\(\left(x-3\right)\left(x+\frac{3}{4}\right)>0\)<=> \(x-3\) và \(x+\frac{3}{4}\) khác dấu

\(\left(+\right)\hept{\begin{cases}x-3>0\\x+\frac{3}{4}>0\end{cases}=>\hept{\begin{cases}x>3\\x>-\frac{3}{4}\end{cases}=>x>3}}\)

\(\left(+\right)\hept{\begin{cases}x-3< 0\\x+\frac{3}{4}< 0\end{cases}=>\hept{\begin{cases}x< 3\\x< -\frac{3}{4}\end{cases}=>x< -\frac{3}{4}}}\)

Vậy \(x>3\) hoặc \(x< -\frac{3}{4}\) thì \(\left(x-3\right)\left(x+\frac{3}{4}\right)>0\)

Ta có : \(4^{15}.5^{30}=4^{15}.5^{2.15}=4^{15}.\left(5^2\right)^{15}=4^{15}.25^{15}=\left(25.4\right)^{15}=100^{15}\)

Vậy : \(4^{15}.5^{30}=100^{15}\)

bang 2 chu bang gi