Question

IfX+1/x =1, then x^2009+1/x^2009 equals ti ?
.

Answer 1

Answer:

x²⁰⁰⁹+1/x²⁰⁰⁹

(x+1/x)²⁰⁰⁹

we know

x+1/x=1

(1)²⁰⁰⁹=1

therefore, 1 is your answer.

Answer 2

Answer:

The value of $x^{2009}+\dfrac{1}{x^{2009}}$ equals to $\bold{-2}$

Step-by-step explanation:

Given:

$x+\dfrac{1}{x}=1$

To Find:

$x^{2009}+\dfrac{1}{x^{2009}}=?$

Solution:

Given that $x+\dfrac{1}{x}=1...(1)$

Now squaring equation $(1)$ on both sides, we get

$x^2+(\dfrac{1}{x})^2=1^2\\x^2+(\dfrac{1}{x})^2=1...(2)$

Now we know that  $\bold{(a+b)^2=a^2+b^2+2ab}$

By applying this identity in the above equation:

$x^2+(\dfrac{1}{x})^2+2.x.\dfrac{1}{x}=1\\x^2+(\dfrac{1}{x})^2+2=1\\x^2+(\dfrac{1}{x})^2=1-2\\x^2+(\dfrac{1}{x})^2=-1$

Now take cube in equation $(1)$ on both sides, we get

$(x+\dfrac{1}{x})^3=1^3\\(x+\dfrac{1}{x})^3=1...(3)$

Now we know that  $\bold{(a+b)^3=a^3+b^3+3ab(a+b)}$

Now apply this identity in the above equation:

$(x+\frac{1}{x})^3=1\\x^3+\dfrac{1}{x^3}+3.x.\dfrac{1}{x}(x+\dfrac{1}{x})=1\\x^3+\dfrac{1}{x^3}+3\times1=1(\because x+\dfrac{1}{x}=1)\\x^3+\dfrac{1}{x^3}=1-3\\x^3+\dfrac{1}{x^3}=-2$

Now take squares in equation $(2)$ on both sides:

$((x+\dfrac{1}{x})^2)^2=1^2\\(x^2+(\dfrac{1}{x^2})+2.x.\dfrac{1}{x})^2=1\\x^4+\dfrac{1}{x^4}+2=1\\x^4+\dfrac{1}{x^4}=-1$

Now by squaring equation $(3)$  on both sides:

$(x^3+\dfrac{1}{x^3}))^2=1\\x^9+\dfrac{1}{x^9}+3(-2)=-8\\x^9+\dfrac{1}{x^9}=-8+6\\x^9+\dfrac{1}{x^9}=-2$

Now from the above calculation, we observe that odd powers give $-1$ and even powers give $-2$

Therefore, the value of $\bold{x^{2009}+\dfrac{1}{x^{2009}}=-2}$  as power $\bold{2009}$  is an odd number.

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