Đáp án:
$\begin{array}{l}
1)\left( {3{x^2} - x + 1} \right)\left( {x - 1} \right) + {x^2}\left( {4 - 3x} \right) = 5\\
\Leftrightarrow 3{x^3} - 3{x^2} - {x^2} + x + x - 1\\
+ 4{x^2} - 3{x^3} = 5\\
\Leftrightarrow 2x - 1 = 5\\
\Leftrightarrow 2x = 6\\
\Leftrightarrow x = 3\\
Vậy\,x = 3\\
B3)a)\\
x\left( {2x + 1} \right) - {x^2}\left( {x + 2} \right) + {x^3} - x + 3\\
= 2{x^2} + x - {x^3} - 2{x^2} + {x^3} - x + 3\\
= 3\left( {dpcm} \right)\\
b)\\
a = 5k + 2\\
b = 5h + 3\\
\Leftrightarrow a.b = \left( {5k + 2} \right)\left( {5.h + 3} \right)\\
= 25.k.h + 15.k + 10.h + 6\\
= 25kh + 15k + 10h + 5 + 1\\
= 5.\left( {5kh + 3k + 2h + 1} \right) + 1
\end{array}$
Vậy a.b chia 5 dư 1
$\begin{array}{l}
c)n\left( {2n - 3} \right) - 2n\left( {n + 1} \right)\\
= 2{n^2} - 3n - 2{n^2} - 2n\\
= - 5n \vdots 5\left( {khi:n \in Z} \right)
\end{array}$