Question

Выполните действия:

Answer 1

$\frac{x+4}{x-1}-\frac{37x-12}{4x^2-3x-1}=\frac{x+4}{x-1}-\frac{37x-12}{(4x+1)(x-1)}=\frac{(x+4)(4x+1)-(37x-12)}{(4x+1)(x-1)}=\\=\frac{4x^2+17x+4-37x+12}{(4x+1)(x-1)}=\frac{4x^2-20x+20}{(4x+1)(x-1)}$

Answer 2

$\frac{x+4}{x-1} - \frac{37x-12}{4x^2-3x-1} = \frac{x+4}{x-1} - \frac{37x-12}{4x^2-4x+x-1}= \frac{x+4}{x-1} - \frac{37x-12}{(x-1)(4x+1)}= \\\ = \frac{(x+4)(4x+1)-37x+12}{(x-1)(4x+1)} = \frac{4x^2+x+16x+4-37x+12}{(x-1)(4x+1)} = \frac{4x^2-20x+16}{(x-1)(4x+1)} =\\\ = \frac{4(x^2-5x+4)}{(x-1)(4x+1)} = \frac{4(x^2-x-4x+4)}{(x-1)(4x+1)} = \frac{4(x-4)(x-1)}{(x-1)(4x+1)} =\frac{4(x-4)}{4x+1} =\frac{4x-16}{4x+1}$