Đáp án:

$\begin{array}{l}
1)A = {x^2} + 2x + 2\\
 = {x^2} + 2x + 1 + 1\\
 = {\left( {x + 1} \right)^2} + 1 \ge 1 > 0\forall x\\
2)B = {x^2} + 4x + 6\\
 = {x^2} + 4x + 4 + 2\\
 = {\left( {x + 2} \right)^2} + 2 \ge 2 > 0\forall x\\
3)C = {x^2} - x + 1\\
 = {x^2} - 2.x.\frac{1}{2} + \frac{1}{4} + \frac{3}{4}\\
 = {\left( {x - \frac{1}{2}} \right)^2} + \frac{3}{4} \ge \frac{3}{4} > 0\forall x\\
4)D = {x^2} + x + 1\\
 = {x^2} + 2.x.\frac{1}{2} + \frac{1}{4} + \frac{3}{4}\\
 = {\left( {x + \frac{1}{2}} \right)^2} + \frac{3}{4} \ge \frac{3}{4} > 0\forall x\\
5)E = {x^2} + 3x + 3\\
 = {x^2} + 2.x.\frac{3}{2} + \frac{9}{4} + \frac{3}{4}\\
 = {\left( {x + \frac{3}{2}} \right)^2} + \frac{3}{4} > 0\forall x\\
6)F = 2{x^2} + 4x + 3\\
 = 2.\left( {{x^2} + 2x + 1} \right) + 1\\
 = 2{\left( {x + 1} \right)^2} + 1 > 0\forall x\\
7)\\
G = 3{x^2} - 5x + 3\\
 = 3.\left( {{x^2} - \frac{5}{3}x} \right) + 3\\
 = 3.\left( {{x^2} - 2.x.\frac{5}{6} + \frac{{25}}{{36}}} \right) - 3.\frac{{25}}{{36}} + 3\\
 = 3.{\left( {x - \frac{5}{6}} \right)^2} + \frac{{11}}{{12}} > 0\forall x\\
8)H = 4{x^2} + 4x + 2\\
 = 4{x^2} + 4x + 1 + 1\\
 = {\left( {2x + 1} \right)^2} + 1 > 0\forall x\\
9)\\
K = 4{x^2} + 3x + 2\\
 = 4{x^2} + 2.2x.\frac{3}{4} + \frac{9}{{16}} + \frac{{23}}{{16}}\\
 = {\left( {2x + \frac{3}{4}} \right)^2} + \frac{{23}}{{16}} > 0\forall x\\
10)L = 2{x^2} + 3x + 4\\
 = 2.\left( {{x^2} + \frac{3}{2}x} \right) + 4\\
 = 2.\left( {{x^2} + 2.\frac{3}{4}x + \frac{9}{{16}}} \right) - 2.\frac{9}{{16}} + 4\\
 = 2.{\left( {x + \frac{3}{4}} \right)^2} + \frac{{23}}{{16}} > 0\forall x
\end{array}$