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why does the problem [tex]\[\frac{y^2-5}{y^4-81}+\frac{4}{y^4-81}\][/tex] equal 1/y^2+9?
After multiplying the second fraction by -1 to make the denominators equal, then adding, and then simplified the y^2 on the top and the y^4 on the bottom you get 1/y^2-81. How does the 81 turn into a 9? and once it is a nine, why don't you simplify it to (y-3)(y+3)?

why does the problem texfracy25y481frac4y481tex equal 1y29 After multiplying the second fraction by 1 to make the denominators equal then adding and then simpli class=

Respuesta :

[tex]\bf \cfrac{y^2-5}{y^4-81}+\cfrac{4}{y^4-81}\implies \cfrac{y^2-5+4}{y^4-81}\implies \cfrac{y^2-1}{y^4-81}\\\\ -----------------------------\\\\ \textit{now recall your }\textit{difference of squares} \\ \quad \\ (a-b)(a+b) = a^2-b^2\qquad \qquad a^2-b^2 = (a-b)(a+b)\\\\ \textit{and keep in mind that } \begin{cases} 1^2=1\\ 1^3=1\\ 1^{1,000,000}=1\\ ----------\\ 81=9^2\\ 9=3^2 \end{cases}\\\\ [/tex]

[tex]\bf -----------------------------\\\\ thus \\\\\\ \cfrac{y^2-1^2}{(y^2)^2-9^2}\implies \cfrac{(y-1)(y+1)}{(\underline{y^2-9})(y^2+9)} \\\\\\ \cfrac{(y-1)(y+1)}{\underline{(y-3)(y+3)}(y^2+9)}[/tex]

would be a factored version of it

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