Question

Dau coroană............................urgent....o

Answer 1

Răspuns:

$a)\:\frac{2x+3}{x-4}\times\frac{x-1}{x+4} =\frac{(2x+3)(x-1)}{(x-4)(x+4)} =\frac{2x^{2} -2x+3x-3}{x^{2} -4^{2} } =\frac{2x^{2} +x-3}{x^{2} -16}$

$b)\:\frac{x+3}{x-5} \times\frac{x+3}{1-4x} =\frac{(x+3)^{2} }{(x-5)(1-4x)} = \frac{x^{2}+6x+9 }{x-4x^{2} -5+20x} = \frac{x^{2}+6x+9 }{21x-4x^{2} -5} = \frac{x^{2}+6x+9 }{-4x^{2} +21x-5}$