Упр.101 Вариант 2 Дидактические материалы ГДЗ Мерзляк Полонский 8 класс (Алгебра)
- Выполните умножение:
1) $$(\sqrt{63}-\sqrt{28})\cdot\sqrt{7}$$;
2) $$(7\sqrt{3}+\sqrt{48}-\sqrt{75})\cdot\sqrt{3}$$;
3) $$(6-\sqrt{5})(2+7\sqrt{5})$$;
4) $$(5\sqrt{2}+6\sqrt{3})(6\sqrt{2}-5\sqrt{3})$$;
5) $$(\sqrt{17}-\sqrt{11})(\sqrt{17}+\sqrt{11})$$;
6) $$(2\sqrt{x}-5\sqrt{y})(2\sqrt{x}+5\sqrt{y})$$;
7) $$(\sqrt{6}-2)^2$$;
8) $$(3\sqrt{7}-2\sqrt{3})^2$$.
$$\left(\sqrt{63}-\sqrt{28}\right)\cdot\sqrt{7}=\left(\sqrt{9\cdot 7}-\sqrt{4\cdot 7}\right)\cdot\sqrt{7}=\left(3\sqrt{7}-2\sqrt{7}\right)\cdot\sqrt{7}=\sqrt{7}\cdot\sqrt{7}=7.$$
$$\left(7\sqrt{3}+\sqrt{48}-\sqrt{75}\right)\cdot\sqrt{3}=\left(7\sqrt{3}+4\sqrt{3}-5\sqrt{3}\right)\cdot\sqrt{3}=6\sqrt{3}\cdot\sqrt{3}=6\cdot 3=18.$$
$$\left(6-\sqrt{5}\right)\left(2+7\sqrt{5}\right)=12+42\sqrt{5}-2\sqrt{5}-7\sqrt{25}=12+40\sqrt{5}-35=40\sqrt{5}-23.$$
$$\left(5\sqrt{2}+6\sqrt{3}\right)\left(6\sqrt{2}-5\sqrt{3}\right)=30\sqrt{2}\sqrt{2}-25\sqrt{2}\sqrt{3}+36\sqrt{3}\sqrt{2}-30\sqrt{3}\sqrt{3}$$
$$=30\cdot 2-25\sqrt{6}+36\sqrt{6}-30\cdot 3=60+11\sqrt{6}-90=11\sqrt{6}-30.$$
$$\left(\sqrt{17}-\sqrt{11}\right)\left(\sqrt{17}+\sqrt{11}\right)=\left(\sqrt{17}\right)^2-\left(\sqrt{11}\right)^2=17-11=6.$$
$$\left(2\sqrt{x}-5\sqrt{y}\right)\left(2\sqrt{x}+5\sqrt{y}\right)=\left(2\sqrt{x}\right)^2-\left(5\sqrt{y}\right)^2=4x-25y.$$
$$\left(\sqrt{6}-2\right)^2=\left(\sqrt{6}\right)^2-2\cdot \sqrt{6}\cdot 2+4=6-4\sqrt{6}+4=10-4\sqrt{6}.$$
$$\left(3\sqrt{7}-2\sqrt{3}\right)^2=\left(3\sqrt{7}\right)^2-2\cdot 3\sqrt{7}\cdot 2\sqrt{3}+\left(2\sqrt{3}\right)^2$$
$$=9\cdot 7-12\sqrt{21}+4\cdot 3=63-12\sqrt{21}+12=75-12\sqrt{21}.$$
