We have a = 2 + 3. What is the value of a-aa+aa-a+1-3?

=>

Since 1< √3 < 2, we have 3 < 2 + √3 < 4 (by adding 2 to each term) and [a] = [2 + √3] = 3.
\(\left[\frac{a-\left[a\right]}{\left[a\right]}+\frac{\left[a\right]}{a-\left[a\right]+1}-\sqrt{3}\right]\)
= \(\left[\frac{2+\sqrt{3}-3}{3}+\frac{3}{2+\sqrt{3}-3+1}-\sqrt{3}\right]\) (substituting a = [2 + √3] and [a] = 3
= \(\left[\frac{\sqrt{3}-1}{3}+\frac{3}{\sqrt{3}}-\sqrt{3}\right]\) (simplifying by adding like terms)
= \(\left[\frac{\sqrt{3}-1}{3}+\frac{3\sqrt{3}}{\sqrt{3}\sqrt{3}}-\frac{3\sqrt{3}}{3}\right]\) (getting a common denominator)
= \(\left[\frac{\sqrt{3}-1}{3}+\frac{3\sqrt{3}}{3}-\frac{3\sqrt{3}}{3}\right]\) (simplifying)

= \(\left[\frac{\sqrt{3}-1}{3}\right]\) (adding fractions together)
= 0.

Therefore, C is the answer.
Answer: C

$$a=2+\sqrt{3}$$
$$\left[a\right]=\left[2+\sqrt{3}\right]$$
$$\left[a\right]=\left[2+1.7\right]=\left[3.7\right]$$
Given that [x] denites the greatest integer less than or equal to x.
[a] = 3
$$\left[\frac{a-\left[a\right]}{\left[a\right]}+\frac{\left[a\right]}{a-\left[a\right]+1}-\sqrt{3}\right] = 0$$
$$\left[\frac{2+\sqrt{3}-3}{3}+\frac{3}{2+\sqrt{3}-3+1}-\sqrt{3}\right] = 0$$
$$\left[\frac{\sqrt{3}-1}{3}+\frac{3}{\sqrt{3}}-\sqrt{3}\right] = 0$$
$$Let's\ solve\ \frac{3}{\sqrt{3}};\ \frac{3}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}=\frac{3\sqrt{3}}{\sqrt{9}} = 0$$
$$\frac{3\sqrt{3}}{3}=\sqrt{3} = 0$$

$$Therefore, \left[\frac{\sqrt{3}-1}{3}+\sqrt{3}-\sqrt{3}\right] = 0$$
$$\left[\frac{\sqrt{3}-1}{3}\right] = 0$$
$$\left[\frac{0.7}{3}\right]=\left[0.23\right] = 0$$
Given that [x] denotes the greatest integer less than or equal to x.
[0.23] = 0

Answer = Option C