Question

помогите решить первый номер​

Answer 1

$\frac{7x-3y}{2x-y}-\frac{2y-3x}{2x-y}=\frac{7x-3y-(2y-3x)}{2x-y}=\frac{7x-3y-2y+3x}{2x-y}=\frac{10x-5y}{2x-y}=\frac{5\cdot(2x-y)}{2x-y}=5\\ \\\frac{4x-3y}{6x-3y}-\frac{2x-y}{2x+5y}=\frac{(4x-3y)(2x+5y)+(2x-3y)(6x-3y)}{(6x-3y)(2x+5y)}=\frac{8x^2-6xy+20xy-15y^2+12x^2-18xy-6xy+9y^2}{(6x-3y)(2x+5y)}=\\\\=\frac{20x^2-10xy-6y^2}{(6x-3y)(2x+5y)}$

$x^2-9x-\frac{x(x-3)(x+3)}{x+7}=\frac{(x^2-9x)(x+7)-x(x-3)(x+3)}{x+7}=\frac{x^3-9x^2+7x^2-63x-x^3+9x}{x+7}=\frac{-2x^2-54x}{x+7}=-2\cdot\frac{x(x+27)}{x+7}$