[2024-04-04]二元系数求解

\[y_i=ax_i^2+bx_i \]

\[y_j=ax_j^2+bx_j \]

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\[y_ix_j=ax_i^2x_j+bx_ix_j \]

\[y_jx_i=ax_j^2x_i+bx_ix_j \]

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\[y_ix_j-y_jx_i=a(x_i^2x_j-x_j^2x_i) \]

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\[a=\frac{y_ix_j-y_jx_i}{x_i^2x_j-x_j^2x_i} \]

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\[y_ix_j^2=ax_i^2x_j^2+bx_ix_j^2 \]

\[y_jx_i^2=ax_i^2x_j^2+bx_jx_i^2 \]

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\[y_ix_j^2-y_jx_i^2=b(x_ix_j^2-x_i^2x_j) \]

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\[b=\frac{y_ix_j^2-y_jx_i^2}{x_ix_j^2-x_i^2x_j} \]

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