Ajutor! Ex 1 punctul b va rogggggggggg. Dau coroanaa

√(x²-2x+5)+√(y²+4y+13)=5
√[( x²-2x+1)+4]+√[(y²+4y+4)+9]=5
√[(x-1)²+2²]+√[(y+2)²+3²]=5
(x-1)²≥0 si (y+2)²≥0
√[(x-1)²+2²]≥2
√[(y+2)²+3²]≥3
Suma lor fiind 5,
=>√[(x-1)²+2²]=2 =>(x-1)²=0; x=1
√[(y+2)²+3²]=3=>(y+2)²=0; y= -2