Question

alguém sabe produtos notáveis​

Answer 1

Explicação passo-a-passo:

1)

a) $\sf (5a+7)^2=(5a)^2+2\cdot5a\cdot7+7^2$

$\sf (5a+7)^2=25a^2+70a+49$

b) $\sf (2n-1)^2=(2n)^2-2\cdot2n\cdot1+1^2$

$\sf (2n-1)^2=4n^2-4n+1$

c) $\sf (7x-a)^2=(7x)^2-2\cdot7x\cdot a+a^2$

$\sf (7x-a)^2=49x^2-14ax+a^2$

d) $\sf (4x+9)^2=(4x)^2+2\cdot4x\cdot9+9^2$

$\sf (4x+9)^2=16x^2+72x+81$

e) $\sf (3x+2y)^2=(3x)^2+2\cdot3x\cdot2y+(2y)^2$

$\sf (3x+2y)^2=9x^2+12xy+4y^2$

f) $\sf (3a^2+1)^2=(3a^2)^2+2\cdot3a^2\cdot1+1^2$

$\sf (3a^2+1)^2=9a^4+6a^2+1$

g) $\sf (2x^2-5)^2=(2x^2)^2-2\cdot2x^2\cdot5+5^2$

$\sf (2x^2-5)^2=4x^4-20x^2+25$

h) $\sf (8x-7a)^2=(8x)^2-2\cdot8x\cdot7a+(7a)^2$

$\sf (8x-7a)^2=64x^2-112ax+49a^2$

i) $\sf (6-a^3)^2=6^2-2\cdot6\cdot a^3+(a^3)^2$

$\sf (6-a^3)^2=36-12a^3+a^6$

j) $\sf (3a^2+1)^2=(3a^2)^2+2\cdot3a^2\cdot1+1^2$

$\sf (3a^2+1)^2=9a^4+6a^2+1$

l) $\sf (10p+3q)^2=(10p)^2+2\cdot10p\cdot3q+(3q)^2$

$\sf (10p+3q)^2=100p^2+60pq+9q^2$

m) $\sf (1+pq)^2=1^2+2\cdot1\cdot pq+(pq)^2$

$\sf (1+pq)^2=1+2pq+p^2q^2$

2)

a) $\sf (1+x)\cdot(1-x)=1^2-x^2$

$\sf (1+x)\cdot(1-x)=1-x^2$

b) $\sf (a-3m)\cdot(a+3m)=a^2-(3m)^2$

$\sf (a-3m)\cdot(a+3m)=a^2-9m^2$

c) $\sf (r+3s)\cdot(r-3s)=r^2-(3s)^2$

$\sf (r+3s)\cdot(r-3s)=r^2-9s^2$

d) $\sf (a^2-8)\cdot(a^2+8)=(a^2)^2-8^2$

$\sf (a^2-8)\cdot(a^2+8)=a^4-64$

e) $\sf (2x^3-1)\cdot(2x^3+1)=(2x^3)^2-1^2$

$\sf (2x^3-1)\cdot(2x^3+1)=4x^6-1$

f) $\sf (m^3-8)\cdot(m^3+8)=(m^3)^2-8^2$

$\sf (m^3-8)\cdot(m^3+8)=m^6-64$

g) $\sf (3xy+z)\cdot(3xy-z)=(3xy)^2-z^2$

$\sf (3xy+z)\cdot(3xy-z)=9x^2y^2-z^2$

h) $\sf (a^2b^4-1)\cdot(a^2b^4+1)=(a^2b^4)^2-1^2$

$\sf (a^2b^4-1)\cdot(a^2b^4+1)=a^4b^8-1$

Answer 2

Resposta:.

Explicação passo-a-passo: