Đáp án:
\(\begin{array}{l}
b)\,\,\left\{ \begin{array}{l}
a = 1\\
b = 8
\end{array} \right.\\
c)\,\,\left\{ \begin{array}{l}
a = 3\\
b = - 4
\end{array} \right.\\
d)\,\,\left[ \begin{array}{l}
\left\{ \begin{array}{l}
a = 2\\
b = 2
\end{array} \right.\\
\left\{ \begin{array}{l}
a = - 2\\
b = - 2
\end{array} \right.
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
b) \(\,a{x^3} + b{x^2} + 5x - 50\) chia hết cho \({x^2} + 3x - 10\)
Ta có:
\(\begin{array}{*{20}{c}}
- \\
{}\\
- \\
{}
\end{array}\begin{array}{*{20}{c}}
{a{x^3} + b{x^2} + 5x - 50}\\
{{{\underline {a{x^3} + 3a{x^2} - 10ax} }^{\,\,\,\,\,}}}\\
{\left( {b - 3a} \right){x^2} + \left( {5 + 10a} \right)x - {{50}^{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}}\\
{\underline {^{}\left( {b - 3a} \right){x^2} + \left( {3b - 9a} \right)x - 10b + 30a} }\\
{\left( {5 - 3b + 19a} \right)x - 50 + 10b - 30a}
\end{array}\begin{array}{*{20}{c}}
\vline& {{x^2} + 3x - 10}\\
\hline
\vline& {ax + \left( {b - 3a} \right)}\\
\vline& {}\\
\vline& {}\\
\vline& {}
\end{array}\)
\( \Rightarrow a{x^3} + b{x^2} + 5x - 50\) chia hết cho \({x^2} + 3x - 10\)
\(\begin{array}{l} \Leftrightarrow \left( {5 - 3b + 19a} \right)x - 50 + 10b - 30a = 0\\ \Leftrightarrow \left\{ \begin{array}{l}5 - 3b + 19a = 0\\ - 50 + 10b - 30a = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 1\\b = 8\end{array} \right..\end{array}\)
c) \(a{x^4} + b{x^3} + 1\) chia hết cho \({\left( {x - 1} \right)^2}\)
Ta có: \({\left( {x - 1} \right)^2} = {x^2} - 2x + 1\)
\(\begin{array}{*{20}{c}}
- \\
{}\\
- \\
{}\\
- \\
{}
\end{array}\begin{array}{*{20}{c}}
{a{x^4} + b{x^3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 1}\\
{{{\underline {a{x^4} - 2a{x^3} + a{x^2}} }^{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}}\\
{\left( {b + 2a} \right){x^3} - a{x^2} + 1}\\
{\underline {\left( {b + 2a} \right){x^3} - \left( {2b + 4a} \right){x^2} + \left( {b + 2a} \right)x} }\\
{\left( {2b + 3a} \right){x^2} - \left( {b + 2a} \right)x + 1}\\
{\underline {\left( {2b + 3a} \right){x^2} - \left( {4b + 6a} \right)x + 2b + 3a} }\\
{\left( {4a + 3b} \right)x + 1 - 2b - 3a}
\end{array}\begin{array}{*{20}{c}}
\vline& {{x^2} - 2x + 1}\\
\hline
\vline& {a{x^2} + \left( {2a + b} \right)x + 3a + 2b}\\
\vline& {}\\
\vline& {}\\
\vline& {}\\
\vline& {}\\
\vline& {}
\end{array}\)
\(\Rightarrow a{x^4} + b{x^3} + 1\) chia hết cho \({\left( {x - 1} \right)^2}\)
\( \Leftrightarrow \left\{ \begin{array}{l}4a + 3b = 0\\1 - 2b - 3a = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 3\\b = - 4\end{array} \right..\)
d) \({x^4} + 4\) chia hết cho \({x^2} + ax + b.\)
Ta có:
\(\begin{array}{*{20}{c}}
- \\
{}\\
- \\
{}\\
- \\
{}
\end{array}\begin{array}{*{20}{c}}
{{x^{4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}} + 4}\\
{\underline {{x^4} + a{x^3} + b{x^2}} }\\
{ - a{x^3} - b{x^{2\,\,\,\,\,\,}} + 4}\\
{\underline { - a{x^3} - {a^2}{x^2} - abx} }\\
{\left( {{a^2} - b} \right){x^2} + abx + 4}\\
{\underline {\left( {{a^2} - b} \right){x^2} + a\left( {{a^2} - b} \right)x + b\left( {{a^2} - b} \right)} }\\
{\left( { - {a^3} + 2ab} \right)x + 4 - b{a^2} + {b^2}}
\end{array}\begin{array}{*{20}{c}}
\vline& {{x^2} + ax + b}\\
\hline
\vline& {{x^2} - ax + {a^2} - b}\\
\vline& {}\\
\vline& {}\\
\vline& {}\\
\vline& {}\\
\vline& {}
\end{array}\)
\( \Rightarrow \) \({x^4} + 4\) chia hết cho \({x^2} + ax + b\)
\(\begin{array}{l} \Leftrightarrow \left\{ \begin{array}{l} - {a^3} + 2ab = 0\\4 - b{a^2} + {b^2} = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}a = 0\\{a^2} - 2b = 0\end{array} \right.\\{b^2} - {a^2}b + 4 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}a = 0\\{b^2} + 4 = 0\,\,\left( {VN} \right)\end{array} \right.\\\left\{ \begin{array}{l}{a^2} = 2b\\{b^2} - 2{b^2} + 4 = 0\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}{a^2} = 2b\\{b^2} = 4\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{a^2} = 2b\\\left[ \begin{array}{l}b = 2\\b = - 2\end{array} \right.\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{a^2} = 4\\b = 2\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}a = 2\\b = 2\end{array} \right.\\\left\{ \begin{array}{l}a = - 2\\b = - 2\end{array} \right.\end{array} \right..\end{array}\)
