Đáp án:
$\begin{array}{l}
3)\\
Q = \left[ {5{x^2} - a.\left( {x + a} \right)} \right] - \left[ {3\left( {{a^2} - {x^2}} \right) + 2a.x} \right]\\
+ \left[ {2a.x - 4\left( {a + 2a.{x^2}} \right)} \right]\\
= 5{x^2} - a.x - {a^2} - 3{a^2} + 3{x^2} - 2a.x\\
+ 2a.x - 4a - 8a.{x^2}\\
= 8{x^2} - 4{a^2} - a.x - 8a.{x^2}\\
= \left( {8 - 8a} \right).{x^2} - a.x - 4{a^2}\\
\Leftrightarrow He\,so:{x^2} = 8 - 8a\\
4)2 - {x^2}\left( {{x^2} + x + 1} \right) = - {x^4} - {x^3} - {x^2} + m\\
\Leftrightarrow 2 - {x^4} - {x^3} - {x^2} = - {x^4} - {x^3} - {x^2} + m\\
\Leftrightarrow m = 2\\
Vậy\,m = 2\\
5)A = a\left( {2b + 1} \right) - b\left( {2a - 1} \right)\\
= 2ab + a - 2ab + b\\
= a + b\\
= 10 + \left( { - 5} \right)\\
= 5\left( {do:a = 10;b = - 5} \right)\\
6)\\
10\left( {3x - 2} \right) - 3\left( {5x + 2} \right) + 5\left( {11 - 4x} \right) = 25\\
\Leftrightarrow 30x - 20 - 15x - 6 + 55 - 20x = 25\\
\Leftrightarrow \left( {30x - 15x - 20x} \right) = 25 + 20 + 6 - 55\\
\Leftrightarrow - 5x = - 4\\
\Leftrightarrow x = \dfrac{4}{5}\\
Vậy\,x = \dfrac{4}{5}\\
B5)\\
1)\left( { - {a^4}{x^5}} \right).\left( { - {a^6}x + 2{a^3}{x^2} - 11a.{x^5}} \right)\\
= {a^{10}}.{x^6} - 2{a^7}{x^7} + 11{a^5}.{x^{10}}\\
2)A = mx\left( {x - y} \right) + {y^3}.\left( {x + y} \right)\\
= m.\left( { - 1} \right).\left( { - 1 - 1} \right) + 1.\left( { - 1 + 1} \right)\\
= 2m\\
3)\\
8.\left( {x - 2} \right) - 2.\left( {3x - 4} \right) = 2\\
\Leftrightarrow 8x - 16 - 6x + 8 = 2\\
\Leftrightarrow 2x = 10\\
\Leftrightarrow x = 5\\
Vậy\,x = 5\\
4)\\
Q = 5x\left( {3{x^2} - x + 2} \right) - 2{x^2}\left( {x - 2} \right) + 15\left( {x - 1} \right)\\
= 15{x^3} - 5{x^2} + 10x - 2{x^3} + 4{x^2} + 15x - 15\\
= 13{x^3} - {x^2} + 25x - 15\\
\Leftrightarrow He\,so\,{x^2} = - 1
\end{array}$
