Question

DAU COROANAAA URGENTTT VA ROOOGGG ​

Answer 1

a)

$E(x) = x^{2} +4x + 5 = x^{2} +4x + 4 + 1 = (x + 2)^{2} + 1$

$(x + 2)^{2} \geq 0 \implies E(x) \geq 1$

b)

$E(a) +E(b) + E(c) = 3$

$(a + 2)^{2} + 1 + (b + 2)^{2} + 1 + (c + 2)^{2} + 1 = 3$

$(a + 2)^{2} + (b + 2)^{2} + (c + 2)^{2} = 0$

​o sumă de pătrate nu poate fi nulă decât dacă fiecare termen este nul

$\left\{\begin{matrix} (a + 2)^{2} = 0 \\(b + 2)^{2} = 0 \\(c + 2)^{2} = 0 \end{matrix}\right. \iff \left\{\begin{matrix} a + 2 = 0 \\b + 2 = 0 \\c + 2 = 0 \end{matrix}\right. \implies \left\{\begin{matrix} a = -2 \\b = -2 \\c = -2 \end{matrix}\right.$