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Simplify: x^2-16/ 4x divided by (x-4)
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We know that \[a^2-b^2 = (a-b)(a+b)\]So x^2 - 16 would be \[x^2 - 16 = (x-4)(x+4)\]
as @zepp said: \[\LARGE \frac{(x-4)(x+4)}{x-4}=\frac{\cancel{(x-4)}(x+4)}{\cancel{x-4}}\]
^ There, I suck at typing in LaTeX :(
Wait, @Kreshnik, you forgot the 4x
\[\frac{(x-4)(x+4)}{4x(x-4)}\]\[\frac{(x+4)}{4x}\]
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auff.. LOL I guess I didn't read the question properly \[\LARGE \left[\frac{(x-4)(x+4)}{4x}\right] \div (x-4)=\] \[\LARGE \frac{(x-4)(x+4)}{4x} \times \frac{1}{(x-4)}=\] \[\LARGE \frac{\cancel{(x-4)}(x+4)}{4x} \times \frac{1}{\cancel{(x-4)}}=\] \[\LARGE \frac{x+4}{4x }\]
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