Solve number 2

Step 1: Understand the given functions. We are given two functions: $g(x) = \frac{x}{3x+2}$ $h(x) = \sqrt{x^2 - 1}$

Step 2: Solve part a). a) Find $g(1) + h(2)$. First, evaluate $g(1)$: $g(1) = \frac{1}{3(1)+2} = \frac{1}{3+2} = \frac{1}{5}$ Next, evaluate $h(2)$: $h(2) = \sqrt{(2)^2 - 1} = \sqrt{4 - 1} = \sqrt{3}$ Now, add the results: $g(1) + h(2) = \frac{1}{5} + \sqrt{3}$ The value is $\boxed{\frac{1}{5} + \sqrt{3}}$.

Step 3: Solve part b). b) Find $g(3) h(1)$. First, evaluate $g(3)$: $g(3) = \frac{3}{3(3)+2} = \frac{3}{9+2} = \frac{3}{11}$ Next, evaluate $h(1)$: $h(1) = \sqrt{(1)^2 - 1} = \sqrt{1 - 1} = \sqrt{0} = 0$ Now, multiply the results: $g(3) h(1) = \frac{3}{11} \times 0 = 0$ The value is $\boxed{0}$.

Step 4: Solve part c). c) Find $\frac{g(5)}{h(2)}$. First, evaluate $g(5)$: $g(5) = \frac{5}{3(5)+2} = \frac{5}{15+2} = \frac{5}{17}$ From part (a), we know $h(2) = \sqrt{3}$. Now, divide $g(5)$ by $h(2)$: $\frac{g(5)}{h(2)} = \frac{\frac{5}{17}}{\sqrt{3}} = \frac{5}{17\sqrt{3}}$ To rationalize the denominator, multiply the numerator and denominator by $\sqrt{3}$: $\frac{5}{17\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{5\sqrt{3}}{17 \times 3} = \frac{5\sqrt{3}}{51}$ The value is $\boxed{\frac{5\sqrt{3}}{51}}$.

Step 5: Solve part d). d) Find $g(a - 1) + h(a + 1)$. First, evaluate $g(a-1)$: $g(a-1) = \frac{a-1}{3(a-1)+2} = \frac{a-1}{3a-3+2} = \frac{a-1}{3a-1}$ Next, evaluate $h(a+1)$: $h(a+1) = \sqrt{(a+1)^2 - 1}$ Expand the term inside the square root: $h(a+1) = \sqrt{a^2 + 2a + 1 - 1} = \sqrt{a^2 + 2a}$ Now, add the results: $g(a-1) + h(a+1) = \frac{a-1}{3a-1} + \sqrt{a^2 + 2a}$ The value is $\boxed{\frac{a-1}{3a-1} + \sqrt{a^2 + 2a}}$.

Step 6: Solve part e). e) Find $g(a^2) h(a^2)$. First, evaluate $g(a^2)$: $g(a^2) = \frac{a^2}{3(a^2)+2} = \frac{a^2}{3a^2+2}$ Next, evaluate $h(a^2)$: $h(a^2) = \sqrt{(a^2)^2 - 1} = \sqrt{a^4 - 1}$ Now, multiply the results: $g(a^2) h(a^2) = \frac{a^2}{3a^2+2} \cdot \sqrt{a^4 - 1} = \frac{a^2\sqrt{a^4 - 1}}{3a^2+2}$ The value is $\boxed{\frac{a^2\sqrt{a^4 - 1}}{3a^2+2}}$.