Potentiation and Radicalization Exercise

VIEW:425 DATA:2020-03-20

When trying to solve this simplification, we must first add or subtract the fraction. The basic rule of adding and subtracting fractions is in the definition.
Using the definition we have the problem with two types of products, one being the simple product and the other the product of difference.
The product of difference is a great tool for taking a square root, from a sum or subtraction that one or both are square root in the sum or subtraction.
following the definition we have, that the divisor becomes the square root of 1, which is the value 1, that way we can stop writing the divisor.
And now to take the square root of the values, we can square and take the square root.
It seems like a logical subterfuge, but this suberfuge brings us out of a negative square root.
So using the definition we have, the square of a subtraction, which is also a remarkable product.
Using the definitions, we arrive at a simplification, in which the result is square root of 4, which generates the value 2.
And so we managed to solve the simplification.

Ex.sqrt{\frac{3-2sqrt{2}}{17-12sqrt{2}}}-sqrt{\frac{3+2sqrt{2}}{17+12sqrt{2}}}\frac{a}{b}-\frac{c}{d}=\frac{a.d-c.b}{b.d}\frac{sqrt{(3-2sqrt{2}).(17+12sqrt{2})}-sqrt{(3+2sqrt{2}).(17-12sqrt{2})}}{sqrt{(17-12sqrt{2}).(17+12sqrt{2})}}(a-b)(c+d)=a.c+a.d-b.c-bd(a-b)(a+b)=a^{2}-b^{2}\frac{sqrt{(3.17+3.12sqrt{2}-2sqrt{2}.17-2sqrt{2}.12sqrt{2})}-sqrt{(3.17-3.12sqrt{2}+2sqrt{2}.17-2sqrt{2}.12sqrt{2})}}{sqrt{(17^{2}-(12sqrt{2})^{2})}}\frac{sqrt{(51+36sqrt{2}-34sqrt{2}-48)}-sqrt{(51-36sqrt{2}+34sqrt{2}-48)}}{sqrt{289-288}}\frac{sqrt{(3+2sqrt{2})}-sqrt{(3-2sqrt{2})}}{sqrt{1}}sqrt{(3+2sqrt{2})}-sqrt{(3-2sqrt{2})}sqrt{(a^{2})}=asqrt{left ( sqrt{(3+2sqrt{2})}-sqrt{(3-2sqrt{2}} \right )^{2}}(a-b)^{2}=a^{2}-2a.b+b^{2}sqrt{left ( left (sqrt{(3+2sqrt{2})}\right )^{2}-2sqrt{(3+2sqrt{2})}.sqrt{(3-2sqrt{2}})+left (sqrt{3-2sqrt{2}}\right )^{2} \right )}(a-b)(a+b)=a^{2}-b^{2}(sqrt{a})^{2}=asqrt{left ( (3+2sqrt{2}-2sqrt{(9-8)})+3-2sqrt{2} \right )}=sqrt{6-2}=sqrt{4}=2

FACEBOOK

Participe de nossa rede facebook.com/osreformadoresdasaude

Novidades, e respostas das perguntas de nossos colaboradores