What is the simplified form of x minus 5 over x squared minus 3x minus 10 ⋅ x plus 2 over x squared plus x minus 12?

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What is the simplified form of x minus 5 over x squared minus 3x minus 10 ⋅ x plus 2 over x squared plus x minus 12?

Mathematics
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|dw:1437150497955:dw|
I got 1/(x+3)(x-2) as an answer

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Other answers:

is your expression like this: \[\frac{{x - 5}}{{{x^2} - 3x - 10}}:\frac{{x + 2}}{{{x^2} + x - 12}}\]
x-5/(x-5)(x+2) timex x+2/(x+4)(x-3)
ok! sorry
we have these steps:
ok
\[\begin{gathered} \frac{{x - 5}}{{{x^2} - 3x - 10}} \cdot \frac{{x + 2}}{{{x^2} + x - 12}} = \hfill \\ \hfill \\ = \frac{{x - 5}}{{\left( {x - 5} \right)\left( {x + 2} \right)}} \cdot \frac{{x + 2}}{{\left( {x - 3} \right)\left( {x + 4} \right)}} = ...? \hfill \\ \end{gathered} \]
Was I wrong?
what's going on? @Michele_Laino
after a simplification I got this: \[\frac{1}{{\left( {x - 3} \right)\left( {x + 4} \right)}}\]
Thanks :) Can I ask another?
ok!
What is the simplified form of the quantity x over 4 plus y over 3 all over the quantity x over 3 minus y over 4?
|dw:1437151350249:dw|
i got 3x+4y/ 4y-3x
do you mean this: \[\frac{{\frac{x}{4} + \frac{y}{3}}}{{\frac{x}{4} - \frac{y}{3}}}\]
\[\Large \frac{{\frac{x}{4} + \frac{y}{3}}}{{\frac{x}{4} - \frac{y}{3}}}\]
no the bottom one is x/3 - y/4
ok! we have to compute the least common multiple at numerator and at denominator, of your main fraction, which is 12, so we get: \[\Large \frac{{\frac{x}{4} + \frac{y}{3}}}{{\frac{x}{3} - \frac{y}{4}}} = \frac{{\frac{{3x + 4y}}{{12}}}}{{\frac{{4x - 3y}}{{12}}}}\]
So we can get rid of the denominator of 12
yes!
:) YAY! Thanks!
:)
Last one?
What is the oblique asymptote of the function f(x) = the quantity x squared minus 5x plus 6 over the quantity x minus 4?
ok! :)
I got x-1 as the oblique asymptote
your function is like this: \[\frac{{\sqrt {{x^2} - 5x + 6} }}{{x - 4}}\]
No. It is without the radical sign
no, sorry, is like this? \[\frac{{{x^2} - 5x + 6}}{{x - 4}}\]
Yup:)
I got x-1.
If I write the asymptote like this: y=mx+q then we have: \[\Large \begin{gathered} m = \mathop {\lim }\limits_{x \to \infty } \frac{{f\left( x \right)}}{x} = \mathop {\lim }\limits_{x \to \infty } \frac{{{x^2} - 5x + 6}}{{x\left( {x - 4} \right)}} = \hfill \\ \hfill \\ = \mathop {\lim }\limits_{x \to \infty } \frac{{{x^2}\left( {1 - \frac{5}{x} + \frac{6}{{{x^2}}}} \right)}}{{{x^2}\left( {1 - \frac{4}{{{x^2}}}} \right)}} = \hfill \\ \end{gathered} \]
Ohh okay:) Kinda confused but I'm getting it
can't you just use synthetic divison?
that limit is 1, so your slope is right!
Thanks:)
I have to go but I need more help. When will you be back?
I stay here for at least 2 hours
Ok:) Thank you. U gtg now!
wait I have to compute the quantity q
oops..please wait...
we have this formula: \[\large \begin{gathered} q = \mathop {\lim }\limits_{x \to \infty } \left\{ {f\left( x \right) - mx} \right\} = \mathop {\lim }\limits_{x \to \infty } \left\{ {f\left( x \right) - x} \right\} = \hfill \\ \hfill \\ = \mathop {\lim }\limits_{x \to \infty } \left\{ {\frac{{{x^2} - 5x + 6}}{{x - 4}} - x} \right\} = \mathop {\lim }\limits_{x \to \infty } \left\{ {\frac{{{x^2} - 5x + 6 - {x^2} + 4x}}{{x - 4}}} \right\} = \hfill \\ \hfill \\ = \mathop {\lim }\limits_{x \to \infty } \left\{ {\frac{{6 - x}}{{x - 4}}} \right\} = \mathop {\lim }\limits_{x \to \infty } \left\{ { - 1 + \frac{2}{{x - 4}}} \right\} = ...? \hfill \\ \end{gathered} \]
that limit is -1, so I confirm your answer, namely the equation of your asymptote is: y=x-1

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