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1)

\(\Leftrightarrow\dfrac{3-2x}{5}-\dfrac{2-x}{3}>0\)

\(\dfrac{9-6x}{15}-\dfrac{10-5x}{15}>0\)

\(\Leftrightarrow9-6x-10+5x>0\)

\(\Leftrightarrow-x-1>0\)

=> -x > 1

=> x > -1

\(\left(1-\frac{3}{4}\right)\left(1-\frac{3}{7}\right)\cdot\cdot\cdot\left(1-\frac{3}{100}\right)\)

\(=\frac{1}{4}\cdot\frac{4}{7}\cdot\frac{7}{10}\cdot\cdot\cdot\frac{97}{100}\)

\(=\frac{1.4.7.10...97}{4.7.10.13...100}\)

\(=\frac{1}{100}\)

Còn có cách khác không bạn

\(=\left[20,12.10+20,12.47+20,12.100\right]+\left[0,9.\left(9+1-10\right)\right]\\ =\left[20,12.\left(10+47+100\right)\right]+0,9.0\\ =20,12.157=3158,84\)

\(\dfrac{13}{2}\) : 4\(\dfrac{2}{3}\): 2

= \(\dfrac{13}{2}\): \(\dfrac{14}{3}\):2

= \(\dfrac{13}{2}\) \(\times\) \(\dfrac{3}{14}\):2

=  \(\dfrac{39}{28}\) : 2

= \(\dfrac{39}{28}\) \(\times\) \(\dfrac{1}{2}\)

= \(\dfrac{39}{56}\)

\(\frac{\left(-16\right)^3x15^2}{12^3x10^4}\) =\(\frac{-4}{75}\)\(x\)là nhân nha

b, \(\dfrac{3}{5}\) - \(\dfrac{3}{8}\)

= \(\dfrac{24}{40}\) - \(\dfrac{15}{40}\)

= \(\dfrac{9}{40}\)

c, \(\dfrac{1}{4}\) + \(\dfrac{5}{6}\) 

= \(\dfrac{6}{24}\) + \(\dfrac{20}{24}\) 

= \(\dfrac{26}{24}\) 

= \(\dfrac{13}{12}\)

a)\(\dfrac{6}{7}+\dfrac{5}{8}=\dfrac{6.8}{7.8}+\dfrac{5.7}{8.7}=\dfrac{48}{56}+\dfrac{35}{56}=\dfrac{83}{56}\)

75

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75 năm