Question

\(\frac{1-4x^2}{x^2+4x}\):\(\frac{2-4x}{3x}\)

Answer 1

\(\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3x}\)

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

=\(\frac{\left(1-4x^2\right)3x}{\left(x^2+4x\right)\left(2-4x\right)}=\frac{3x-12x^3}{2x^2-4x^3+8x-16x^2}\)

Answer 2

Thuc hien phep tinh giup minh

 

Answer 3

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