Question

Помогите пожалуйста ​

Answer 1

$\dfrac{(b-a)^2}{a+b}:\bigg(\dfrac{2ab}{b^2-a^2}+\dfrac a{a+b}-\dfrac b{b-a}\bigg)=$

$=\dfrac{(b-a)^2}{a+b}:\bigg(\dfrac{2ab}{(b-a)(b+a)}+\dfrac{a}{a+b}-\dfrac b{b-a}\bigg)=$

$=\dfrac{(b-a)^2}{a+b}:\dfrac{2ab+a(b-a)-b(a+b)}{(b-a)(a+b)}=\dfrac{(b-a)^2}{a+b}\cdot\dfrac{(b-a)(a+b)}{2ab+ab-a^2-ab-b^2}=$

$=\dfrac{(b-a)^2}{1}\cdot\dfrac{b-a}{-(a^2-2ab+b^2)}=(a-b)^2\cdot\dfrac{b-a}{-(a-b)^2}=\dfrac{b-a}{-1}=-(b-a)=a-b$

При  a = 8,4;  b = -0,6

$8.4-(-0.6)=8.4+0.6=9$

Ответ:  9

Answer 2

Ответ:

$\frac{ {(b - a)}^{2} }{a + b} \div ( \frac{2ab}{ {b}^{2} - {a}^{2} } + \frac{a}{a + b} - \frac{b}{b - a} ) = \frac{ {(b - a)}^{2} }{a + b} \div \frac{2ab + a \times (b - a) - b \times (b + a)}{(b - a)(b + a) } = \frac{ {(b - a)}^{2} }{a + b} \div \frac{2ab + ab - {a}^{2} - {b}^{2} - ab }{(b - a)(b + a)} = \frac{ {(b - a)}^{2} }{a + b} \div \frac{2ab - {a}^{2} - {b}^{2} }{(b - a)(b + a)} = \frac{ {(b - a)}^{2} }{a + b} \times \frac{(b - a)(b + a)}{ - {(b - a)}^{2} } = \frac{b - a}{ - 1} = a - b$

$a - b = 8.4 - ( - 0.6) = 8.4 + 0.6 = 9$