Question

VĂ ROG SĂ MĂ AJUTAȚI

DAU COROANĂ


IMI TREBUIE URGENT!!​

Answer 1

Explicație pas cu pas:

Pct a):

$\sqrt{x^2-2x+10}+\sqrt{y^2-4y+20}+\sqrt{z^2-6z+90}=16\\\sqrt{x^2-2x+1+9}+\sqrt{y^2-4y+4+16}+\sqrt{z^2-6z+9+81}=16\\\sqrt{(x-1)^2+9}+\sqrt{(y-2)^2+16}+\sqrt{(z-3)^2+81}=16$

Obervam ca:

$\sqrt{9}+\sqrt{16}+\sqrt{81}=\\=3+4+9=\\=7+9=\\=16$

Asadar, trebuie ca fiecare patrat de sub radical sa fie 0:

$(x-1)^2=0 => x-1=0 =>x=1\\(y-2)^2=0 => y-2=0 => y=2\\(z-3)^2=0 => z-3=0 => z=3$

Pct b):

$\sqrt{x^2+2x+5}+\sqrt{y^2-6y+13}+\sqrt{4z^2-4z+5}\leq 6\\\sqrt{x^2+2x+1+4}+\sqrt{y^2-6y+9+4}+\sqrt{4z^2-4z+1+4}\leq 6\\\sqrt{(x+1)^2+4}+\sqrt{(y-3)^2+4}+\sqrt{(2z-1)^2+4}\leq 6$

Dar, cum $\sqrt4+\sqrt4+\sqrt4=\\=2+2+2=\\=6$, fapt ce se intampla cand fiecare patrat de sub radical este 0, atunci expresia noastra este:

$\sqrt{(x+1)^2+4}+\sqrt{(y-3)^2+4}+\sqrt{(2z-1)^2+4}\geq 6$

Si cum, $\sqrt{(x+1)^2+4}+\sqrt{(y-3)^2+4}+\sqrt{(2z-1)^2+4}\leq 6$, atunci obtinem:

$\sqrt{(x+1)^2+4}+\sqrt{(y-3)^2+4}+\sqrt{(2z-1)^2+4}=6$

Asadar, trebuie ca fiecare patrat de sub radical sa fie 0:

$(x+1)^2=0 => x+1=0 => x=-1\\(y-3)^2=0 => y-3=0 => y=3\\(2z-1)^2=0 => 2z-1=0 => 2z=1 => z=\frac{1}{2}$