Explicație pas cu pas:
La ex de genul acesta trebuie sa te legi de acea ecuatie
x-7y+3=0 => x=7y-3
E=V(x²+y²+6x+9)+V(x²+y²-8x-2y+17)
=> E=V[(x+3)²+y²] + V(x²-8x+16+y²-2y+1) =>
E=V[(7y-3+3)²+y²]+V[(x-4)²+(y-1)²] =>
E=V[(7y)²+y²] +V[(7y-7)²+(y-1)²] =>
E=V50y² + V[7²(y-1)²+(y-1)²] =>
E=|y|V50 + V50(y-1)² =>
E=|y|V50+|y-1|V50 => E=V50(|y|+|y-1|)
Acum sa vedem in ce interval se afla y
x€[-3,4] => -3<=x<=4 => -3<=7y-3<=4 => 0<=7y<=7 =>
0<=y<=1 => y-1<=0
Deci E=V50(y+1-y) => E=V50=5V2 => valoare constanta
y=(x+3)/5 => 5y=x+3 => x=5y-3
E=V(x²+y²+6x+9)+V(x²+y²-4x-2y+5) =>
E=V[(x+3)²+y²] +V(x²-4x+4+y²-2y+1) =>
E=V[(5y-3+3)+y²]+V[(x-2)²+(y-1)²] =>
E=V[(5y)²+y²]+V[(5y-3-2)²+(y-1)²] =>
E=V26y²+V[(5y-5)²+(y-1)²] =>
E=|y|V26 + V[25(y-1)²+(y-1)²] =>
E=|y|V26+V26(y-1)² =>
E=|y|V26 +|y-1|V26 => E=V26(|y|+|y-1|)
x€[-3,2] =>-3<=x<=2 => -3<=5y-3<=2 => 0<=5y<=5 =>
0<=y<=1 => y-1<=0 => E=V26(y+1-y) =V26=>valoare constanta
