Answer :

Sure, let's simplify the expression [tex]\( x\left(1-\frac{1}{\sqrt{x}}\right)^2 \)[/tex] step by step.

1. Start with the original expression:
[tex]\[ x\left(1-\frac{1}{\sqrt{x}}\right)^2 \][/tex]

2. Simplify the term inside the parentheses:
[tex]\[ 1 - \frac{1}{\sqrt{x}} \][/tex]
To make it easier, let's express [tex]\(\frac{1}{\sqrt{x}}\)[/tex] as [tex]\(x^{-\frac{1}{2}}\)[/tex]. So, we have:
[tex]\[ 1 - x^{-\frac{1}{2}} \][/tex]

3. Square the expression inside the parentheses:
[tex]\[ \left(1 - x^{-\frac{1}{2}}\right)^2 \][/tex]
Using the formula for the square of a binomial [tex]\((a - b)^2 = a^2 - 2ab + b^2\)[/tex], where [tex]\(a = 1\)[/tex] and [tex]\(b = x^{-\frac{1}{2}}\)[/tex]:
[tex]\[ \left(1 - x^{-\frac{1}{2}}\right)^2 = 1^2 - 2 \cdot 1 \cdot x^{-\frac{1}{2}} + \left(x^{-\frac{1}{2}}\right)^2 \][/tex]
simplifying:
[tex]\[ = 1 - 2x^{-\frac{1}{2}} + x^{-1} \][/tex]

4. Substitute [tex]\( \left(1 - x^{-\frac{1}{2}}\right)^2 \)[/tex] back into the original expression:
[tex]\[ x\left(1 - 2x^{-\frac{1}{2}} + x^{-1}\right) \][/tex]

5. Distribute [tex]\( x \)[/tex] through the terms inside the parentheses:
[tex]\[ x \cdot 1 - x \cdot 2x^{-\frac{1}{2}} + x \cdot x^{-1} \][/tex]
simplifying each term individually:
[tex]\[ = x - 2x^{1 - \frac{1}{2}} + x^{1 - 1} \][/tex]
[tex]\[ = x - 2x^{\frac{1}{2}} + x^0 \][/tex]

6. Simplify the expression [tex]\( x^0 = 1 \)[/tex], so we have:
[tex]\[ x - 2x^{\frac{1}{2}} + 1 \][/tex]

7. Rewrite [tex]\( 2x^{\frac{1}{2}} \)[/tex] as [tex]\(2\sqrt{x}\)[/tex] to match the form of the result we are aiming for:
[tex]\[ x - 2\sqrt{x} + 1 \][/tex]

8. Notice that this can be rewritten as a square of a binomial:
[tex]\[ (\sqrt{x} - 1)^2 \][/tex]

So, the simplified form of the expression [tex]\( x\left(1 - \frac{1}{\sqrt{x}}\right)^2 \)[/tex] is:
[tex]\[ (\sqrt{x} - 1)^2 \][/tex]