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adell [148]
3 years ago
9

Expand [x+2/x-3] [x-2/x-3]

Mathematics
2 answers:
Tomtit [17]3 years ago
7 0

Answer:

=  \frac{ {x}^{2} - 4 }{ {x }^{2}-6x+9 }  \\

Step-by-step explanation:

\frac{x + 2}{x -  3}  \times  \frac{x - 2}{x - 3}  \\  \frac{(x + 2)(x - 2)}{(x - 3)(x - 3)}  \\  \frac{x(x  - 2) + 2(x - 2)}{ {(x - 3)}^{2} }  \\  \frac{ {x}^{2} - 2x + 2x - 4 }{ {x}^{2} -  6x+9}  \\  =  \frac{ {x}^{2} - 4 }{ {x }^{2} -6x+9 }

zimovet [89]3 years ago
4 0

Answer:

  \dfrac{x^2-4}{x^2-6x+9}

Step-by-step explanation:

We assume you want to expand ...

  \dfrac{x+2}{x-3}\cdot\dfrac{x-2}{x-3}=\dfrac{(x+2)(x-2)}{(x-3)^2}=\boxed{\dfrac{x^2-4}{x^2-6x+9}}

_____

In each case, the product of the factors is ...

  (x +a)(x +b) = x² +(a+b)x +ab

For the numerator, you have (a, b) = (2, -2).

For the denominator, you have (a, b) = (-3, -3).

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3 years ago
How are the graphs of the function f(x)= square root 16^x and g(x)=3 square root 64^x related
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A is the correct awner

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Answer:

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Read 2 more answers
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