Respuesta :
Answer:
(a) [tex]x+25[/tex]
(b) [tex]x^2-x+1[/tex]
(c) [tex](x^2+\frac{x}{2}+\frac{1}{4})[/tex]
(d) x-0.1
Step-by-step explanation:
We have to compute
(a) [tex]\frac{x^2-625}{x-25}[/tex]
We know the algebraic equation [tex]a^2-b^2=(a+b)(a-b)[/tex]
So [tex]\frac{x^2-625}{x-25}=\frac{x^2-25^2}{x-25}=[tex]x+25[/tex]
(b) We have to compute [tex]\frac{x^3+1}{x+1}[/tex]
We know the algebraic equation [tex]a^3+b^3=(a+b)(a^2-ab+b^2)[/tex]
So [tex]\frac{x^3+1}{x+1}=\frac{(x+1)(x^2-x+1)}{x+1}=x^2-x+1[/tex]
(c) We have to compute [tex]\frac{x^3-\frac{1}{8}}{x-\frac{1}{2}}[/tex]
We know the algebraic equation [tex]a^3-b^3=(a-b)(a^2+ab+b^2)[/tex]
[tex]\frac{x^3-\frac{1}{8}}{x-\frac{1}{2}}=\frac{(x-\frac{1}{2})(x^2+\frac{x}{2}+\frac{1}{4})}{x-\frac{1}{2}}[/tex]
[tex](x^2+\frac{x}{2}+\frac{1}{4})[/tex]
(d) We have to compute [tex]\frac{x^2-0.01}{x+0.1}[/tex]
We know the algebraic equation [tex]a^2-b^2=(a+b)(a-b)[/tex]
[tex]\frac{x^2-0.01}{x+0.1}=\frac{(x+0.1)(x-0.1)}{x+0.1}=x-0.1[/tex]
