Dạng 4: Bài luyện tập dạng 1: Sử dụng quy tắc chia để thực hiện phép tính có đáp án

$\frac{15x}{7y^3}:\frac{x^2}{2y^2}=\frac{15x}{7y^3}.\frac{2y^2}{x^2}=\frac{30}{7xy}$

$\left(-\frac{3x^2}{8y}\right):\frac{11x^4}{4y^2}=\frac{-3x^2}{8y}.\frac{4y^2}{11x^4}=\frac{-3y}{22x^2}$

$\left(-\frac{20x}{3y^2}\right):\left(-\frac{4x^3}{5y}\right)=\frac{-20x}{3y^2}.\frac{-5y}{4x^3}=\frac{25}{3x^{2}y}$

$-\frac{25x^2{y}^5}{3x}:15x{y}^2=-\frac{25x^2{y}^5}{3x}.\frac{1}{15x{y}^2}=\frac{-5y^3}{9}$

$\frac{4x+12}{{\left(x+4\right)}^2}:\frac{3\left(x+3\right)}{x+4}=\frac{4\left(x+3\right)}{{\left(x+4\right)}^2}.\frac{x+4}{3\left(x+3\right)}=\frac{4}{3\left(x+4\right)}$

$\frac{7x+2}{3x{y}^3}:\frac{14x+4}{x^{2}y}=\frac{7x+2}{3x{y}^3}.\frac{x^{2}y}{2\left(7x+2\right)}=\frac{x}{6y^2}$

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

$\frac{5xy}{2x-3}:\frac{15x{y}^3}{12-8x}=\frac{5xy}{2x-3}.\frac{12-8x}{15x{y}^3}=\frac{5xy}{2x-3}.\frac{4\left(3-2x\right)}{15x{y}^3}=\frac{5xy}{2x-3}.\frac{-4\left(2x-3\right)}{15x{y}^3}=\frac{-4}{3y^2}$

$\frac{5x-10}{x^2+7}:\left(2x-4\right)=\frac{5\left(x-2\right)}{x^2+7}.\frac{1}{2\left(x-2\right)}=\frac{5}{2\left(x^2+7\right)}$

$\left(x^2-25\right):\frac{2x+10}{3x-7}=\left(x-5\right)\left(x+5\right).\frac{3x-7}{2x+10}=\left(x-5\right)\left(x+5\right).\frac{3x-7}{2\left(x+5\right)}=\frac{\left(x-5\right)\left(3x-7\right)}{2}$

$\left(4x^2-16\right):\frac{3x+6}{7x-2}=4\left(x^2-4\right).\frac{7x-2}{3\left(x+2\right)}=4\left(x-2\right)\left(x+2\right).\frac{7x-2}{3\left(x+2\right)}=\frac{4\left(x-2\right)\left(7x-2\right)}{3}$

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

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

$\frac{8xy}{3x-1}:\frac{12x{y}^3}{5-15x}=\frac{8xy}{3x-1}.\frac{5\left(1-3x\right)}{12x{y}^3}=\frac{8xy}{3x-1}.\frac{-5\left(3x-1\right)}{12x{y}^3}=\frac{-10}{3y^2}$

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

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

$\frac{27-a^3}{5a+10}:\frac{a-3}{3a+6}=\frac{\left(3-a\right)\left(9-3a+a^2\right)}{5\left(a+2\right)}.\frac{3\left(a+2\right)}{a-3}=\frac{-3\left(a^2-3a+9\right)}{5}$

$\left(2b^2-32\right):\frac{b+4}{7b-2}=2\left(b^2-16\right).\frac{7b-2}{b+4}=2\left(b-4\right)\left(7b-2\right)$

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

$\frac{x^2+x}{5x^2-10x+5}:\frac{3x+3}{5x-5}=\frac{x\left(x+1\right)}{5\left(x^2-2x+1\right)}.\frac{5\left(x-1\right)}{3\left(x+1\right)}=\frac{x}{3(x-1)}$