Question

Va rog fros exercitiile 2,3 .

Answer 1

$2)a) {x}^{2} + 4x + 4 = {(x + 2)}^{2}$

$b) {y}^{2} - 36y + 324 = {(y - 18)}^{2}$

$c) {a}^{2} - 6ab + 9 {b}^{2} = {(a - 3b)}^{2}$

$d) {x}^{4} + 12 {x}^{2} + 36 = {( {x}^{2} + 6)}^{2}$

$e)16 {y}^{2} - 24y + 9 = {(4y - 3)}^{2}$

$f) {a}^{2} - \frac{2}{3} a + \frac{1}{9} = {(a - \frac{1}{3} )}^{2}$

$g)25 - 10ab + {a}^{2} {b}^{2} = {(5 - ab)}^{2}$

$h) {a}^{2} - 2 \sqrt{3} a + 3 = {(a - \sqrt{3}) }^{2}$

$i) {a}^{2} - 4 \sqrt{7} a + 28 = {(a - 2 \sqrt{7}) }^{2}$

$j) {x}^{2} + 14x + 49 = {(x + 7)}^{2}$

$3)a) {x}^{2} - 4x + 4 = {(x - 2)}^{2}$

$b) {x}^{2} - 8x + 16 = {(x - 4)}^{2}$

$c)16 {x}^{2} - 8x + 1 = {(4x - 1)} ^{2}$

$d) {x}^{2} + 4 + 4x = {(x + 2)}^{2}$

$e) \frac{1}{2} {x}^{2} + 9 + 3 \sqrt{2} x = {( \frac{ \sqrt{2} }{2} x + 3)}^{2}$

$f)9 {x}^{2} - 9x + \frac{9}{4} = {(3x - \frac{3}{2} )}^{2}$

$g) {x}^{2} + 4 + 4x= {(x + 2) }^{2}$

$h) {x}^{2} + \frac{1}{16} + \frac{1}{2} x = {(x + \frac{1}{4}) }^{2}$

$i) {x}^{2} + 4xy + 4 {y}^{2} = {(x + 2y)}^{2}$

$j)2 {x}^{2} + 1 + 2 \sqrt{2}x = { (\sqrt{2} x + 1)}^{2}$

$k)9 {x}^{2} - 2x + \frac{1}{9} = {(3x - \frac{1}{3}) }^{2}$

$l) {x}^{2} + 3 + 2 \sqrt{3} x = {( x+ \sqrt{3} )}^{2}$

$m)18 {x}^{2} - 6x + \frac{1}{2} = {(3 \sqrt{2}x - \frac{ \sqrt{2} }{2} )}^{2}$

$n)45 {x}^{4} + 0.2 = 45 {x}^{4} + \frac{2}{10} = 45 {x}^{4} + \frac{1}{5} + 6{x}^{2} = {(3 \sqrt{5} {x}^{2} + \frac{ \sqrt{5} }{5} )}^{2}$

Answer 2

sper sa te ajute.......