Question

Fatore: diferença entre dois quadrados

a) 81a 4 – b 6 =

b) 4x 2 – 1 =

c) x 4 – y 4 =

d) x 2 y 2 – 16a 2 b 2 =

e) 1/25 - a^2/4 =

f) (b^2 - c^2/16) =

g) 16x 2 – 9y 2 =

h) 1 – m 2 n 2 =

i) x 10 – 100 =

j) 49h 2 – 81p 2 =

Answer 1

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$\boxed{\begin{array}{l}\tt a)~\sf 81a^4-b^6=(9a^2-b^3)\cdot(9a^2+b^3)\\\tt b)~\sf4x^2-1=(2x-1)\cdot(2x+1)\\\tt c)~\sf x^4-y^4=(x^2-y^2)\cdot(x^2+y^2)\\\sf=(x-y)\cdot(x+y)\cdot(x^2+y^2)\\\tt d)~\sf x^2y^2-16a^2b^2=(xy-4ab)\cdot(xy+4ab)\\\tt e)~\sf\dfrac{1}{25}-\dfrac{a^2}{4}=\bigg(\dfrac{1}{5}-\dfrac{a}{2}\bigg)\cdot\bigg(\dfrac{1}{5}+\dfrac{a}{2}\bigg)\end{array}}$

$\boxed{\begin{array}{l}\tt f)~\sf\bigg(b^2-\dfrac{c^2}{16}\bigg)=\bigg(b-\dfrac{c}{4}\bigg)\cdot\bigg(b+\dfrac{c}{4}\bigg)\\\tt g)~\sf16x^2-9y^2=(4x-3y)\cdot(4x+3y)\\\tt h)~\sf 1-m^2n^2=(1-mn)\cdot(1+mn)\\\tt i)~\sf x^{10}-100=(x^5-10)\cdot(x^5+10)\\\tt j)~\sf 49h^2-81p^2=(7h-9p)\cdot(7h+9p)\end{array}}$