Question

aflati valorile ale lui t pentru care: diferenta rapoartelor algebrice 3t+1 supra 1-2t si t-1 supra t+1 este egala cu 1supra 2

Answer 1

[tex]\frac{3t+1}{1-2t}-\frac{t-1}{t+1}=\frac{1}{2}\\ 2(3t+1)(t+1)-2(t-1)(1-2t)=(1-2t)(t+1)\\ 2(3t^2+3t+t+1)-2(t-2t^2-1+2t)=(t-2t^2-2t+1)\\ 2(3t^2+4t+1)-2(-2t^2+3t-1)=-2t^2-t+1\\ 6t^2+8t+2+4t^2-6t+2=-2t^2-t+1\\ 12t^2+3t+3=0\\ 4t^2+t+1=0\\ \Delta=1-16=-15\ \textless \ 0\\ Deci\ t=\oslash(multimea\ vida)[/tex]