Question

a. Giải hệ pt sau bằng phương pháp thế.

X-2y =-1

3x-4y =9

b. Giải pt bậc hai một ẩn : X-8x+16=0

Answer 1

a)

\(\left\{{}\begin{matrix}x-2y=-1\\3x-4y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2y-1\\3\left(2y-1\right)-4y=9\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}x=2y-1\\y=6\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}x=11\\y=6\end{matrix}\right.\)

Answer 2

\(\left\{{}\begin{matrix}x-2y=-1\\3x-4y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2y-1\\3\left(2y-1\right)-4y=9\end{matrix}\right.\left\{{}\begin{matrix}x=11\\y=6\end{matrix}\right.\)

Answer 3

\(x^2-8x+16=0\Leftrightarrow\left(x-4\right)^2=0\Leftrightarrow x=4\)