Question

Упростите выражение; 1) 2y^2-5xy/x^2-4y^2-x/2y-x-y/x+2y; 2) x-1/2x-6-1/x-3x-3/2x^2-6x/

Answer 1

$\tt\displaystyle \frac{2y^2-5xy}{x^2-4y^2}-\frac{x}{2y-x}-\frac{y}{x+2y}=\frac{2y^2-5xy}{(x-2y)(x+2y)}+\frac{x}{x-2y}-\frac{y}{x+2y}= \\\\\\=\frac{2y^2-5xy+x(x+2y)-y(x-2y)}{(x-2y)(x+2y)}=\frac{2y^2-5xy+x^2+2yx-yx+2y^2}{(x-2y)(x+2y)}=\\\\\\=\frac{4y^2-4xy+x^2}{(x-2y)(x+2y)}=\frac{(x-2y)^2}{(x-2y)(x+2y)}=\bold{\frac{x-2y}{x+2y}}$

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$\tt\displaystyle \frac{x-1}{2x-6} -\frac{1}{x} -\frac{3x-3}{2x^2-6x}=\frac{x-1}{2(x-3)} -\frac{1}{x} -\frac{3x-3}{2x(x-3)}=\\\\\\\frac{x(x-1)-1(2(x-3))-3x+3}{2x(x-3)} =\frac{x^2-x-1(2x-6)-3x+3}{2x(x-3)} =\\\\\\=\frac{x^2-x-2x+6-3x+3}{2x(x-3)} =\frac{x^2-6x+9}{2x(x-3)} =\frac{(x-3)^2}{2x(x-3)}=\bold{\frac{x-3}{2x}}$