Упр.22.10 ГДЗ Мордкович Семенов 8 класс (Алгебра)

Задача

а) v(16a) — v(36a) + v(49 г) 0,5v(12x) + 5v(3x) — 100v(0,03x);
б) v(5b) + v(80b) — v(60b); д) 2/3 v(252a^3) — 3/4 av(448a) + 3/5 v(175a^3);
в) 2v(3y) + v(27y) — 0,1v(300y); е) 3/4 yv(128y^3) + 3/5 v(50y^5) — 1/3 v(242y^5).

Подробный ответ

а)
$$\sqrt{16a}-\sqrt{36a}+\sqrt{49a}=4\sqrt{a}-6\sqrt{a}+7\sqrt{a}=5\sqrt{a}.$$
б)
$$\sqrt{5b}+\sqrt{80b}-\sqrt{60b}=\sqrt{5b}+\sqrt{16\cdot 5b}-\sqrt{4\cdot 15b}$$
$$=\sqrt{5b}+4\sqrt{5b}-2\sqrt{15b}=5\sqrt{5b}-2\sqrt{15b}.$$
в)
$$2\sqrt{3y}+\sqrt{27y}-0{,}1\sqrt{300y}=2\sqrt{3y}+\sqrt{9\cdot 3y}-0{,}1\sqrt{100\cdot 3y}$$
$$=2\sqrt{3y}+3\sqrt{3y}-0{,}1\cdot 10\sqrt{3y}=5\sqrt{3y}-\sqrt{3y}=4\sqrt{3y}.$$
г)
$$0{,}5\sqrt{12x}+5\sqrt{3x}-100\sqrt{0{,}03x}=0{,}5\sqrt{4\cdot 3x}+5\sqrt{3x}-100\sqrt{0{,}01\cdot 3x}$$
$$=0{,}5\cdot 2\sqrt{3x}+5\sqrt{3x}-100\cdot 0{,}1\sqrt{3x}$$
$$=\sqrt{3x}+5\sqrt{3x}-10\sqrt{3x}=-4\sqrt{3x}.$$
д)
$$\frac{2}{3}\sqrt{252a^3}-\frac{3}{4}a\sqrt{448a}+\frac{3}{5}\sqrt{175a^3}$$
$$=\frac{2}{3}\sqrt{36\cdot 7\cdot a^2\cdot a}-\frac{3}{4}a\sqrt{64\cdot 7\cdot a}+\frac{3}{5}\sqrt{25\cdot 7\cdot a^2\cdot a}$$
$$=\frac{2}{3}\cdot 6a\sqrt{7a}-\frac{3}{4}a\cdot 8\sqrt{7a}+\frac{3}{5}\cdot 5a\sqrt{7a}$$
$$=4a\sqrt{7a}-6a\sqrt{7a}+3a\sqrt{7a}=a\sqrt{7a}.$$
е)
$$\frac{3}{4}y\sqrt{128y^3}+\frac{3}{5}\sqrt{50y^5}-\frac{1}{3}\sqrt{242y^5}$$
$$=\frac{3}{4}y\sqrt{64\cdot 2\cdot y^2\cdot y}+\frac{3}{5}\sqrt{25\cdot 2\cdot y^4\cdot y}-\frac{1}{3}\sqrt{121\cdot 2\cdot y^4\cdot y}$$
$$=\frac{3}{4}y\cdot 8y\sqrt{2y}+\frac{3}{5}\cdot 5y^2\sqrt{2y}-\frac{1}{3}\cdot 11y^2\sqrt{2y}$$
$$=6y^2\sqrt{2y}+3y^2\sqrt{2y}-\frac{11}{3}y^2\sqrt{2y}$$
$$=9y^2\sqrt{2y}-\frac{11}{3}y^2\sqrt{2y}=\frac{16}{3}y^2\sqrt{2y}.$$