Question

The expansion of (x+1/x)^2 will be (x^2+2x+1/x^2) Please help....I will mark as Brainliest ​

Answer 1

The statement is false:

The expansion of $\sf{(x+\dfrac{1}{x} )^2}$ will be $\sf{x^2+2+\dfrac{1}{x^2} }$.

Proof:

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

By use of distributive law

$\sf{=x(x+\dfrac{1}{x})+\dfrac{1}{x}(x+\dfrac{1}{x})}$

Expansion

$\sf{=x^2+\dfrac{\cancel{x}}{\cancel{x}}+\dfrac{\cancel{x}}{\cancel{x}}+\dfrac{1}{x^2}}$

$\sf{=x^2+2+\dfrac{1}{x^2} }$

Therefore the statement was proven false.

Informations:

• True statement

For a statement to be true, it is required to be true all over the cases.

One exception will prove the claim false.

• Components of Statements

$\boxed{\sf{Universal\:Set\:\mathbb{U}}}$ : which contains all the cases. It can be real numbers, natural numbers, or a defined set.

$\boxed{\sf{Truth\:Set}}$ : conditions where the statement is true.

In conclusion:

(Where x=1)

• $\sf{(x+\dfrac{1}{x} )^2=(1+\dfrac{1}{1} )^2=4}$
• $\sf{x^2+2x+\dfrac{1}{x^2}=1^2+2(1)+\dfrac{1}{1^2} =4}$

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

So in conclusion, not only we can show the statement is false, it is true when the universal set $\mathbb{U}$ is defined as $\sf{\mathbb{U}=\{1\}}$,

though it is universally taken that set $\mathbb{U}$ is over all the numbers.