Question

Помогите. 13.14.15.......

Answer 1

№13
((5.7*5.7-4.3*4.3)(8.1*8.1-1.9*1.9))/((2.2*2.2-7,8*7.8)(3,7*3,7-6,3*6,3)=((32.49-18,49)(65,61-3.61))/((4.84-60.84)(13.69-39.69))=(14*62)/(-56*(-26))=868/1456=424/738=212/369
№14
((7.29-1.96)(7.29+3.78+1.96))/(19.683-2.744)=(5.33*13.03)/16.939=69.4499/16.939=4.1
№15
((53,29-27.04)(53,29+37.96+27.04))/(389.017-140.608)=(26.25*118.29)/248.409=3105.1125/248.409=12.5

Answer 2

Условие переписывать не буду..
13) по ФСУ :  а² - b² = (a-b)(a+b)

$...= frac{(5,7-4,3)(5,7+4,3)(8,1-1,9)(8,1+1,9)}{(2,2-7,8)(2,2+7,8)(3,7-6,3)(3,7+6,3)} = frac{1,4*10*6,2*10}{(-5,6)*10 *(-2,6)*10} = \ \ = frac{14*62}{(-56)*(-26)} = frac{1*31}{4*13} = frac{31}{52}$

14) и 15 )  используем две формулы:
a²-b² = (a-b)(a+b)
а³ - b³ = (a - b)(a²+ab+b²)
$14) ...= frac{(2,7 + 1,4)(2,7 - 1.4)(2.7^2+2.7*1.4+1.4^2)}{2.7^3 - 1.4^3} = frac{4.1(2.7^3 - 1.4^3)}{2.7^3 - 1.4^3} = frac{4.1}{1} =4.1$

$15) ... = frac{(7.3+5.2)(7.3-5.2)(7.3^2+7.3*5.2+5.2^2)}{7.3^3-5.2^3} = frac{12.5*(7.3^3 -5.2^3)}{7.3^3-5.2^3} = frac{12.5}{1} =12.5$