Question

$x,y,z \in R. \\ Daca~(2x-1)^2+(y-2)^2+(3z-1)^2=9,~de~ce~Xmin=-1? \\ Xmin=minimul~de~x. \\ Eu~l-am~determinat~ca~fiind~ \frac{1}{2}$

Answer 1

$\displaystyle E~destul~de~evdent. \\ \\ 9=(2x-1)^2+(y-2)^2+(3z-1)^2 \ge(2x-1)^2+0+0=(2x-1)^2. \\ \\ Deci~9 \ge (2x-1)^2 \Leftrightarrow -3 \le 2x-1 \le 3 \Leftrightarrow -1 \le x \le 2. \\ \\ Rezulta~x_{min}=-1. \\ \\ Evident,~valoarea~poate~fi~atinsa.$