Question

alguém me ajuda por favor ​

Answer 1

Resposta:

d) $bx^2-2by+5x^2-10y$

$=b\left(x^2-2y\right)+5x^2-10y\\=b\left(x^2-2y\right)+5\left(x^2-2y\right)\\=\left(x^2-2y\right)\left(b+5\right)$

e) $2b^2+2-b^2k-k$

$=\left(2b^2+2\right)+\left(-b^2k-k\right)\\=2\left(b^2+1\right) -k\left(b^2+1\right)\\=\left(b^2+1\right)\left(-k+2\right)\\=-\left(b^2+1\right)\left(k-2\right)$

f) $5y^3-4y^2+10y-8$

$=\left(5y^3-4y^2\right)+\left(10y-8\right)\\=y^2\left(5y-4\right)+2\left(5y-4\right)\\=\left(y^2+2\right)\left(5y-4\right)$

g) $a^{12}+a^8-a^4-1$

Considerando a² = u,

$=u^6+u^4-u^2-1$

Considerando u² = t,

$=t^3+t^2-t-1\\=\left(t^3+t^2\right)+\left(-t-1\right)\\=t^2\left(t+1\right)-\left(t+1\right)\\=\left(t+1\right)\left(t^2-1\right)\\=\left(t+1\right)\left(t+1\right)\left(t-1\right)$

Voltando t = u²,

$=\left(u^2+1\right)\left(u^2+1\right)\left(u^2-1\right)\\=\left(u^2+1\right)\left(u^2+1\right)\left(u+1\right)\left(u-1\right)$

Voltando u = a²,

$=\left(a^4+1\right)\left(a^4+1\right)\left(a^2+1\right)\left(a+1\right)\left(a-1\right)\\=\left(a^4+1\right)^2\left(a^2+1\right)\left(a+1\right)\left(a-1\right)\\=\left(a^4+1\right)^2\left(a^2+1\right)\left(a+1\right)^2$