Đáp án:

 \(\begin{array}{l}
2.6\\
a,\\
x = \dfrac{9}{{10}}\\
b,\\
x = \dfrac{7}{5}\\
c,\\
x = \dfrac{{10}}{9}\\
2.7\\
{x^2} + {y^2} = 28202\\
2.8\\
A = 392\\
2.9\\
a,\\
x\left( {x - 6} \right) + 10 > 0,\,\,\,\forall x\\
b,\\
\left( {x - 3} \right)\left( {x - 5} \right) + 3 > 0,\,\,\forall x\\
c,\\
{x^2} + x + 1 > 0,\,\,\,\forall x\\
2.10\\
a,\\
\left\{ \begin{array}{l}
x = 1\\
y = 2
\end{array} \right.\\
b,\\
\left\{ \begin{array}{l}
x = \dfrac{5}{2}\\
y = 1
\end{array} \right.\\
c,\\
\left\{ \begin{array}{l}
x = \dfrac{2}{3}\\
y = \dfrac{1}{2}
\end{array} \right.
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
2.6\\
a,\\
{\left( {x + 2} \right)^2} + {\left( {x + 3} \right)^2} - 2.\left( {x - 2} \right)\left( {x - 3} \right) = 19\\
 \Leftrightarrow \left( {{x^2} + 2.x.2 + {2^2}} \right) + \left( {{x^2} + 2.x.3 + {3^2}} \right) - 2.\left( {{x^2} - 3x - 2x + 6} \right) = 19\\
 \Leftrightarrow \left( {{x^2} + 4x + 4} \right) + \left( {{x^2} + 6x + 9} \right) - 2.\left( {{x^2} - 5x + 6} \right) = 19\\
 \Leftrightarrow {x^2} + 4x + 4 + {x^2} + 6x + 9 - 2{x^2} + 10x - 12 = 19\\
 \Leftrightarrow \left( {{x^2} + {x^2} - 2{x^2}} \right) + \left( {4x + 6x + 10x} \right) + \left( {4 + 9 - 12} \right) = 19\\
 \Leftrightarrow 20x + 1 = 19\\
 \Leftrightarrow 20x = 18\\
 \Leftrightarrow x = \dfrac{9}{{10}}\\
b,\\
\left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right) - x\left( {{x^2} - 5} \right) = 15\\
 \Leftrightarrow \left( {x + 2} \right).\left( {{x^2} - x.2 + {2^2}} \right) - \left( {{x^3} - 5x} \right) = 15\\
 \Leftrightarrow {x^3} + {2^3} - \left( {{x^3} - 5x} \right) - 15 = 0\\
 \Leftrightarrow {x^3} + 8 - {x^3} + 5x - 15 = 0\\
 \Leftrightarrow 5x - 7 = 0\\
 \Leftrightarrow 5x = 7\\
 \Leftrightarrow x = \dfrac{7}{5}\\
c,\\
{\left( {x - 1} \right)^3} + \left( {2 - x} \right).\left( {4 + 2x + {x^2}} \right) + 3x.\left( {x + 2} \right) = 17\\
 \Leftrightarrow \left( {{x^3} - 3.{x^2}.1 + 3.x{{.1}^2} - {1^3}} \right) + \left( {2 - x} \right).\left( {{2^2} + 2.x + {x^2}} \right) + \left( {3{x^2} + 6x} \right) = 17\\
 \Leftrightarrow \left( {{x^3} - 3{x^2} + 3x - 1} \right) + \left( {{2^3} - {x^3}} \right) + \left( {3{x^2} + 6x} \right) - 17 = 0\\
 \Leftrightarrow {x^3} - 3{x^2} + 3x - 1 + 8 - {x^3} + 3{x^2} + 6x - 17 = 0\\
 \Leftrightarrow \left( {{x^2} - {x^3}} \right) + \left( { - 3{x^2} + 3{x^2}} \right) + \left( {3x + 6x} \right) + \left( { - 1 + 8 - 17} \right) = 0\\
 \Leftrightarrow 9x - 10 = 0\\
 \Leftrightarrow 9x = 10\\
 \Leftrightarrow x = \dfrac{{10}}{9}\\
2.7\\
{x^2}y + x{y^2} + x + y = 2016\\
 \Leftrightarrow \left( {{x^2}y + x{y^2}} \right) + \left( {x + y} \right) = 2016\\
 \Leftrightarrow xy\left( {x + y} \right) + \left( {x + y} \right) = 2016\\
 \Leftrightarrow \left( {x + y} \right).\left( {xy + 1} \right) = 2016\\
xy = 11 \Rightarrow \left( {x + y} \right).\left( {11 + 1} \right) = 2016\\
 \Rightarrow x + y = \dfrac{{2016}}{{12}} = 168\\
{x^2} + {y^2} = \left( {{x^2} + 2xy + {y^2}} \right) - 2xy = {\left( {x + y} \right)^2} - 2xy\\
 = {168^2} - 2.11 = 28202\\
2.8\\
A = {a^2}\left( {a + 1} \right) - {b^2}\left( {b - 1} \right) - 3ab\left( {a - b + 1} \right) + ab\\
 = \left( {{a^3} + {a^2}} \right) - \left( {{b^3} - {b^2}} \right) - \left( {3{a^2}b - 3a{b^2} + 3ab} \right) + ab\\
 = {a^3} + {a^2} - {b^3} + {b^2} - 3{a^2}b + 3a{b^2} - 3ab + ab\\
 = {a^3} + {a^2} - {b^3} + {b^2} - 3{a^2}b + 3a{b^2} - 2ab\\
 = \left( {{a^3} - 3{a^2}b + 3a{b^2} - {b^3}} \right) + \left( {{a^2} - 2ab + {b^2}} \right)\\
 = {\left( {a - b} \right)^3} + {\left( {a - b} \right)^2}\\
a - b = 7 \Rightarrow A = {7^3} + {7^2} = 392\\
2.9\\
a,\\
x\left( {x - 6} \right) + 10 = {x^2} - 6x + 10 = \left( {{x^2} - 6x + 9} \right) + 1\\
 = \left( {{x^2} - 2.x.3 + {3^2}} \right) + 1 = {\left( {x - 3} \right)^2} + 1\\
{\left( {x - 3} \right)^2} \ge 0,\,\,\,\forall x\\
 \Rightarrow {\left( {x - 3} \right)^2} + 1 \ge 1 > 0,\,\,\,\forall x\\
 \Rightarrow x\left( {x - 6} \right) + 10 > 0,\,\,\,\forall x\\
b,\\
\left( {x - 3} \right)\left( {x - 5} \right) + 3 = \left[ {\left( {x - 4} \right) + 1} \right].\left[ {\left( {x - 4} \right) - 1} \right] + 3\\
 = \left[ {{{\left( {x - 4} \right)}^2} - {1^2}} \right] + 3 = {\left( {x - 4} \right)^2} - 1 + 3 = {\left( {x - 4} \right)^2} + 2\\
{\left( {x - 4} \right)^2} \ge 0,\,\,\,\forall x\\
 \Rightarrow {\left( {x - 4} \right)^2} + 2 \ge 2 > 0,\,\,\,\,\forall x\,\\
 \Rightarrow \left( {x - 3} \right)\left( {x - 5} \right) + 3 > 0,\,\,\forall x\\
c,\\
{x^2} + x + 1 = \left( {{x^2} + x + \dfrac{1}{4}} \right) + \dfrac{3}{4}\\
 = \left[ {{x^2} + 2.x.\dfrac{1}{2} + {{\left( {\dfrac{1}{2}} \right)}^2}} \right] + \dfrac{3}{4} = {\left( {x + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4}\\
{\left( {x + \dfrac{1}{2}} \right)^2} \ge 0,\,\,\forall x\\
 \Rightarrow {\left( {x + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} \ge \dfrac{3}{4} > 0,\,\,\forall x\\
 \Rightarrow {x^2} + x + 1 > 0,\,\,\,\forall x\\
2.10\\
a,\\
{x^2} - 2x + 5 + {y^2} - 4y = 0\\
 \Leftrightarrow \left( {{x^2} - 2x + 1} \right) + \left( {{y^2} - 4y + 4} \right) = 0\\
 \Leftrightarrow \left( {{x^2} - 2.x.1 + {1^2}} \right) + \left( {{y^2} - 2.y.2 + {2^2}} \right) = 0\\
 \Leftrightarrow {\left( {x - 1} \right)^2} + {\left( {y - 2} \right)^2} = 0\,\,\,\,\,\left( 1 \right)\\
\left. \begin{array}{l}
{\left( {x - 1} \right)^2} \ge 0,\,\,\,\forall x\\
{\left( {y - 2} \right)^2} \ge 0,\,\,\forall y\,
\end{array} \right\} \Rightarrow {\left( {x - 1} \right)^2} + {\left( {y - 2} \right)^2} \ge 0,\,\,\forall x,y\\
\left( 1 \right) \Rightarrow \left\{ \begin{array}{l}
{\left( {x - 1} \right)^2} = 0\\
{\left( {y - 2} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x - 1 = 0\\
y - 2 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 1\\
y = 2
\end{array} \right.\\
b,\\
4{x^2} + {y^2} - 20x - 2y + 26 = 0\\
 \Leftrightarrow \left( {4{x^2} - 20x + 25} \right) + \left( {{y^2} - 2y + 1} \right) = 0\\
 \Leftrightarrow \left[ {{{\left( {2x} \right)}^2} - 2.2x.5 + {5^2}} \right] + \left( {{y^2} - 2.y.1 + {1^2}} \right) = 0\\
 \Leftrightarrow {\left( {2x - 5} \right)^2} + {\left( {y - 1} \right)^2} = 0\,\,\,\,\,\left( 2 \right)\\
\left. \begin{array}{l}
{\left( {2x - 5} \right)^2} \ge 0,\,\,\forall x\\
{\left( {y - 1} \right)^2} \ge 0,\,\,\forall y
\end{array} \right\} \Rightarrow {\left( {2x - 5} \right)^2} + {\left( {y - 1} \right)^2} \ge 0,\,\,\,\forall x,y\\
\left( 2 \right) \Rightarrow \left\{ \begin{array}{l}
{\left( {2x - 5} \right)^2} = 0\\
{\left( {y - 1} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
2x - 5 = 0\\
y - 1 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{5}{2}\\
y = 1
\end{array} \right.\\
c,\\
9{x^2} + 4{y^2} + 4y - 12x + 5 = 0\\
 \Leftrightarrow \left( {9{x^2} - 12x + 4} \right) + \left( {4{y^2} + 4y + 1} \right) = 0\\
 \Leftrightarrow \left[ {{{\left( {3x} \right)}^2} - 2.3x.2 + {2^2}} \right] + \left[ {{{\left( {2y} \right)}^2} + 2.2y.1 + {1^2}} \right] = 0\\
 \Leftrightarrow {\left( {3x - 2} \right)^2} + {\left( {2y - 1} \right)^2} = 0\,\,\,\,\,\left( 3 \right)\\
\left. \begin{array}{l}
{\left( {3x - 2} \right)^2} \ge 0,\,\,\forall x\\
{\left( {2y - 1} \right)^2} \ge 0,\,\,\forall y
\end{array} \right\} \Rightarrow {\left( {3x - 2} \right)^2} + {\left( {2y - 1} \right)^2} \ge 0,\,\,\forall x,y\\
\left( 3 \right) \Rightarrow \left\{ \begin{array}{l}
{\left( {3x - 2} \right)^2} = 0\\
{\left( {2y - 1} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
3x - 2 = 0\\
2y - 1 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{2}{3}\\
y = \dfrac{1}{2}
\end{array} \right.
\end{array}\)