2. Вычислите: 1) √16 − √25; √25 · √4; √49 : √400. 2) √1225 − √2116; 1/√289 + √2,25; 3/√1089 − √(49/121).

1)

1. \(\sqrt{16}-\sqrt{25}=4-5=-1\);
2. \(\sqrt{25}\cdot\sqrt{4}=5\cdot2=10\);
3. \(\sqrt{49}:\sqrt{400}=7:20=\frac{7}{20}\).

2)

1. \(\sqrt{1225}-\sqrt{2116}=35-46=-11\);
2. \(\frac{1}{\sqrt{289}}+\sqrt{2.25}=\frac{1}{17}+1.5=\frac{1}{17}+\frac{3}{2}=\frac{61}{34}\);
3. \(\frac{3}{\sqrt{1089}}-\sqrt{\frac{49}{121}}=\frac{3}{33}-\frac{7}{11}=\frac{1}{11}-\frac{7}{11}=-\frac{6}{11}\).

Ответ: 1) \(-1; 10; \frac{7}{20}\). 2) \(-11; \frac{61}{34}; -\frac{6}{11}\).