Bài 101984

Question

Với \(a\neq b, p\neq q \) hãy tính nghiệm của HPT sau:
\(
\begin{cases}\frac{x}{a-p}+\frac{y}{b-p}=1 \\ \frac{x}{a-q}+\frac{y}{b-q}= 1\end{cases}
\)

score 0 · Answer 1

\(
D=\frac{1}{\left ( a-p \right )}.\frac{1}{\left ( b-q \right )}-\frac{1}{\left ( a-q \right )}.\frac{1}{\left ( b-p \right )}
\)
\(
=\frac{\left ( a-q \right ).\left ( b-p \right )-\left ( a-p \right ).\left ( b-q \right )}{\left ( a-p \right ).\left ( b-q \right ).\left ( a-q \right ).\left ( b-p \right )}
\)
\(
=\frac{ab-ap-bq+pq-ab+aq+bp-pq}{\left ( a-p \right ).\left ( b-q \right ).\left ( a-q \right ).\left ( b-p \right )}
\)
\(
=\frac{-a\left ( p-q \right )+b\left ( p-q \right )}{\left ( a-p \right ).\left ( b-q \right ).\left ( a-q \right ).\left ( b-p \right )}=\frac{\left ( p-q \right )\left ( b-a \right )}{\left ( a-p \right ).\left ( b-q \right ).\left ( a-q \right ).\left ( b-p \right )}
\)
\(
D_{x}=\frac{1}{b-q}-\frac{1}{b-p}=\frac{b-p-b+q}{\left ( b-p \right )\left ( b-q \right )}=\frac{q=p}{\left ( b-q \right )\left ( b-p \right )}
\)
\(
D_{y}=\frac{1}{a-p}-\frac{1}{a-q}=\frac{a-q-q+p}{\left ( p-q \right )\left ( b-a \right )}=\frac{p-q}{\left ( a-p \right )\left ( b-q \right )}
\)
\(
\Rightarrow x=\frac{\left ( q-p \right )}{\left ( b-q \right )\left ( b-p \right )}\times \frac{\left ( a-p \right )\left ( b-q \right )\left ( a-q \right )\left ( b-p \right )}{\left ( p-q \right )\left ( b-a \right )}=\frac{\left ( a-p \right )\left ( a-q \right )}{a-b}
\)
\(
y=\frac{\left ( p-q \right )}{\left ( a-p \right )\left ( a-q \right )}\times \frac{\left ( a-p \right )\left ( b-q \right )\left ( a-q \right )\left ( b-p \right )}{b-a}=\frac{\left ( b-p \right )\left ( b-q \right )}{b-a}
\)

Bài 101984

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Bài 101984

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