Question

va rog rapid 100pct !!!​

Answer 1

Explicație pas cu pas:

b > 0; a < 0 => |a| = - a

$\sqrt{ \frac{16 {a}^{2} }{9 {b}^{4} } } + \sqrt{ \frac{64 {a}^{4} }{9 {b}^{4} } } = \sqrt{ \frac{ {4}^{2}{a}^{2} }{ {3}^{2} {( {b}^{2} )}^{2} } } + \sqrt{ \frac{ {8}^{2} {( {a}^{2} )}^{2} }{ {3}^{2} {( {b}^{2} )}^{2}} } = \\ = \sqrt{ {\Big( \frac{4a}{3 {b}^{2} } \Big)}^{2} } + \sqrt{ {\Big( \frac{8 {a}^{2} }{3 {b}^{2} } \Big)}^{2} } = \frac{4 |a| }{3 {b}^{2} } + \frac{8 {a}^{2} }{3 {b}^{2} } \\ = \frac{8 {a}^{2} - 4a}{3 {b}^{2}} = \frac{4a(2a - 1)}{3 {b}^{2}}$