Question

Срочно. 10.11.12.С решением

Answer 1

$10) \frac{657^3 +3*657^2*343+3*657*343^2+343^3}{0.5*49000000 -0.5*9000000} - 35 = \frac{(657 + 343)^3}{0.5 * 10^7*(4.9-0.9)} -35= \\ \\ = \frac{1000^3}{0.5*4*10^7} -35 = \frac{(10^3)^3}{2*10^7} - 35 = \frac{10^9}{2*10^7} -35 = \frac{10^2}{2} -35 = 50-35 = 15$

По формуле сокращенного умножения:
а²  - b²  = (a-b)(a+b)
$11) \frac{(6.3^2-3.7^2)(4.2^2-5.8^2)}{(3.3^2-6.7^2)(4.8^2-5.2^2)} = \frac{(6.3-3.7)(6.3+3.7)(4.2-5.8)(4.2+5.8)}{(3.3-6.7)(3.3+6.7)*(4.8-5.2)(4.8+5.2)} = \\ \\ = \frac{2.6 * 10 *(-1.6)*10 }{(-3.4)*10*(-0.4)*10} = \frac{26*(-16)}{(-34)*(-4)} = \frac{- (13 *4)}{17*1} = - \frac{52}{17} =-3 \frac{1}{17}$

$12) \frac{(8.7^2 - 1.3^2)(5.1^2-4.9^2)}{(3.9^2-6.1^2)(2.7^2-7.3^2)} = \frac{(8.7 -1.3)(8.7+1.3)(5.1-4.9)(5.1+4.9)}{(3.9-6.1)(3.9+6.1)(2.7-7.3)(2.7+7.3)} = \\ \\ = \frac{7.4*10 * 0.2*10}{-2.2*10*(-4.6)*10}= \frac{74*2}{(-22)*(-46)} = \frac{37*1}{11*23} = \frac{37}{253}$