Упр.101 Вариант 1 Дидактические материалы ГДЗ Мерзляк Полонский 8 класс (Алгебра)

1) (v99-v44)•v11;
2) (4v6-v54+v24)•v6;
3) (12-v7)(3+2v7);
4) (2v3+3v5)(3v3-2v5);
5) (v14-v10)(v14+v10);
6) (3va+7vb)(3va-7vb);
7) (v7+1)^2;
8) (4v5-5v2)^2.

1. $$\left(\sqrt{99}-\sqrt{44}\right)\cdot \sqrt{11}=\left(\sqrt{9\cdot 11}-\sqrt{4\cdot 11}\right)\cdot \sqrt{11}$$
   $$=(3\sqrt{11}-2\sqrt{11})\cdot \sqrt{11}=\sqrt{11}\cdot \sqrt{11}=11.$$
2. $$\left(4\sqrt{6}-\sqrt{54}+\sqrt{24}\right)\cdot \sqrt{6}=\left(4\sqrt{6}-\sqrt{9\cdot 6}+\sqrt{4\cdot 6}\right)\cdot \sqrt{6}$$
   $$=\left(4\sqrt{6}-3\sqrt{6}+2\sqrt{6}\right)\cdot \sqrt{6}=3\sqrt{6}\cdot \sqrt{6}=3\cdot 6=18.$$
3. $$\left(12-\sqrt{7}\right)\left(3+2\sqrt{7}\right)=12\cdot 3+12\cdot 2\sqrt{7}-3\sqrt{7}-2\sqrt{7}\cdot \sqrt{7}$$
   $$=36+24\sqrt{7}-3\sqrt{7}-14=22+21\sqrt{7}.$$
4. $$\left(2\sqrt{3}+3\sqrt{5}\right)\left(3\sqrt{3}-2\sqrt{5}\right)=2\sqrt{3}\cdot 3\sqrt{3}-2\sqrt{3}\cdot 2\sqrt{5}$$
   $$+3\sqrt{5}\cdot 3\sqrt{3}-3\sqrt{5}\cdot 2\sqrt{5}$$
   $$=6\cdot 3-4\sqrt{15}+9\sqrt{15}-6\cdot 5=5\sqrt{15}-12.$$
5. $$\left(\sqrt{14}-\sqrt{10}\right)\left(\sqrt{14}+\sqrt{10}\right)=\left(\sqrt{14}\right)^2-\left(\sqrt{10}\right)^2=14-10=4.$$
6. $$\left(3\sqrt{a}+7\sqrt{b}\right)\left(3\sqrt{a}-7\sqrt{b}\right)=\left(3\sqrt{a}\right)^2-\left(7\sqrt{b}\right)^2=9a-49b.$$
7. $$\left(\sqrt{7}+1\right)^2=\left(\sqrt{7}\right)^2+2\sqrt{7}\cdot 1+1^2=7+2\sqrt{7}+1=8+2\sqrt{7}.$$
8. $$\left(4\sqrt{5}-5\sqrt{2}\right)^2=\left(4\sqrt{5}\right)^2-2\cdot 4\sqrt{5}\cdot 5\sqrt{2}+\left(5\sqrt{2}\right)^2$$
   $$=16\cdot 5-40\sqrt{10}+25\cdot 2=80-40\sqrt{10}+50=130-40\sqrt{10}.$$

Ответ

1) $$11$$; 2) $$18$$; 3) $$22+21\sqrt{7}$$; 4) $$5\sqrt{15}-12$$; 5) $$4$$; 6) $$9a-49b$$; 7) $$8+2\sqrt{7}$$; 8) $$130-40\sqrt{10}$$.