Question

2= ao quadrado// me ajudemm prv


1. Calcule:

(5a+4b)2=

(11-9p)2=

(4b-1)2=

(2a3+2a)2=

(2b+4).(2b-4)=

(3x2 y+2x).(3x2 y-2x)=

(5xy+4b)2=

(2-4a)2=

(3a-2b).(3a+2b)=

Answer 1

Resposta:

(5a+4b)2=

$25 {a}^{2} + 40ab + 16 {b}^{2}$

(11-9p)2=

$121 - 198p + {81p}^{2}$

(4b-1)2=

${16b}^{2} - 8b + 1$

(2a3+2a)2=

${4a}^{6} + 8 {a}^{3}a + 4 {a}^{2}$

(2b+4).(2b-4)=

${4b}^{2} - 16$

(3x2 y+2x).(3x2 y-2x)=

$9 {x}^{4} {y}^{2} - 4 {x}^{2}$

(5xy+4b)2=

$25 {x}^{2} {y }^{2} + 40bxy + 16 {b}^{2}$

(2-4a)2=

$4 - 16a + 16 {a}^{2}$

(3a-2b).(3a+2b)=

$9 {a}^{2} - 4 {b}^{2}$