Упр.17.22 ГДЗ Мерзляк Поляков 8 класс (Алгебра)

1) (3v2+1)(v8-2);
2) (3-2v7)^2+(3+2v7)^2;
3) (10-4v6) (2+v6)^2;
4) (v(9-4v2) +v(9+4v2) )^2.

1) $$\left(3\sqrt{2}+1\right)\left(\sqrt{8}-2\right)=\left(3\sqrt{2}+1\right)\left(2\sqrt{2}-2\right)$$

$$=3\sqrt{2}\cdot 2\sqrt{2}-3\sqrt{2}\cdot 2+1\cdot 2\sqrt{2}-2$$

$$=12-6\sqrt{2}+2\sqrt{2}-2=10-4\sqrt{2}.$$

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

$$= \left(9-12\sqrt{7}+28\right)+\left(9+12\sqrt{7}+28\right)$$

$$=9+28+9+28=74.$$

3) $$\left(10-4\sqrt{6}\right)\left(2+\sqrt{6}\right)^2$$

$$\left(2+\sqrt{6}\right)^2=4+4\sqrt{6}+6=10+4\sqrt{6}$$

Тогда

$$\left(10-4\sqrt{6}\right)\left(10+4\sqrt{6}\right)=100-(4\sqrt{6})^2=100-96=4.$$

4) $$\left(\sqrt{9-4\sqrt{2}}+\sqrt{9+4\sqrt{2}}\right)^2$$

$$= \left(9-4\sqrt{2}\right)+2\sqrt{\left(9-4\sqrt{2}\right)\left(9+4\sqrt{2}\right)}+\left(9+4\sqrt{2}\right)$$

$$=18+2\sqrt{81-(4\sqrt{2})^2}=18+2\sqrt{81-32}=18+2\sqrt{49}=18+14=32.$$

Ответ

1) $$10-4\sqrt{2}$$; 2) $$74$$; 3) $$4$$; 4) $$32$$.