Rewrite the given condition as

\sqrt{a}+\dfrac{1}{\sqrt{a} } =3

It's found it is the form of Nike function but the variable in the form of square root.

For a binomial of a variable plus its reciprocal, squaring of it will raise the power of variable and remain the form of Nike function.

Then, let's square both sides of equation (1) to remove the radical

a+\dfrac{1}{a} +2=9

Combine the constent terms

a+\dfrac{1}{a} =7

Again square above equation to raise the degree of leading term,

a^2+\dfrac{1}{a^2} =45

Now we are close to the solution

Plug in equation (1) and (2) to the asking expression, we get

\dfrac{a^2+a^{-2}+1}{a+a^{-1}-1}=\dfrac{45+1}{7-1} = \dfrac{23}{3}

Steven Zheng posted 1 year ago

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