a) Solve the equation:

$$
x+\sqrt{(x+1)(x+2)}=3
$$

b) Solve the equation:

$$
x+\sqrt{(x-1) x}+\sqrt{x(x+1)}+\sqrt{(x+1)(x-1)}=3
$$
a) It suffices to isolate the square root and square it:

$\sqrt{(x+1)(x+2)}=3-x$ implies $x^{2}+3 x+2=9-6 x+x^{2}$, which simplifies to $9 x=7, x=\frac{7}{9}$. We verify that this solution satisfies the initial equation: for $x=\frac{7}{9}$, $\sqrt{(x+1)(x+2)}=\frac{20}{9}$ and $\frac{7+20}{9}=3$.

b) For this second, more difficult equation, there were at least three distinct solutions.

The first solution involves isolating two of the square roots on one side:

$\sqrt{(x-1) x}+\sqrt{x(x+1)}=3-x-\sqrt{(x+1)(x-1)}$ so that when squaring, there is only one square root on each side: $(x-1) x+x(x+1)+$ $2 x \sqrt{(x+1)(x-1)}=\left(9-6 x+x^{2}\right)+(x+1)(x-1)-2(3-x) \sqrt{(x+1)(x-1)}$. This is precisely the same $\sqrt{(x+1)(x-1)}$ on both sides, so we can group these two square roots on the left side and everything else on the right side: $6 \sqrt{(x+1)(x-1)}=8-6 x$. We simplify by 2 and square it: $9\left(x^{2}-1\right)=16-24 x+9 x^{2}$, which simplifies to $24 x=25: x=\frac{25}{24}$. We verify that this solution satisfies the initial equation: for $x=\frac{25}{24}, \sqrt{(x-1) x}=\frac{5}{24}, \sqrt{x(x+1)}=\frac{35}{24}$ and $\sqrt{(x+1)(x-1)}=\frac{7}{24}$, we indeed have: $\frac{25+5+35+7}{24}=3$. Note that we have the choice of which of the three square roots to move to the right side: each of the three variations ultimately leads to the same equation.

The second solution leaves three terms on the left side:

$x+\sqrt{(x-1) x}+\sqrt{x(x+1)}=3-\sqrt{(x+1)(x-1)}$. Certainly, when squaring, there will be three square roots on the left side due to the double products: $x^{2}+(x-1) x+x(x+1)+2 x(\sqrt{(x-1) x}+\sqrt{x(x+1)}+\sqrt{(x+1)(x-1)})=9+(x+$ 1) $(x-1)-6 \sqrt{(x-1)(x+1)}$, but we recognize the three square roots whose sum, according to the initial equation, is equal to $3-x$, so the equation implies: $3 x^{2}+$ $2 x(3-x)=8+x^{2}-6 \sqrt{(x-1)(x+1)}$, which reduces to the equation from the first method.

The third method involves factoring the left side:

$(\sqrt{x}+\sqrt{x-1})(\sqrt{x}+\sqrt{x+1})=3$. Now, $(\sqrt{x}+\sqrt{x-1})(\sqrt{x}-\sqrt{x-1})=1$ and $(\sqrt{x}+\sqrt{x+1})(\sqrt{x}-\sqrt{x+1})=-1$. Therefore: $(\sqrt{x}+\sqrt{x-1})=3(-\sqrt{x}+\sqrt{x+1})$ and $(\sqrt{x}+\sqrt{x+1})=3(\sqrt{x}-\sqrt{x-1})$. We eliminate $\sqrt{x+1}$ between these two equations: $3 \sqrt{x+1}=4 \sqrt{x}+\sqrt{x-1}=3(2 \sqrt{x}-3 \sqrt{x-1})$, from which $\sqrt{x}=5 \sqrt{x-1}$, $x=25(x-1)$, which simplifies to $x=\frac{25}{24}$.