Решите пожалуйста только с 1 по 4 задания

1.
$\frac{4x+y}{10}+ \frac{6x-5y}{4}-x= \frac{2(4x+y)}{20}+ \frac{5(6x-5y)}{20}- \frac{20x}{20}= \\ \frac{2(4x+y)+5(6x-5y)-20x}{20}= \frac{8x+2y+30x-25y-20x}{20}= \frac{18x-23y}{20}$

2.
$\frac{4a-k}{15k}+ \frac{k-3a}{25k} = \frac{5(4a-k)}{75k}+ \frac{3(k-3a)}{75k} =\frac{5(4a-k)+3(k-3a)}{75k} =\frac{20a-5k+3k-9a}{75k} = \\ \frac{11a-2k}{75k}$

3.
$\frac{5a^2}{ab-2b^2} - \frac{10a}{a-2b} = \frac{5a^2}{b(a-2b)} - \frac{10a}{a-2b} =\frac{5a^2}{b(a-2b)} - \frac{10ab}{b(a-2b)} = \\ =\frac{5a^2-10ab}{b(a-2b)} =\frac{5a(a-2b)}{b(a-2b)} = \frac{5a}{b}$

4.
$\frac{x-12y}{x^2-16y^2}- \frac{4y}{4xy-x^2} = \frac{x-12y}{x^2-(4y)^2}- \frac{4y}{x(4y-x)} = \\ \frac{x-12y}{(x-4y)(x+4y)}- \frac{4y}{x(4y-x)} =\frac{x(x-12y)}{x(x-4y)(x+4y)}+ \frac{4y(x+4y)}{x(x-4y)(x+4y)} = \\ \frac{x(x-12y)+4y(x+4y)}{x(x-4y)(x+4y)} = \frac{x^2-12xy+4xy+16y^2}{x(x-4y)(x+4y)} =\frac{x^2-8xy+16y^2}{x(x-4y)(x+4y)} = \\ =\frac{x^2-2*x*4y+(4y)^2}{x(x-4y)(x+4y)} = \frac{(x-4y)^2}{x(x-4y)(x+4y)} = \frac{x-4y}{x(x+4y)} = \frac{x-4y}{x^2+4xy}$