Đáp án:
$\begin{array}{l}
2a)A = 2{x^2} + \left| {7x - 1} \right| - \left( {5 - x - 2{x^2}} \right)\\
= 2{x^2} + \left| {7x - 1} \right| - 5 + x + 2{x^2}\\
= 4{x^2} + x - 5 + \left| {7x - 1} \right|\\
+ Khi:x \ge \dfrac{1}{7} \Leftrightarrow \left| {7x - 1} \right| = 7x - 1\\
\Leftrightarrow A = 4{x^2} + x - 5 + 7x - 1\\
= 4{x^2} + 8x - 6\\
+ Khi:x < \dfrac{1}{7}\\
A = 4{x^2} + x - 5 + 1 - 7x\\
= 4{x^2} - 6x - 4\\
b)A = 2\\
+ Khi:x \ge \dfrac{1}{7}\\
\Leftrightarrow 4{x^2} + 8x - 6 = 2\\
\Leftrightarrow 4{x^2} + 8x - 8 = 0\\
\Leftrightarrow {x^2} + 2x - 2 = 0\\
\Leftrightarrow {\left( {x + 1} \right)^2} = 3\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 1 + \sqrt 3 \left( {tm} \right)\\
x = - 1 - \sqrt 3 \left( {ktm} \right)
\end{array} \right.\\
+ Khi:x < \dfrac{1}{7}\\
\Leftrightarrow 4{x^2} - 6x - 4 = 2\\
\Leftrightarrow 4{x^2} - 6x - 6 = 0\\
\Leftrightarrow {x^2} - \dfrac{3}{2}x - \dfrac{3}{2} = 0\\
\Leftrightarrow {x^2} - 2.x.\dfrac{3}{4} + \dfrac{9}{{16}} = \dfrac{9}{{16}} + \dfrac{3}{2}\\
\Leftrightarrow {\left( {x - \dfrac{3}{4}} \right)^2} = \dfrac{{33}}{{16}}\\
\Leftrightarrow x = \dfrac{{ - \sqrt {33} + 3}}{4}\left( {do:x < \dfrac{1}{7}} \right)\\
Vậy\,x = - 1 + \sqrt 3 ;x = \dfrac{{ - \sqrt {33} + 3}}{4}\\
4)a)M = 7x - 7y + 4a.x - 4a.y - 5\\
= 7.\left( {x - y} \right) + 4a.\left( {x - y} \right) + 5\\
= 7.0 + 4.a.0 + 5\left( {do:x - y = 0} \right)\\
= 5\\
b)N = x\left( {{x^2} + {y^2}} \right) - y\left( {{x^2} + {y^2}} \right) + 3\\
= \left( {{x^2} + {y^2}} \right).\left( {x - y} \right) + 3\\
= \left( {{x^2} + {y^2}} \right).0 + 3\\
= 3
\end{array}$
