Question

EXERCÍCIOS
Utilizando as regras dos produtos notáveis, calcule:


b) ( x – y)² =
c) (9x + 1)(9x – 1) =
d) (3y – 5)² =
e) (a – 4y)² =
f) (ab + a) (ab – a) =
g) (2x + 3xy)² =
h) ( 10x – ab)² =
i) (x – 4)(x +4) =
j) (x + 2)(x – 2) =

Answer 1

Resposta:

$b) {(x - y)}^{2} = {x}^{2} - 2 \times x \times y + {y}^{2}$

${x}^{2} - 2xy + {y}^{2}$

$c)(9x + 1)(9x - 1) = 81 {x}^{2} - 1$

$d) {(3y - 5)}^{2} = {(3y)}^{2} - 2 \times 3y \times5+{5}^{2} = 9 {y}^{2} - 30y + 25$

$e) {(a - 4y)}^{2} = {a}^{2} - 2 \times a \times 4y + {(4y)}^{2} = {a}^{2} - 8ay + 16 {y}^{2}$

$f)(ab + a)(ab - a) = {a}^{2} {b}^{2} - {a}^{2}$

$g) {(2x + 3xy)}^{2} = {(2x)}^{2} + 2 \times 2x \times 3xy + {(3xy)}^{2} = 4 {x}^{2} + 12 {x}^{2} y + 9 {x}^{2} {y}^{2}$

$h) {(10x - ab)}^{2} = {(10x)}^{2} - 2 \times 10x \times ab + {(ab)}^{2} = {100}^{2} - 20xab + {a}^{2} {b}^{2}$

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

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

Espero ter te ajudado.