Question

Представьте в виде дроби выражение 5b/28 + 9b/28 9m/7n - 2m/7n 5x-3y/8z + 3x-13y/8z 4c-3d/cd - c-3d/cd 6x/x²-16 - 24/x²-16 m²+10m/9-m2 - 4m-9/9-m²

Answer 1

$\displaystyle \frac{5b}{28}+\frac{9b}{28}=\frac{5b+9b}{28}=\frac{14b}{14*2}=\frac{b}{2} \\\\ \frac{9m}{7n}-\frac{2m}{7n}=\frac{9m-2m}{7n}=\frac{7m}{7n}=\frac{m}{n}\\\\\frac{5x-3y}{8z}+\frac{3x-13y}{8z}=\frac{5x-3y+3x-13y}{8z}=\frac{8x-16y}{8z}=\frac{8(x-2y)}{8z}=\frac{x-2y}{z} \\\\\frac{4c-3d}{cd}-\frac{c-3d}{cd}=\frac{(4c-3d)-(c-3d)}{cd}=\frac{4c-3d-c+3d}{cd}=\frac{3c}{cd}=\frac{3}{d}\\\\\frac{6x}{x^2-16}-\frac{24}{x^2-16}=\frac{6x-24}{x^2-16}=\frac{6(x-4)}{(x-4)(x+4)}=\frac{6}{x+4}$

$\displaystyle \frac{m^2+10m}{9-m^2}-\frac{4m-9}{9-m^2}=\frac{m^2+10m-4m+9}{(3-m)(3+m)}=\frac{m^2+6m+9}{(3-m)(3+m)}=\\\\=\frac{(m+3)^2}{(3-m)(3+m)}=\frac{m+3}{3-m}$