Упр.20.5 ГДЗ Мордкович Семенов 8 класс (Алгебра)

Задача

а) v(81/49 a^6 b^4) при a = 7, b = 3;
б) 0,5v(1 24/25 x^4 y^2) при x = 3, y = -2;
в) -3v(49a^2/64b^4) при a = 1/7, b = 3/4;
г) v(25/64 x^4 y^8) при x = 4, y = -5;
д) 1/4 v(5 11/49 a^4 b^6) при a = -2, b = 3,5;
е) -2v((0,49x^4)/(0,81y^2)) при x = 5/7, y = -2/9.

Подробный ответ

а) $$\sqrt{\frac{81}{49}a^6b^4}=\sqrt{\frac{81}{49}}\cdot\sqrt{(a^3)^2}\cdot\sqrt{(b^2)^2}=\frac{9}{7}|a^3|b^2.$$

При $$a=7,\ b=3$$:

$$\frac{9}{7}\cdot 7^3\cdot 3^2=\frac{9\cdot 343\cdot 9}{7}=3969.$$

б) $$0{,}5\sqrt{1\frac{24}{25}x^4y^2}=0{,}5\sqrt{\frac{49}{25}}\cdot\sqrt{(x^2)^2}\cdot\sqrt{y^2}=0{,}5\cdot\frac{7}{5}\cdot x^2|y|=0{,}7x^2|y|.$$

При $$x=3,\ y=-2$$:

$$0{,}7\cdot 3^2\cdot |-2|=0{,}7\cdot 9\cdot 2=12{,}6.$$

в) $$-3\sqrt{\frac{49a^2}{64b^4}}=-3\cdot\frac{\sqrt{49}\cdot\sqrt{a^2}}{\sqrt{64}\cdot\sqrt{(b^2)^2}}=-3\cdot\frac{7|a|}{8b^2}.$$

При $$a=\frac17,\ b=\frac34$$:

$$-3\cdot\frac{7\cdot \frac17}{8\cdot \left(\frac34\right)^2}=-3\cdot\frac{1}{8\cdot \frac{9}{16}}=-3\cdot\frac{2}{9}=-\frac23.$$

г) $$\sqrt{\frac{25}{64}x^4y^8}=\sqrt{\frac{25}{64}}\cdot\sqrt{(x^2)^2}\cdot\sqrt{(y^4)^2}=\frac{5}{8}x^2y^4.$$

При $$x=4,\ y=-5$$:

$$\frac{5}{8}\cdot 4^2\cdot (-5)^4=\frac{5}{8}\cdot 16\cdot 625=6250.$$

д) $$\frac14\sqrt{5\frac{11}{49}a^4b^6}=\frac14\sqrt{\frac{256}{49}}\cdot\sqrt{(a^2)^2}\cdot\sqrt{(b^3)^2}=\frac14\cdot\frac{16}{7}\cdot a^2|b^3|=\frac47a^2|b^3|.$$

При $$a=-2,\ b=3{,}5=\frac72$$:

$$\frac47\cdot (-2)^2\cdot \left|\frac72\right|^3=\frac47\cdot 4\cdot \frac{343}{8}=98.$$

е) $$-2\sqrt{\frac{0{,}49x^4}{0{,}81y^2}}=-2\cdot\frac{\sqrt{0{,}49}\cdot\sqrt{(x^2)^2}}{\sqrt{0{,}81}\cdot\sqrt{y^2}}=-2\cdot\frac{0{,}7x^2}{0{,}9|y|}.$$

При $$x=\frac57,\ y=-\frac29$$:

$$-2\cdot\frac{0{,}7\left(\frac57\right)^2}{0{,}9\left|-\frac29\right|}=-2\cdot\frac{\frac{7}{10}\cdot \frac{25}{49}}{\frac{9}{10}\cdot \frac29}=-2\cdot\frac{25}{14}=-\frac{25}{7}=-3\frac47.$$