Question

$x-(2- \frac{1}{x+3} ):[1-( \frac{x+3}{x+2} )^{-2}$ va rog.

Answer 1

[tex]x-(2- \frac{1}{x+3}) :[1-( \frac{x+3}{x+2} ) ^{-2} ] \\ x-2+ \frac{1}{x+3} :[1-( \frac{x+2}{x+3} )^{2}] \\ x-2+ \frac{1}{x+3} :[1- \frac{(x+2) ^{2} }{(x+3) ^{2} } ] \\ x-2+ \frac{1}{x+3} : \frac{(x+3) ^{2}-(x+2) ^{2} }{(x+3) ^{2} } \\ x-2+ \frac{1}{x+3} * \frac{(x+3) ^{2} }{(x+3-x-2)(x+3+x+2)} \\ x-2+ \frac{x+3}{2x+5} \\ \frac{(x-2)(2x+5)+x+3}{2x+5} \\ \frac{2 x^{2} +5x-4x-10+x+3}{2x+5} \\ \frac{2 x^{2} +2x-7}{2x+5} \\ [/tex]
Nu gasesc o cale sa merg mai departe