Задача 75098 окажите помощь с алгеброй!...
Условие
Решение
б) [m]\sqrt{1 \frac{13}{36} \cdot 2 \frac{1}{4}} = \sqrt{\frac{1 \cdot 36+13}{36} \cdot \frac{2 \cdot 4+1}{4}} = \sqrt{\frac{49}{36} \cdot \frac{9}{4}} = \frac{7}{6} \cdot \frac{3}{2} = \frac{7}{4}[/m]
в) sqrt(8,1*250) = sqrt(81*25) = 9*5 = 45
г) sqrt(26^2 - 10^2) = sqrt(676 - 100) = sqrt(576) = 24
д) sqrt((-5)^6) = sqrt(5^6) = 5^3 = 125
2) а) sqrt(8)*sqrt(32) = sqrt(8*32) = sqrt(2^3*2^5) = sqrt(2^8) = 2^4 = 16
б) sqrt(0,7)*sqrt(630) = sqrt(0,7*630) = sqrt(7*63) = sqrt(7*7*9) = 7*3 = 21
в) [m]\frac{\sqrt{12}}{\sqrt{75}} = \frac{\sqrt{3}\sqrt{4}}{\sqrt{3}\sqrt{25}}= \frac{\sqrt{4}}{\sqrt{25}} = \frac{2}{5} = 0,4[/m]
г) [m]\frac{\sqrt{80} \cdot \sqrt{0,3}}{\sqrt{6}} = \frac{\sqrt{80 \cdot 0,3}}{\sqrt{2 \cdot 3}} = \frac{\sqrt{8 \cdot 3}}{\sqrt{2 \cdot 3}} = \frac{\sqrt{8} \cdot \sqrt{3}}{\sqrt{2} \cdot \sqrt{3}} = \frac{\sqrt{8}}{\sqrt{2}} = \sqrt{4} = 2[/m]
3) а) sqrt(a^8) = a^4
б) sqrt(121b^6), если b < 0, то -b > 0
sqrt(121b^6) = -11b^3
в) sqrt(x^(10)y^(14)), если x < 0, y < 0, то -x > 0, -y > 0
sqrt(x^(10)y^(14)) = (-x^5)(-y^7) = x^5y^7
г) [m]\frac{0,49a^2}{b^{18}}[/m], если a < 0, b > 0, то -a > 0
[m]\sqrt{\frac{0,49a^2}{b^{18}}} = \frac{-0,7a}{b^{9}} = -\frac{7a}{10b^{9}}[/m]