Срочно!!!!! 155 Б и Г, 156 весь пожалуйста помогите!

Question 1

Question 2

155.
б)
$\frac{xy-1.2y^2}{x^2-1.2xy} + \frac{x^2+0.4xy}{xy+0.4y^2} = \frac{y(x-1.2y)}{x(x-1.2y)} + \frac{x(x+0.4y)}{y(x+0.4y)} = \frac{y} {x}+ \frac{x}{y}= \frac{x^2+y^2}{xy}$

г) смотри скан, возможно, я где-то ошиблась, т.к. длинновата концовка

156
а)
$\frac{1}{(a-b)(a-c)}+ \frac{1}{(b-c)(b-a)}+ \frac{1}{(c-a)(c-b)}= \\ \frac{1}{(a-b)(a-c)}- \frac{1}{(b-c)(a-b)}-\frac{1}{(a-c)(c-b)}= \\ \frac{1}{(a-b)(a-c)}- \frac{1}{(b-c)(a-b)}+\frac{1}{(a-c)(b-c)}= \\ \frac{b-c}{(a-b)(a-c)(b-c)}- \frac{a-c}{(a-b)(a-c)(b-c)}+\frac{a-b}{(a-b)(a-c)(b-c)}= \\ \frac{b-c-a+c+a-b}{(a-b)(a-c)(b-c)}= \frac{0}{(a-b)(a-c)(b-c)}= 0$

б)
$\frac{a}{(a-b)(a-c)}+ \frac{b}{(b-c)(b-a)}+ \frac{c}{(c-a)(c-b)}= \\ \frac{a}{(a-b)(a-c)}- \frac{b}{(b-c)(a-b)}-\frac{c}{(a-c)(c-b)}= \\ \frac{a}{(a-b)(a-c)}- \frac{b}{(b-c)(a-b)}+\frac{c}{(a-c)(b-c)}= \\ \frac{a(b-c)}{(a-b)(a-c)(b-c)}- \frac{b(a-c)}{(a-b)(a-c)(b-c)}+\frac{c(a-b)}{(a-b)(a-c)(b-c)}= \\ \frac{a(b-c)-b(a-c)+c(a-b)}{(a-b)(a-c)(b-c)}=\frac{ab-ac-ab+bc+ac-bc}{(a-b)(a-c)(b-c)}= \\ \frac{0}{(a-b)(a-c)(b-c)}= 0$

в)
$\frac{a^2}{(a-b)(a-c)}+ \frac{b^2}{(b-c)(b-a)}+ \frac{c^2}{(c-a)(c-b)}= \\ \frac{a^2}{(a-b)(a-c)}- \frac{b^2}{(b-c)(a-b)}-\frac{c^2}{(a-c)(c-b)}= \\ \frac{a^2}{(a-b)(a-c)}- \frac{b^2}{(b-c)(a-b)}+\frac{c^2}{(a-c)(b-c)}= \\ \frac{a^2(b-c)}{(a-b)(a-c)(b-c)}- \frac{b^2(a-c)}{(a-b)(a-c)(b-c)}+\frac{c^2(a-b)}{(a-b)(a-c)(b-c)}= \\ \frac{a^2(b-c)-b^2(a-c)+c^2(a-b)}{(a-b)(a-c)(b-c)}=\frac{a^2b-a^2c-ab^2+b^2c+ac^2-bc^2}{(a-b)(a-c)(b-c)}= \\ \frac{a^2b-a^2c-ab^2+b^2c+ac^2-bc^2}{(a^2-ac-ab+bc)(b-c)}= \\$
$\frac{a^2b-a^2c-ab^2+b^2c+ac^2-bc^2}{a^2b-a^2c-abc+ac^2-ab^2+abc+b^2c-bc^2}= \\ \frac{a^2b-a^2c-ab^2+b^2c+ac^2-bc^2}{a^2b-a^2c+ac^2-ab^2+b^2c-bc^2} \\ \frac{a^2b-a^2c-ab^2+b^2c+ac^2-bc^2}{a^2b-a^2c-ab^2+b^2c+ac^2-bc^2}=1$