Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(^{\dfrac{y+z+t-nx}{x}=\dfrac{z+t+x-ny}{y}=\dfrac{t+x+y-nz}{z}=\dfrac{x+y+z-nt}{t}}\)
\(\Rightarrow\dfrac{y+z+t}{x}-n=\dfrac{z+t+x}{y}-n=\dfrac{t+x+y}{z}-n=\dfrac{x+y+z}{t}-n\)
\(\Rightarrow\dfrac{y+z+t}{x}=\dfrac{z+t+x}{y}=\dfrac{t+x+y}{z}=\dfrac{x+y+z}{t}\)
\(\Rightarrow\dfrac{y+z+t}{x}+1=\dfrac{z+t+x}{y}+1=\dfrac{t+x+y}{z}+1=\dfrac{x+y+z}{t}+1\)
\(\Rightarrow\dfrac{x+y+z+t}{x}=\dfrac{x+y+z+t}{y}=\dfrac{x+y+z+t}{z}=\dfrac{x+y+z+t}{t}\)
\(\Rightarrow\dfrac{2012}{x}=\dfrac{2012}{y}=\dfrac{2012}{z}=\dfrac{2012}{t}\)
\(\Rightarrow x=y=z=t\)
Kết hợp \(x+y+z+t=2012\Leftrightarrow x=y=z=t=503\)
\(P=x+2y-3z+t=x+2x-3x+x=x=503\)
vậy....
\(x+y+z+t=2019\Rightarrow\left\{{}\begin{matrix}x+y+z=2019-t\\x+y+t=2019-z\\x+z+t=2019-y\\y+z+t=2019-x\end{matrix}\right.\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{y+z+t-nx}{x}=\dfrac{x+z+t-ny}{y}...=\dfrac{\left(3-n\right)\left(x+y+z+t\right)}{x+y+z+t}=3-n\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{y+z+t-nx}{x}=3-n\\\dfrac{x+z+t-ny}{y}=3-n\\\dfrac{x+y+t-nz}{z}=3-n\\\dfrac{x+y+z-nt}{t}=3-n\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2019-x-nx}{x}=3-n\\\dfrac{2019-y-ny}{y}=3-n\\\dfrac{2019-z-nz}{z}=3-n\\\dfrac{2019-t-nt}{t}=3-n\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2019-\left(n+1\right)x=\left(3-n\right)x\\2019-\left(n+1\right)y=\left(3-n\right)y\\2019-\left(n+1\right)z=\left(3-n\right)z\\2019-\left(n+1\right)t=\left(3-n\right)t\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{2019}{3-n+n+1}=\dfrac{2019}{4}\\y=\dfrac{2019}{3-n+n+1}=\dfrac{2019}{4}\\z=\dfrac{2019}{3-n+n+1}=\dfrac{2019}{4}\\t=\dfrac{2019}{3-n+n+1}=\dfrac{2019}{4}\end{matrix}\right.\)
\(\Rightarrow x=y=z=t\Rightarrow P=x+2x-3x+x=x=\dfrac{2019}{4}\)
=y+z+t/x - n.x/x=z+t+x/y - n.y/y=t+x+y/z - n.z/z=x+y+z/t - n.t/t
=y+z+t/x - n=z+t+x/y - n=t+x+y/z - n=x+y+z/t - n
=y+z+t/x=z+t+x/y=t+x+y/z=x+y+z/t
áp dụng tính chất của dãy tỉ số bằng nhau ta có:
y+z+t/x=z+t+x/y=t+x+y/z=x+y+z/t=y+z+t+z+t+x+t+x+y+x+y+z/x+y+z+t=3.(x+y+z+t)/x+y+z+t=3
ok bạn tiếp tục làm được nhé cho mih nha
\(\dfrac{y+z+t-nx}{x}=\dfrac{z+t+x-ny}{y}=\dfrac{t+x+y-nz}{z}=\dfrac{x+y+z-nt}{t}\)
\(=\dfrac{y+z+t-nx+z+t+x-ny+t+x+y-nz+x+y+z-nt}{x+y+z+t}\)
\(=\dfrac{3x+3y+3z+3t-n\left(x+y+z+t\right)}{x+y+z+t}\)
\(=\dfrac{3\left(x+y+z+t\right)-n\left(x+y+z+t\right)}{x+y+z+t}=\dfrac{\left(3-n\right)\left(x+y+z+t\right)}{x+y+z+t}=3-n\)
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
