Question

y=\(2x-1\left(d_1\right)\)

y=\(\left(2n-1\right)x+\dfrac{3}{2}\left(d_2\right)\)

y=\(-x+3\left(d_3\right)\)

Tìm n đồng quy

Answer 1

Tọa độ giao điểm của (d1) và (d3) là:

\(\left\{{}\begin{matrix}2x-1=-x+3\\y=-x+3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x=4\\y=-x+3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=-\dfrac{4}{3}+3=\dfrac{5}{3}\end{matrix}\right.\)

Thay x=4/3 và y=5/3 vào (d2), ta được:

\(\dfrac{4}{3}\left(2n-1\right)+\dfrac{3}{2}=\dfrac{5}{3}\)

=>\(\dfrac{8}{3}n-\dfrac{4}{3}+\dfrac{3}{2}=\dfrac{5}{3}\)

=>\(\dfrac{8}{3}n=\dfrac{5}{3}+\dfrac{4}{3}-\dfrac{3}{2}=\dfrac{3}{2}\)

=>\(n=\dfrac{3}{2}:\dfrac{8}{3}=\dfrac{3}{2}\cdot\dfrac{3}{8}=\dfrac{9}{16}\)