Question

ДАЮ МАКС.БАЛЛ
ПОЖАЛУЙСТА СРОЧНО

Answer 1

Ответ: 1)

а) $\frac{x}{3} + \frac{x-2}{5} = \frac{5x+3(x-2)}{15} = \frac{5x+3x-6}{15} = \frac{8x-6}{15}$

б) $\frac{3y-2}{6} - \frac{y+1}{4} = \frac{2(3y-2)-3(y+1)}{12} = \frac{6y-4-3y-3}{12} = \frac{3y-7}{12}$

в) $-\frac{b-c}{7} + \frac{3b-c}{14} = \frac{-2(b-c)+3b-c}{14} = \frac{-2b+2c+3b-c}{14} = \frac{b+c}{14}$

г) $\frac{1}{a^2} + \frac{a-2}{a} = \frac{1+a*(a-2)}{a^2} = \frac{1+a^2-2a}{a^2} = \frac{a^2-2a+1}{a^2}$

д) $\frac{3x-5}{x} - \frac{3y-2}{y} = \frac{y*(3x-5)-x*(3y-2)}{xy} = \frac{3xy-5y-3xy+2x}{xy} = \frac{-5y+2x}{xy}$

е) $\frac{b-a}{ab} - \frac{a-b}{b^2} = \frac{b*(b-a)-a*(a-b)}{ab^2} = \frac{b^2-ab-a^2+ab}{ab^2} = \frac{b^2-a^2}{ab^2}$

2)

a) $\frac{(x+y)^2}{6y} + \frac{(x-y)^2}{12y} - \frac{x^2-y^2}{4y} = \frac{x^2+2xy+y^2}{6y} + \frac{x^2-2xy+y^2}{12y} - \frac{x^2-y^2}{4y}$

$\frac{2(x^2+2xy+y^2)+x^2-2xy+y^2-3(x^2-y&2)}{12y} = \frac{2x^2+4xy+2y&2+x^2-2xy+y^2-3x^2+3y^2}{12y}$

$\frac{0+2xy+6y^2}{12y} = \frac{2y*(x+3y)}{12y} = \frac{x+3y}{6}$

б) $\frac{3a+1}{7a} - \frac{7a+b}{14ab} - \frac{b-1}{2b} = \frac{2b*(3a+1)0&a+b)-7a*(b-1)}{14ab}$

$\frac{6ab+2b-7a-b-7ab+7a}{14ab} = \frac{-ab+b}{14ab} = \frac{b*(-a+1)}{14ab} = \frac{-a+1}{14a}$

Answer 2

решение на фотографиях