Question

Determinati cea mai mica valoare a numarului:

A=radical din X^2+10x+34 + radical din y^2-4x+8

Unde x si y sunt numere reale.

Answer 1

Daca numarul A are cea mai mica valoare, A fiind numar natural inseamna ca A=0

$A=\sqrt{x^2+10x+34}+\sqrt{y^2-4x+8}=0 \\\\ \sqrt{x^2+10x+25+9}+\sqrt{y^2-4x+4+4}=0 \\\\ \sqrt{(x+5)^2+9}+\sqrt{(y-4)^2+4}=0 \\\\ ||x+5|+3|+||y-4|+4|=0 \\\\ ||x+5|+3|=0 \\\\ 1) |x+5|+3=0 \\\\ |x+5|=-3 \ \ \ "F" \\\\ 2)|x+5|-3=0 \\\\ |x+5|=3 \\\\ a) |x+5|=3 \\\\ x+5=3 \\\\ x=3-5 \\\\ x=-2 \\\\ b)|x+5|=-3 \ \ \ "F" \\\\\\ ||y-4|+4|=0 \\\\ 1)|y-4|+4=0 \\\\ |y-4|=-4 \ \ \ "F" \\\\ 2)|y-4|-4=0 \\\\ |y-4|=4 \\\\ a) |y-4|=4 \\\\ y-4=4 \\\\ y=8 \\\\ b)|y-4|=-4 \ \ \ "F"$

Solutiile sunt:

x= -2   si   y=8