Question

ХЭЛП ПЖ СРОЧНО АЛГЕБРА!!!!!!!!!!!!!

Answer 1

Ответ:

7

Объяснение:

(2ad/(a²-d²)+(a-d)/(2a+2d))*14a/(a+d)+7d/(d-a)

(2ad(2a+2d)+(a-d)(a²-d²))/((a²-d²)(2a+2d))*14a/(a+d)+7d/(d-a)

(4a²d+4ad²+a³-ad²-a²d+d³)/((a-d)(a+b)(2a+2d))*14a/(a+d)+7d/(d-a)

(a³+3a²d+3ad²+d³)/((a-d)(a+b)(2a+2d))*14a/(a+d)+7d/(d-a)

(a+d)³/((a-d)(a+b)(2a+2d))*14a/(a+d)+7d/(d-a)

(a+d)²/(2a²-2d²)*14a/(a+d)+7d/(d-a)

(a+d)²/(2(a²-d²))*14a/(a+d)+7d/(d-a)

(a+d)²/(2(a-d)(a+b))*14a/(a+d)+7d/(d-a)

(a+d)/(2(a-d))*14a/(a+d)+7d/(d-a)

(14a(a+d))/(2*(a-d)(a+d))+7d/(d-a)

14a/(2(a-d))+7d/(d-a)

7a/(a-d)+7d/(d-a)

(7a(d-a)+7d(a-d))/((a-d)(d-a))

7(ad-a²+ad-d²)/(2ad-a²-d²)

7((2ad-a²-d²)/(2ad-a²-d²)

7*1=7

Answer 2

$\displaystyle\bf\\1)\\\\\frac{2ad}{a^{2} -d^{2} } +\frac{a-d}{2a+2d}=\frac{2ad}{(a+d)(a-d)} +\frac{a-d}{2(a+d)} =\\\\\\=\frac{2ad\cdot 2+(a-d)\cdot(a-d)}{2(a+d)(a-d)} =\frac{4ad+a^{2}-2ad+d^{2} }{2(a+d)(a-d)} =\\\\\\=\frac{a^{2}+2ad+d^{2} }{2(a+d)(a-d)} =\frac{(a+d)^{2} }{2(a+d)(a-d)} =\frac{a+d}{2(a-d)} \\\\\\2)\\\\\frac{a+d}{2(a-d)} \cdot\frac{14a}{a+d} =\frac{7a}{a-d}$

$\displaystyle\bf\\3)\\\\\frac{7a }{a-d} +\frac{7d}{d-a} =\frac{7a}{a-d} -\frac{7d}{a-d} =\frac{7a-7d}{a-d}=\frac{7(a-d)}{a-d} =7\\\\Otvet: \ 7$

В ответе нет переменных ,  значит значение выражения от них не зависит .