Question

b. Numărul natural n pentru care n<radical din 39 < n + 1 este... ​

Answer 1

$\sqrt{36} < \sqrt{39} < \sqrt{49}$

$6 < \sqrt{39} < 7$

=> n=6

Answer 2

$\bf \: n < \sqrt{39} < n + 1 \: \: \: \: \bigg | ^{2}$

$\bf \: n^{2} < 39 < (n + 1)^{2}$

$\bf \: n^{2} < 39 < n^{2} + 2n + 1$

$\bf \: 6^{2} < 39 < 6^{2} + 2 \cdot6 + 1$

$\bf \: 36< 39 < 36 + 12+ 1$

$\red{ \boxed{\bf \: 36< 39 < 49 \: }}$

$\red{ \boxed{\bf \: n = 6\: }}$