(\dfrac{4}{3}\)
= \(\dfrac{18}{9}\) - \(\dfrac{4}{3}\)
= 2 - \(\dfrac{4}{3}\)
= \(\dfrac{2\times3}{3}\) - \(\dfrac{4}{3}\)
= \(\dfrac{6}{3}\) - \(\dfrac{4}{3}\)
= \(\dfrac{2}{3}\)
\(\frac{4}{9}+\frac{3}{8}+\frac{14}{9}=\left(\frac{4}{9}+\frac{14}{9}\right)+\frac{3}{8}=2+\frac{3}{8}=2\frac{3}{8}\)
4/9 + 3/8 + 14/9
= ( 4/9 + 14/9 ) + 3/8
= 18/9 + 3/8
= 2 + 3/8
= 16/8 + 3/8
= 19/8
a: =6/7-5/9
=54/63-35/63
=19/63
b: =1000,02-725,48
=274,54
C = 68/9 + ( 11/4 + 32/9 )
C = 68/9 + 11/4 + 32/9
C = ( 68/9 + 32/9 ) + 11/4
C = 100/9 + 11/4
C = 499/36.
Vậy C = 499/36.
3/10 × 4/9 = 12/90 = 2/15
\(\dfrac{3}{10}\) x \(\dfrac{4}{9}\) = \(\dfrac{3\times2\times2}{5\times2\times3\times3}\) = \(\dfrac{2}{15}\)