- $$\frac{√30+√35}{√12+√14} = \frac{√(5 \cdot 6) + √(5 \cdot 7)}{√(2 \cdot 6) + √(2 \cdot 7)} = \frac{√5(√6 + √7)}{√2(√6 + √7)} = \frac{√5}{√2} = \frac{√5 \cdot √2}{√2 \cdot √2} = \frac{√10}{2}$$
- $$\frac{√x + √y}{x-y} = \frac{√x + √y}{(√x)^2 - (√y)^2} = \frac{√x + √y}{(√x - √y)(√x + √y)} = \frac{1}{√x - √y}$$
- $$\frac{c+2√{cd}+d}{c-d} = \frac{(√c + √d)^2}{(√c - √d)(√c + √d)} = \frac{√c + √d}{√c - √d}$$
- $$\frac{b-25}{√b+5} = \frac{(√b)^2 - 5^2}{√b+5} = \frac{(√b - 5)(√b + 5)}{√b+5} = √b - 5$$
Ответ: $$\frac{√10}{2}; \frac{1}{√x - √y}; \frac{√c + √d}{√c - √d}; √b - 5$$