Question

B) va rogggg ............

Answer 1

Răspuns:

$\frac{x-1}{x-2} -\frac{x+3}{x+2}+\frac{2}{)x-2)*(x+2)}  \\\frac{(x+2)*(x-1)-(x-2)*(x+3)+2}{(x-2)*(x+2)} \\\frac{x^{2}-x+2x-2-(x^{2} +3x-2x-6)+2 }{x^{2} -4} \\\frac{x^{2} -x+2x-2-(x^{2} +x-6)+2}{x^{2} -4}\\\frac{x^{2} -x+2x-x-x^{2} -x+6}{x^{2} -4} \\\frac{-x+2x-x+6}{x^{2} -4} \\\frac{0+6}{x^{2} -4} \\\frac{6}{x^{2} -4}$

Explicație pas cu pas:

Answer 2

Transformam 4 in 2^2

(x-1/x-2)-(x+3/x+2)+2/(x^2-2^2)

Acum  avem formula a^2-b^2=(a-b)(a+b)

(x-1/x-2)-(x+3/x+2)+2/((x-2)*(x+2))

Amplificam fractiile pentru a ajunge la acelasi numitor:

((x+2)*(x-1))/((x+2)*(x-2))-((x-2)(x+3)/(x-2)(x+2))+2/(x-2)(x+2)

Scriem toti numaratorii sub acelasi numitor comun (x-2)*(x+2)

((x+2)(x-1)-(x-2)(x+3)+2/(x-2)(x+2))

x^2-x+2x-(x^2+3x-2x-6)+2/x^2-4 //am imultit parantezele si am folosit formula (a-b)(a+b)=a^2-b^2

x^2-x+2x-(x^2+x-6)/x^2-4

Schimbam semnul fiecarui termen din paranteza, deoarece avem "-" in fata

x^2-x+2x-x^2-x+6/x^2-4

Eliminam x^2 si -x^2

-x+2x-x+6/x^2-4

=0+6/x^2-4

=6/x^2-4