Question

produto da soma da diferença de dois termos

A-(y + 1 )(y-1 )
5 5
B-(x-2 ) (x+ 2 )
3 3
C-(y+6 ) (y-6 )
7 7
D-(1+x ) (1-x )
3 3
E-(x-1 )(x + 1 )
5 5
2 2
F-(x - 1)(x + 1)
7 7

G-(x+2 )(x-2 )
4 3 4 3
H-(3+2x) (3-2x)
7 7
I-(3x-a) (3x+a)
4 5 4 5

Me ajudem rápido

Answer 1

a)

$(y+\frac{1}{5})(y-\frac{1}{5})=y^{2}-(\frac{1}{5})^{2}\\\\(y+\frac{1}{5})(y-\frac{1}{5})=y^{2}-\frac{1}{25}\\\\\boxed{\boxed{(y+\frac{1}{5})(y-\frac{1}{5})=\frac{25y^{2}-1}{25}}}$

b)

$(x-\frac{2}{3})(x+\frac{2}{3})=x^{2}-(\frac{2}{3})^{2}\\\\(x-\frac{2}{3})(x+\frac{2}{3})=x^{2}-\frac{4}{9}\\\\\boxed{\boxed{(x-\frac{2}{3})(x+\frac{2}{3})=\frac{9x^{2}-4}{9}}}$

c)

$(y+\frac{6}{7})(y-\frac{6}{7})=y^{2}-(\frac{6}{7})^{2}\\\\(y+\frac{6}{7})(y-\frac{6}{7})=y^{2}-\frac{36}{49}\\\\\boxed{\boxed{(y+\frac{6}{7})(y-\frac{6}{7})=\frac{49y^{2}-36}{49}}}$

d)

$(1+\frac{x}{3})(1-\frac{x}{3})=1^{2}-(\frac{x}{3})^{2}\\\\(1+\frac{x}{3})(1-\frac{x}{3})=1-\frac{x^{2}}{9}\\\\\boxed{\boxed{(1+\frac{x}{3})(1-\frac{x}{3})=\frac{9-x^{2}}{9}}}$

e)

$(\frac{x}{5}-1)(\frac{x}{5}+1)=(\frac{x}{5})^{2}-1^{2}\\\\(\frac{x}{5}-1)(\frac{x}{5}+1)=\frac{x^{2}}{25}-1\\\\\boxed{\boxed{(\frac{x}{5}-1)(\frac{x}{5}+1)=\frac{x^{2}-25}{25}}}$

f)

$(x-\frac{1}{7})(x+\frac{1}{7})=x^{2}-(\frac{1}{7})^{2}\\\\(x-\frac{1}{7})(x+\frac{1}{7})=x^{2}-\frac{1}{49}\\\\\boxed{\boxed{(x-\frac{1}{7})(x+\frac{1}{7})=\frac{49x^{2}-1}{49}}}$

g)

$(\frac{x}{4}+\frac{2}{3})(\frac{x}{4}-\frac{2}{3})=(\frac{x}{4})^{2}-(\frac{2}{3})^{2}\\\\(\frac{x}{4}+\frac{2}{3})(\frac{x}{4}-\frac{2}{3})=\frac{x^{2}}{16}-\frac{4}{9}\\\\(\frac{x}{4}+\frac{2}{3})(\frac{x}{4}-\frac{2}{3})=\frac{9.x^{2}-16.4}{16.9}\\\\\boxed{\boxed{(\frac{x}{4}+\frac{2}{3})(\frac{x}{4}-\frac{2}{3})=\frac{9x^{2}-64}{144}}}$

h)

$(3+\frac{2x}{7})(3-\frac{2x}{7})=3^{2}-(\frac{2x}{7})^{2}\\\\(3+\frac{2x}{7})(3-\frac{2x}{7})=9-\frac{4x^{2}}{49}\\\\(3+\frac{2x}{7})(3-\frac{2x}{7})=\frac{9.49-4x^{2}}{49}\\\\\boxed{\boxed{(3+\frac{2x}{7})(3-\frac{2x}{7})=\frac{441-4x^{2}}{49}}}$

i)

$(\frac{3x}{4}-\frac{a}{5})(\frac{3x}{4}+\frac{a}{5})=(\frac{3x}{4})^{2}-(\frac{a}{5})^{2}\\\\(\frac{3x}{4}-\frac{a}{5})(\frac{3x}{4}+\frac{a}{5})=\frac{9x^{2}}{16}-\frac{a^{2}}{25}\\\\(\frac{3x}{4}-\frac{a}{5})(\frac{3x}{4}+\frac{a}{5})=\frac{25.9x^{2}-16.a^{2}}{16.25}\\\\\boxed{\boxed{(\frac{3x}{4}-\frac{a}{5})(\frac{3x}{4}+\frac{a}{5})=\frac{225x^{2}-16a^{2}}{400}}}$