Question

Va roggggg. Dau coroanaaaa.

Answer 1

[tex]28\\ a^2+4a+4+9b^2-6b+1=0\\ (a+2)^2+(3b-1)^2=0\\ a+2=0,a=-2\\ 3b-1=0,b=\frac{1}{3}\\ 29\\ \sqrt{a^2-4a+4+16}+\sqrt{b^2+16b+64+4}=\\ \sqrt{(a-2)^2+4}+\sqrt{b+8)^2+4}\geq\sqrt{4}+\sqrt{4}=2+2=4\\ [/tex]
Valoarea minima este 4 si se obtine cand  [tex]a=2\\ b=-\frac{1}{3}[/tex]
[tex]30\\ \sqrt{a^2-10a+50}=\sqrt{a^2-10a+25+25}=\sqrt{(a-5)^2+25}\geq\sqrt{25}=5\\ 20-\sqrt{a^2-10a+50}\leq 20-5=15 [/tex]
Valoarea maxima este 15 si se obtine cand a-5=0,a=5