Question

Разложить на множители 16+9x^2/16-9x^2-3x/4-3x

Answer 1

$\frac{16+9x^2}{16-9x^2}-\frac{3x}{4-3x}=\\ =\frac{16+9x^2}{(4-3x)(4+3x)}-\frac{3x}{4-3x}=\\ =\frac{16+9x^2-3x(4+3x)}{(4-3x)(4+3x)}=\\ =\frac{16+9x^2-12x-9x^2}{(4-3x)(4+3x)}=\frac{16-12x}{(4-3x)(4+3x)}=\\ =\frac{4(4-3x)}{(4-3x)(4+3x)}=\frac{4}{4+3x}.$

Answer 2

(16+9x²)/(4-3x)(4+3x)-3x/(4-3x)=(16+9x²-12x-9x²)/(4-3x)94+3x)=
=(16-12x)/(4-3x)(4+3x)=4(4-3x)(4-3x)(4+3x)=4/(4+3x)