Let's solve the problem step by step.
### Statement 1:
We need to find the range of the function \( y = \frac{x^2 - x + 3}{x + 2} \).
1. **Cross-multiply**:
\[
y(x + 2) = x^2 - x + 3
\]
This simplifies to:
\[
x^2 - x - yx + 2y - 3 = 0
\]
Rearranging gives:
\[
x^2 + (-y - 1)x + (2y - 3) = 0
\]
2. **Use the discriminant**:
For \( x \) to be real, the discriminant \( D \) must be non-negative:
\[
D = b^2 - 4ac \geq 0
\]
Here, \( a = 1 \), \( b = -y - 1 \), and \( c = 2y - 3 \):
\[
(-y - 1)^2 - 4(1)(2y - 3) \geq 0
\]
3. **Calculate the discriminant**:
\[
(y + 1)^2 - 8y + 12 \geq 0
\]
Expanding gives:
\[
y^2 - 6y + 13 \geq 0
\]
4. **Finding the roots**:
The roots of the quadratic \( y^2 - 6y + 13 = 0 \) can be found using the quadratic formula:
\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 13}}{2 \cdot 1}
\]
\[
= \frac{6 \pm \sqrt{36 - 52}}{2} = \frac{6 \pm \sqrt{-16}}{2} = \frac{6 \pm 4i}{2} = 3 \pm 2i
\]
Since the discriminant is negative, the quadratic does not cross the x-axis, meaning it is always positive.
5. **Conclusion**:
The quadratic \( y^2 - 6y + 13 \) is always positive, thus the function \( y \) can take all real values except for the interval \( (-11, 1) \). Therefore, Statement 1 is true.
### Statement 2:
We need to analyze the function \( f(x) = x - \lfloor x \rfloor \) and the equation \( f(x) + f(1/x) = 1 \).
1. **Understanding \( f(x) \)**:
The function \( f(x) \) represents the fractional part of \( x \), which is periodic with a period of 1. Hence, \( f(x) \) takes values in the interval \( [0, 1) \).
2. **Analyzing \( f(1/x) \)**:
When \( x \) varies, \( 1/x \) also varies, and since \( f(1/x) \) is also periodic, we can analyze the equation:
\[
f(x) + f(1/x) = 1
\]
This means that for every \( x \), there exists a corresponding \( 1/x \) such that their fractional parts add up to 1.
3. **Finding the roots**:
Since \( f(x) \) and \( f(1/x) \) are continuous and periodic, there are infinitely many values of \( x \) that satisfy the equation \( f(x) + f(1/x) = 1 \).
4. **Conclusion**:
Thus, Statement 2 is true as there are infinitely many real roots.
### Statement 3:
We need to find the value of \( h \) such that the difference of the roots of the equation \( x^2 + hx + 7 = 0 \) is 6.
1. **Using the formula for the difference of roots**:
The difference of the roots \( \alpha \) and \( \beta \) is given by:
\[
\alpha - \beta = \sqrt{D} = 6
\]
where \( D = b^2 - 4ac \).
2. **Setting up the equation**:
Here, \( D = h^2 - 4 \cdot 1 \cdot 7 = h^2 - 28 \).
Therefore, we have:
\[
\sqrt{h^2 - 28} = 6
\]
3. **Squaring both sides**:
\[
h^2 - 28 = 36
\]
\[
h^2 = 64
\]
4. **Finding \( h \)**:
Thus, \( h = 8 \) or \( h = -8 \).
5. **Conclusion**:
Therefore, Statement 3 is true.
### Final Conclusion:
All three statements are true.
