Dạng 1. Sử dụng quy tắc chia để thực hiện phép tính có đáp án

$\frac{7xy}{3x+1}:\frac{14x^{2}y}{6x+2}=\frac{7xy}{3x+1}.\frac{6x+2}{14x^{2}y}=\frac{7xy}{3x+1}.\frac{2.\left(3x+1\right)}{14x^{2}y}=\frac{1}{x}.$

$\frac{34x^2{y}^3}{2x{y}^2+2y^2}:\frac{17xy}{3x+3}=\frac{34x^2{y}^3}{2x{y}^2+2y^2}.\frac{3x+3}{17xy}=\frac{34x^2{y}^3}{2y^2\left(x+1\right)}.\frac{3\left(x+1\right)}{17xy}=\frac{102x^2{y}^3.\left(x+1\right)}{34x{y}^3\left(x+1\right)}=3x$

$\frac{x^3-27}{x+3}:\left(x^2-6x+9\right)=\frac{x^3-3^3}{x+3}.\frac{1}{x^2-6x+9}=\frac{\left(x-3\right)\left(x^2+3x+9\right)}{x+3}.\frac{1}{{\left(x-3\right)}^2}=$$\frac{x^2+3x+9}{\left(x+3\right)\left(x-3\right)}=\frac{x^2+3x+9}{x^2-9}$

$\left(x^2+2x+1\right):\frac{x^2-1}{2x+3}=\left(x^2+2x+1\right).\frac{2x+3}{x^2-1}={\left(x+1\right)}^2.\frac{2x+3}{\left(x-1\right)\left(x+1\right)}=\frac{\left(x+1\right)\left(2x+3\right)}{x-1}$

$\frac{9x^2-4}{3x+1}:\frac{3x+2}{6x^2+2x}=\frac{{\left(3x\right)}^2-2^2}{3x+1}.\frac{6x^2+2x}{3x+2}=\frac{\left(3x-2\right)\left(3x+2\right)}{3x+1}.\frac{2x\left(3x+1\right)}{3x+2}=2x\left(3x-2\right)$

$\frac{5x-15}{x^2-4}:\frac{x-3}{x+2}=\frac{5.\left(x-3\right)}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{x-3}=\frac{5}{x-2}$

$\frac{x^3-8}{x^2-4}:\left(x^2+2x+4\right)=\frac{x^3-2^3}{x^2-2^2}.\frac{1}{x^2+2x+4}=\frac{\left(x-2\right)\left(x^2+2x+4\right)}{\left(x-2\right)\left(x+2\right)}.\frac{1}{x^2+2x+4}=\frac{1}{x+2}$

$\frac{2x+4x^2}{x^2+x}:\frac{4x^2+4x+1}{x+1}=\frac{2x+4x^2}{x^2+x}.\frac{x+1}{4x^2+4x+1}=\frac{2x.\left(1+2x\right)}{x.\left(x+1\right)}.\frac{x+1}{{\left(2x+1\right)}^2}=\frac{2}{2x+1}$

$\left(3x^2-48\right):\frac{2x-8}{9x+6}=\left(3x^2-48\right).\frac{9x+6}{2x-8}=3.\left(x^2-16\right).\frac{3.\left(3x+2\right)}{2.\left(x-4\right)}$$=3.\left(x-4\right)\left(x+4\right).\frac{3.\left(3x+2\right)}{2.\left(x-4\right)}=\frac{9.\left(x+4\right)\left(3x+2\right)}{2}$

$\left(3-6x+3x^2\right):\frac{x^2-1}{x}=3.\left(1-2x+x^2\right).\frac{x}{x^2-1}=3.{\left(x-1\right)}^2.\frac{x}{\left(x-1\right)\left(x+1\right)}=\frac{3x.\left(x-1\right)}{x+1}$

$\frac{x-1}{x-2}:\frac{x-2}{x-3}:\frac{\left(x-1\right)\left(x-3\right)}{x^2-4}=\frac{x-1}{x-2}.\frac{x-3}{x-2}:\frac{\left(x-1\right)\left(x-3\right)}{\left(x-2\right)\left(x+2\right)}$$=\frac{\left(x-1\right)\left(x-3\right)}{{\left(x-2\right)}^2}.\frac{\left(x-2\right)\left(x+2\right)}{\left(x-1\right)\left(x-3\right)}=\frac{x+2}{x-2}$

$\frac{x^3+1}{x-1}:\left(x^2-x+1\right):\frac{x+1}{x-1}=\frac{\left(x+1\right)\left(x^2-x+1\right)}{x-1}.\frac{1}{x^2-x+1}:\frac{x+1}{x-1}=\frac{x+1}{x-1}.\frac{x-1}{x+1}=1$

$\frac{9x^2-6x+1}{x^2+2xy+4y^2}:\frac{12x-4}{4x^3+32y^3}=\frac{{\left(3x-1\right)}^2}{x^2+2xy+4y^2}.\frac{4x^3+32y^3}{12x-4}=\frac{{\left(3x-1\right)}^2}{x^2+2xy+4y^2}.\frac{4.\left(x^3+8y^3\right)}{4.\left(3x-1\right)}$

$=\frac{{\left(3x-1\right)}^2}{x^2+2xy+4y^2}.\frac{4.\left(x+2y\right)\left(x^2+2xy+4y^2\right)}{4.\left(3x-1\right)}=\left(3x-1\right)\left(x+2y\right)$

$\frac{x^2+3x+2}{x^2+x-6}:\frac{x^2+x}{x^2-4x+4}=\frac{x^2+3x+2}{x^2+x-6}.\frac{x^2-4x+4}{x^2+x}=\frac{\left(x+1\right)\left(x+2\right)}{\left(x+3\right)\left(x-2\right)}.\frac{{\left(x-2\right)}^2}{x.\left(x+1\right)}=\frac{\left(x+2\right)\left(x-2\right)}{x.\left(x+3\right)}$

$\frac{x^4-y^4}{4x^2-4x+1}:\frac{3x^{2}y+3x{y}^2}{6-12x}=\frac{x^4-y^4}{4x^2-4x+1}.\frac{6-12x}{3x^{2}y+3x{y}^2}=\frac{\left(x^2-y^2\right)\left(x^2+y^2\right)}{{\left(2x-1\right)}^2}.\frac{6.\left(1-2x\right)}{3xy.\left(x+y\right)}$

$=\frac{\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)}{{\left(1-2x\right)}^2}.\frac{6.\left(1-2x\right)}{3xy.\left(x+y\right)}=\frac{2\left(x-y\right)\left(x^2+y^2\right)}{xy.\left(1-2x\right)}$

$\frac{x^2-4xy+4y^2}{2x^2-2xy+2y^2}:\frac{10x-20y}{5x^3+5y^3}=\frac{{\left(x-2y\right)}^2}{2.\left(x^2-xy+y^2\right)}.\frac{5.\left(x^3+y^3\right)}{10.\left(x-2y\right)}$

$=\frac{{\left(x-2y\right)}^2}{2.\left(x^2-xy+y^2\right)}.\frac{5.\left(x+y\right)\left(x^2-xy+y^2\right)}{10.\left(x-2y\right)}=\frac{\left(x-2y\right)\left(x+y\right)}{4}$