cho A(x)=m+nx+px(x-1).biết A(0)=5,A(1)=-2,A(2)=7 tìm đa thức A(x) 23/08/2021 Bởi Harper cho A(x)=m+nx+px(x-1).biết A(0)=5,A(1)=-2,A(2)=7 tìm đa thức A(x)
Đáp án: $\begin{array}{l}A\left( x \right) = m + n.x + p.x\left( {x – 1} \right)\\\left\{ \begin{array}{l}A\left( 0 \right) = 5\\A\left( 1 \right) = – 2\\A\left( 2 \right) = 7\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}m + n.0 + p.0.\left( { – 1} \right) = 5\\m + n.1 + p.1.\left( {1 – 1} \right) = – 2\\m + n.2 + p.2.\left( {2 – 1} \right) = 7\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}m = 5\\m + n = – 2\\m + 2n + 2p = 7\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}m = 5\\n = – 2 – m\\2p = 7 – m – 2n\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}m = 5\\n = – 2 – 5 = – 7\\2p = 7 – 5 – 2.\left( { – 7} \right)\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}m = 5\\n = – 7\\2p = 16\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}m = 5\\n = – 7\\p = 8\end{array} \right.\\ \Leftrightarrow A\left( x \right) = m + n.x + p.x\left( {x – 1} \right)\\ = 5 + \left( { – 7} \right).x + 8.x.\left( {x – 1} \right)\\ = 5 – 7x + 8{x^2} – 8x\\ = 8{x^2} – 15x + 5\end{array}$ Bình luận
Đáp án:
$\begin{array}{l}
A\left( x \right) = m + n.x + p.x\left( {x – 1} \right)\\
\left\{ \begin{array}{l}
A\left( 0 \right) = 5\\
A\left( 1 \right) = – 2\\
A\left( 2 \right) = 7
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m + n.0 + p.0.\left( { – 1} \right) = 5\\
m + n.1 + p.1.\left( {1 – 1} \right) = – 2\\
m + n.2 + p.2.\left( {2 – 1} \right) = 7
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m = 5\\
m + n = – 2\\
m + 2n + 2p = 7
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m = 5\\
n = – 2 – m\\
2p = 7 – m – 2n
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m = 5\\
n = – 2 – 5 = – 7\\
2p = 7 – 5 – 2.\left( { – 7} \right)
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m = 5\\
n = – 7\\
2p = 16
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m = 5\\
n = – 7\\
p = 8
\end{array} \right.\\
\Leftrightarrow A\left( x \right) = m + n.x + p.x\left( {x – 1} \right)\\
= 5 + \left( { – 7} \right).x + 8.x.\left( {x – 1} \right)\\
= 5 – 7x + 8{x^2} – 8x\\
= 8{x^2} – 15x + 5
\end{array}$