Упр.16.82 ГДЗ Мордкович 8 класс (Алгебра)

б) (корень 5 — корень 2 — 1)2;
в) (корень 6 + корень 2 — 1)2;
г) (корень 3 — корень 2 + 5)2.

1. $$\left(\sqrt{3}+\sqrt{2}+1\right)^2$$
   $$=\left(\sqrt{3}+\sqrt{2}+1\right)\left(\sqrt{3}+\sqrt{2}+1\right)$$
   $$=3+2\sqrt{6}+2\sqrt{3}+2\sqrt{2}+1+2\sqrt{6}+2\sqrt{3}+2\sqrt{2}$$
   $$=6+2\sqrt{6}+2\sqrt{3}+2\sqrt{2}.$$
2. $$\left(\sqrt{5}-\sqrt{2}-1\right)^2$$
   $$=\left(\sqrt{5}-(\sqrt{2}+1)\right)^2$$
   $$=5-2\sqrt{5}(\sqrt{2}+1)+(\sqrt{2}+1)^2$$
   $$=5-2\sqrt{10}-2\sqrt{5}+2+2\sqrt{2}+1$$
   $$=8-2\sqrt{10}-2\sqrt{5}+2\sqrt{2}.$$
3. $$\left(\sqrt{6}+\sqrt{2}-1\right)^2$$
   $$=\left(\sqrt{6}+(\sqrt{2}-1)\right)^2$$
   $$=6+2\sqrt{6}(\sqrt{2}-1)+(\sqrt{2}-1)^2$$
   $$=6+2\sqrt{12}-2\sqrt{6}+2-2\sqrt{2}+1$$
   $$=9+4\sqrt{3}-2\sqrt{6}-2\sqrt{2}.$$
4. $$\left(\sqrt{3}-\sqrt{2}+5\right)^2$$
   $$=\left(\sqrt{3}+ (5-\sqrt{2})\right)^2$$
   $$=3+2\sqrt{3}(5-\sqrt{2})+(5-\sqrt{2})^2$$
   $$=3+10\sqrt{3}-2\sqrt{6}+25-10\sqrt{2}+2$$
   $$=30-2\sqrt{6}+10\sqrt{3}-10\sqrt{2}.$$

Ответ

а) $$6+2\sqrt{6}+2\sqrt{3}+2\sqrt{2}$$
б) $$8-2\sqrt{10}-2\sqrt{5}+2\sqrt{2}$$
в) $$9+4\sqrt{3}-2\sqrt{6}-2\sqrt{2}$$
г) $$30-2\sqrt{6}+10\sqrt{3}-10\sqrt{2}$$