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Alexus [3.1K]
3 years ago
5

Explain why rationalizing the denominator does not change the value of the original expression

Mathematics
2 answers:
anzhelika [568]3 years ago
4 0
Rationalizing is just simpllifying, so the simplified value has the same value as the original expression.
Setler79 [48]3 years ago
4 0

Answer:

Because basically, you are multiplying by 1

Step-by-step explanation:

Let me explain this with an example. Rationalize the following expression:

\frac{5}{\sqrt{7} }

In order to rationalize the denominator, the numerator and denominator of the fraction must be multiplied by the root of the denominator. So, what happen if you do that? Well first of all you aren't altering the expression because:

\frac{\sqrt{7} }{\sqrt{7} } =1

Right? Because a certain quantity divided by itself is always equal to 1. So basically you are doing this because you want to rewrite the expression without altering its original value, it is the same when you do this:

9=3^2=3+3+3=\sqrt{81}

Therefore, the only thing you do when you rationalize is remove radicals from the denominator of a fraction. Take a look:

\frac{5}{\sqrt{7} } *\frac{\sqrt{7} }{\sqrt{7} } =\frac{5*\sqrt{7} }{\sqrt{7}*\sqrt{7}  } =\frac{5*\sqrt{7}}{(\sqrt{7} )^2} =\frac{5*\sqrt{7} }{7}

You can check this new expression is equal to the original using a calculator:

\frac{5}{\sqrt{7} } \approx1.8898\\\\\frac{5*\sqrt{7} }{7} \approx1.8898

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Ayudemen porfavor . Determine 4 ala potencia x + 4 ala potencia -x, si se sabe que 2ala potencia x + 2 ala potencia -x =3
Afina-wow [57]

Answer:

Es igual a 7.

Step-by-step explanation:

Ok, sabemos que:

2^x + 2^{-x} = 3

Queremos calcular:

4^x + 4^{-x}

Tambien sabemos que 4 = 2*2

Entonces:

4^x + 4^{-x} = (2*2)^x + (2*2)^{-x}

y sabemos que 2*2 = 2^2

Entonces:

4^x + 4^{-x} = (2*2)^x + (2*2)^{-x} = (2^2)^x + (2^2)^{-x}

y ahora podemos usar la relación:

(a^n)^m = (a^m)^n = a^{m*n}

entonces:

(2^2)^x + (2^2)^{-x} = (2^x)^2 + (2^{-x})^2

Ahora podemos completar cuadrados sumando y restando el termino:

2*(2^x)*(2^{-x})

Asi tendremos:

(2^x)^2 + (2^{-x})^2 =  (2^x)^2 + (2^{-x})^2 + 2*(2^x)*(2^{-x}) - 2*(2^x)*(2^{-x}) = (2^x + 2^{-x})^2 -  2*(2^x)*(2^{-x})

Entonces de momento tenemos que:

4^x + 4^{-x} = (2^x + 2^{-x})^2 - 2*(2^x)*(2^{-x})

Sabemos que el termino que esta dentro del parentesis es igual a 3.

Y tambien podemos usar la propiedad:

a^n*a^m = a^{n + m}

en el termino de la derecha.

Asi tendremos:

(2^x + 2^{-x})^2 - 2*(2^x)*(2^{-x}) = (3)^2 - 2*2^{x + (-x)}  = 9 - 2 = 7

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