Đáp án:
$\begin{array}{l}
1)a)3{x^2}y\left( {4x{y^2} + 5xy} \right)\\
= 12{x^3}{y^3} + 15{x^3}{y^2}\\
b)\left( {2a - b} \right)\left( {b - 2a} \right)\\
= - \left( {2a - b} \right)\left( {2a - b} \right)\\
= - {\left( {2a - b} \right)^2}\\
= - \left( {4{a^2} - 4ab + {b^2}} \right)\\
= - 4{a^2} + 4ab - {b^2}\\
c)\left( {{x^3} - 27} \right):\left( {9 + 3x + {x^2}} \right)\\
= \left( {x - 3} \right)\left( {{x^2} + 3x + 9} \right):\left( {9 + 3x + {x^2}} \right)\\
= x - 3\\
2)a){x^2} + 2{x^3} + {x^4}\\
= {x^2}\left( {1 + 2x + {x^2}} \right)\\
b)2{x^2} + 3x - 5\\
= 2{x^2} + 5x - 2x - 5\\
= x\left( {2x + 5} \right) - \left( {2x + 5} \right)\\
= \left( {2x + 5} \right)\left( {x - 1} \right)\\
B3)a)4{x^2}\left( {3 - x} \right) = x - 3\\
\Rightarrow \left( {x - 3} \right)\left( {1 + 4{x^2}} \right) = 0\\
\Rightarrow x - 3 = 0\\
\Rightarrow x = 3\\
\text{Vậy}\,x = 3\\
b)\left( {x + 5} \right) - 0,5{x^2} - 2,5x = 0\\
\Rightarrow \left( {x + 5} \right) - 0,5.x\left( {x + 5} \right) = 0\\
\Rightarrow \left( {x + 5} \right)\left( {1 - 0,5.x} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x + 5 = 0\\
1 - 0,5.x = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = - 5\\
x = 2
\end{array} \right.\\
\text{Vậy}\,x = - 5;x = 2\\
B4)\\
a)Dkxd:x \ne 0\\
M = \dfrac{{2x}}{{{x^3} + x}} - \dfrac{1}{{{x^2} + 1}}\\
= \dfrac{2}{{{x^2} + 1}} - \dfrac{1}{{{x^2} + 1}}\\
= \dfrac{1}{{{x^2} + 1}}\\
b)Do:{x^2} > 0\\
\Rightarrow {x^2} + 1 > 1\\
\Rightarrow \dfrac{1}{{{x^2} + 1}} < 1\\
\Rightarrow M < 1
\end{array}$
Vậy ko có đủ dữ kiện để tìm GTLN của M
