Упр.17.21 ГДЗ Мерзляк Поляков 8 класс (Алгебра)

1) (v5-2)^2-(3+v5)^2;
2) v(v17-4)•v(v17+4);
3) (7+4v3) (2-v3)^2;
4) (v(6+2v5) -v(6-2v5) )^2.

1. $$\left(\sqrt{5}-2\right)^2-\left(3+\sqrt{5}\right)^2$$

   $$= \left(5-4\sqrt{5}+4\right)-\left(9+6\sqrt{5}+5\right)$$

   $$= 9-4\sqrt{5}-9-6\sqrt{5}-5=-10\sqrt{5}-5.$$

2. $$\sqrt{\sqrt{17}-4}\cdot \sqrt{\sqrt{17}+4}$$

   $$= \sqrt{(\sqrt{17}-4)(\sqrt{17}+4)}$$

   $$= \sqrt{17-16}=\sqrt{1}=1.$$

3. $$\left(7+4\sqrt{3}\right)\left(2-\sqrt{3}\right)^2$$

   $$= \left(7+4\sqrt{3}\right)\left(4-4\sqrt{3}+3\right)$$

   $$= \left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)$$

   $$= 7^2-\left(4\sqrt{3}\right)^2=49-16\cdot 3=49-48=1.$$

4. $$\left(\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\right)^2$$

   $$= \left(\sqrt{6+2\sqrt{5}}\right)^2-2\sqrt{6+2\sqrt{5}}\cdot \sqrt{6-2\sqrt{5}}+\left(\sqrt{6-2\sqrt{5}}\right)^2$$

   $$= 6+2\sqrt{5}-2\sqrt{(6+2\sqrt{5})(6-2\sqrt{5})}+6-2\sqrt{5}$$

   $$= 12-2\sqrt{36-20}=12-2\sqrt{16}=12-2\cdot 4=4.$$

Ответ

$$1)\,-10\sqrt{5}-5;\quad 2)\,1;\quad 3)\,1;\quad 4)\,4.$$