Question

[Full Môn Học]

3,Cho x,y,z,t \(\ne0\) thoả mãn :

\(\dfrac{y+z+t-nx}{x}=\dfrac{z+t+x-ny}{y}=\dfrac{t+x+y-nz}{z}=\dfrac{x+y+z-nt}{t}\left(n\in N;x+y+z+t=2012\right)\)

Tính : P = x + 2y - 3z + t

Answer 1

[Vũ Thị Huế]

\(\dfrac{y+z+t-nx}{x}=\dfrac{z+t+x-ny}{y}=\dfrac{t+x+y-nz}{z}=\dfrac{x+y+z-nt}{t}\)

\(=\dfrac{y+z+t-nx+z+t+x-ny+t+x+y-nz+x+y+z-nt}{x+y+z+t}\)

\(=\dfrac{3x+3y+3z+3t-n\left(x+y+z+t\right)}{x+y+z+t}\)

\(=\dfrac{3\left(x+y+z+t\right)-n\left(x+y+z+t\right)}{x+y+z+t}=\dfrac{\left(3-n\right)\left(x+y+z+t\right)}{x+y+z+t}=3-n\)

Nên \(\left\{{}\begin{matrix}y+z+t-nx=3x-nx\\z+t+x-ny=3y-ny\\t+x+y-nz=3z-nz\\x+y+z-nt=3t-nt\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}y+z+t=3x\\z+t+x=3y\\t+x+y=3z\\x+y+z=3t\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{y+z+t}{3}\\y=\dfrac{z+t+x}{3}\\z=\dfrac{t+x+y}{3}\\t=\dfrac{x+y+z}{3}\end{matrix}\right.\)

Thay vào \(P\) ta có:

\(P=x+2y-3z+t\)

\(P=\dfrac{y+z+t}{3}+\dfrac{2\left(z+t+x\right)}{3}-\dfrac{3\left(t+x+y\right)}{3}+\dfrac{x+y+z}{3}\)

\(P=\dfrac{y+z+t+2z+t+x-3t-3x-3y+x+y+z}{3}\)

\(P=\dfrac{\left(x+x-3x\right)+\left(y+y-3y\right)+\left(z+z+2z\right)+\left(t+t-3t\right)}{3}\)

\(P=\dfrac{-x-y-z+4t}{3}\)

\(P=\dfrac{-\left(x+y+z+t\right)+5t}{3}\)

\(P=\dfrac{-2012+5t}{3}\)

Tốn sức quá T^T