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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.


Determinati Cea Mai Mica Valoare A Numarului Aradical Din X210x34 Radical Din Y24x8 Unde X Si Y Sunt Numere Reale class=

Răspuns :

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

[tex]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" [/tex]

Solutiile sunt:

x= -2   si   y=8