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puteri [66]
3 years ago
8

Why does the problem

C%5D" id="TexFormula1" title="\[\frac{y^2-5}{y^4-81}+\frac{4}{y^4-81}\]" alt="\[\frac{y^2-5}{y^4-81}+\frac{4}{y^4-81}\]" align="absmiddle" class="latex-formula"> 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)?

Mathematics
1 answer:
adelina 88 [10]3 years ago
8 0
\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}\\\\


\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)}

would be a factored version of it

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