Question

Сократите дробь 4а/16аb -3x^2/9x^3 30a^2c3/48a^3c^2 3c+15d/9(c+5d) x^2+8x+16/2x+8 y^2+y^3 2) 3x/x-4+x + 8/x -4 y + x -3/y^2-9 -x/y^2-9 2x+17/49-x^2-10+x/49-x2 y^2-8y/y-3-9-2y/3-y 29+x^2/(6-x)^2-2(5x-1)/(x-6)^2+5-2x/(6-x)^2

Answer 1

$\frac{4a}{16ab} =\frac{1}{4b} \\ \\ \frac{3x^{2}} {9x^{3} }=\frac{1}{3x} \\ \\ \frac{30a^{2} c^{3}} {48a^{3} c^{2} }=\frac{5c}{7a} \\ \\ \frac{3c+15d}{9(c+5d)} =\frac{3(c+5d)}{9(c+5d)} =\frac{1}{3} \\ \\ \frac{x^{2}+8x+16} {2x+8} =\frac{(x+4)(x+4)}{2(x+4)} =\frac{x+4}{2} \\ \\ y^{2} +y^{3} =y^{2} (1+y)$

$\frac{3x}{x+4} +\frac{x+8}{x+4} =\frac{3x+x+8}{x+4} =\frac{4x+8}{x+4} =\frac{4(x+2)}{x+4} \\ \\ \frac{y+x-3}{y^{2}-9} -\frac{x}{y^{2}-9} =\frac{y+x-3-x}{y^{2}-9} =\frac{y-3}{(y-3)(y+3)} =\frac{1}{y+3} \\ \\ \frac{2x+17}{49-x^{2}}-\frac{10+x}{49-x^{2}} =\frac{2x+17-10-x}{(7+x)(7-x)} =\frac{x+7}{(7+x)(7-x)} =\frac{1}{7-x} \\ \\ \frac{y^{2}-8y}{y-3} -\frac{9-2y}{3-y} =\frac{y^{2}-8y}{y-3} +\frac{9-2y}{y-3}= \frac{y^{2}-10y+9}{y-3} [tex]\frac{29+x^{2}}{(6-x)^{2}} +\frac{5-2x}{(6-x)^{2}} =\frac{x^{2}-2x+34}{36-12x+x^{2}}$