Question

ajutatima,va rog frumos

Answer 1

1.)16=$2^{4}$; 16=$( \frac{1}{2}) ^{-4}$;16=$4^{2}$;
16=$( \frac{1}{4}) ^{-2}$
2.)F pentru ca $\frac{27x}{54y^3} \neq \frac{2x}{y}$
3.)$5-3y \geq 0$⇔$y \leq \frac{5}{3}$
4.)$\sqrt{ 68^{2}-32^2 } = \sqrt{(68-32)(68+32)}= \sqrt{36*100}=6*10=60$