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Tuy không hoàn toàn giống nhưng bạn tham khảo rồi chứng minh tương tự nhé !
https://hoc24.vn/hoi-dap/question/459079.html
\(M=\dfrac{x}{x+y+z}+\dfrac{y}{x+y+t}+\dfrac{z}{y+z+t}+\dfrac{t}{x+z+t}\)
Ta có:
\(\left\{{}\begin{matrix}\dfrac{x}{x+y+z}>\dfrac{x}{x+y+z+t}\\\dfrac{y}{x+y+t}>\dfrac{y}{x+y+z+t}\\\dfrac{z}{y+z+t}>\dfrac{z}{x+y+z+t}\\\dfrac{t}{x+z+t}>\dfrac{t}{x+y+z+t}\end{matrix}\right.\) Cộng theo \(3\) vế ta có:
\(M>\dfrac{x}{x+y+z+t}+\dfrac{y}{x+y+z+t}+\dfrac{z}{x+y+z+t}+\dfrac{t}{x+y+z+t}=1\)
Lại có:
\(\left\{{}\begin{matrix}\dfrac{x}{x+y+z}< \dfrac{x+t}{x+y+z+t}\\\dfrac{y}{x+y+t}< \dfrac{y+z}{x+y+z+t}\\\dfrac{z}{y+z+t}< \dfrac{z+x}{x+y+z+t}\\\dfrac{t}{x+z+t}< \dfrac{t+y}{x+y+z+t}\end{matrix}\right.\)Cộng theo \(3\) vế ta có:
\(M< \dfrac{x+t}{x+y+z+t}+\dfrac{y+z}{x+y+z+t}+\dfrac{z+x}{x+y+z+t}+\dfrac{t+y}{x+y+z+t}=2\)Như vậy \(1< M< 2\Leftrightarrow M\notin N\left(đpcm\right)\)
\(M=\dfrac{x}{x+y+z}=\dfrac{y}{x+y+t}=\dfrac{z}{y+z+t}=\dfrac{z}{x+z+t}\)\(\dfrac{x}{x+y+z}< 1\Rightarrow\dfrac{x+t}{x+y+z+t}>\dfrac{x}{x+y+z}\)
\(Tương\)\(tự\):\(\Rightarrow M< \dfrac{2\left(x+y+z+t\right)}{x+y+z+t}\)
\(Ta\) \(có\):\(2>M>1\)
\(\Rightarrow M\notin N\)\(sao\)
\(A=\dfrac{x}{x+y+z}+\dfrac{y}{x+y+t}+\dfrac{z}{y+z+t}+\dfrac{t}{x+z+t}\)
Giả sử \(A\in N\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{x+y+z}\in N\\\dfrac{y}{x+y+t}\in N\\\dfrac{z}{y+z+t}\in N\\\dfrac{t}{x+z+t}\in N\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x⋮x+y+z\\y⋮x+y+t\\z⋮y+z+t\\t⋮x+z+t\end{matrix}\right.\)
Vì \(x;y;z;t\in N\circledast\) nên:
\(\left\{{}\begin{matrix}x\ge x+y+z\\y\ge x+y+t\\z\ge y+z+t\\t\ge x+z+t\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-x\ge x+y+z-x\\y-y\ge x+y+t-y\\z-z\ge y+z+t-z\\t-t\ge x+z+t-t\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y+z\le0\\x+t\le0\\y+t\le0\\x+z\le0\end{matrix}\right.\)
Vì \(x;y;z;t\in N\circledast\) nên những điều trên không thể xảy ra
\(\Rightarrow\) điều giả sử sai,\(A\notin N\left(đpcm\right)\)
\(M>\dfrac{x}{x+y+z+t}+\dfrac{y}{x+y+z+t}+\dfrac{z}{x+y+z+t}+\dfrac{t}{x+y+z+t}=1\)
\(\dfrac{a}{b}< 1\Rightarrow\) \(\dfrac{a}{b}< \dfrac{a+m}{b+m}\) (Bạn chứng minh qua nhân chéo nhé)
\(\Rightarrow M< \dfrac{x+t}{x+y+z+t}+\dfrac{y+z}{x+y+z+t}+\dfrac{z+x}{x+y+z+t}+\dfrac{t+y}{x+y+z+t}=2\)
Do \(1< M< 2\) mà \(1\) và \(2\) là hai số tự nhiên liên tiếp
\(\Rightarrow M\notin\) N
CM: M>1
\(M=\dfrac{x}{x+y+z}+\dfrac{y}{x+y+t}+\dfrac{z}{y+z+t}+\dfrac{t}{x+z+t}\\ >\dfrac{x}{x+y+z+t}+\dfrac{y}{x+y+z+t}+\dfrac{z}{x+y+z+t}+\dfrac{t}{x+y+z+t}=1\left(\text{đ}pcm\right)\)
cm : M<2
\(M< \dfrac{x}{x+y}+\dfrac{y}{x+y}+\dfrac{z}{z+t}+\dfrac{t}{z+t}=1+1=2\left(\text{đ}pcm\right)\)
Vì 1<M<2 nên M không phải là số tự nhiên
Ta có:
\(\dfrac{x}{y+z+t}=\dfrac{y}{z+t+x}=\dfrac{z}{t+x+y}=\dfrac{t}{x+y+z}\)
\(\Rightarrow\dfrac{x}{y+z+t}+1=\dfrac{y}{z+t+x}+1=\dfrac{z}{t+x+y}+1=\dfrac{t}{x+y+z}+1\)
\(\Rightarrow\dfrac{x+y+z+t}{y+z+t}=\dfrac{x+y+z+t}{z+t+x}=\dfrac{x+y+z+t}{y+x+x}=\dfrac{x+y+z+t}{y+x+z}\)
. Xét TH1: \(x+y+z+t=0\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=-\left(z+t\right)\\y+z=-\left(x+t\right)\\z+t=-\left(x+y\right)\\x+t=-\left(y+z\right)\end{matrix}\right.\)
. Xét TH2: \(x+y+z+t\ne0\)
\(\Rightarrow x=y=z=t\)
\(\Rightarrow A=1\)
\(\Rightarrow\left\{{}\begin{matrix}A=1\\A=-1\end{matrix}\right.\)
Ta có:
\(\dfrac{x}{x+y+z+t}< \dfrac{x}{x+y+z}< \dfrac{x}{x+y}\)
\(\dfrac{y}{x+y+z+t}< \dfrac{y}{x+y+t}< \dfrac{y}{x+y}\)
\(\dfrac{z}{x+y+z+t}< \dfrac{z}{y+z+t}< \dfrac{z}{z+t}\)
\(\dfrac{t}{x+y+z+t}< \dfrac{t}{x+z+t}< \dfrac{t}{z+t}\)
Cộng vế với vế ta được:
\(\Rightarrow\dfrac{x+y+z+t}{x+y+z+t}< \dfrac{x}{x+y+z}+\dfrac{y}{x+y+t}+\dfrac{z}{y+z+t}+\dfrac{t}{x+z+t}< \dfrac{x+y}{x+y}+\dfrac{z+t}{z+t}\)
\(\Rightarrow1< \dfrac{x}{x+y+z}+\dfrac{y}{x+y+t}+\dfrac{z}{y+z+t}+\dfrac{t}{x+z+t}< 2\)
\(\Rightarrow1< M< 2\)
=> M không là số tự nhiên
Ta có: \(\dfrac{x}{y+z+t}=\dfrac{y}{z+t+x}=\dfrac{z}{y+t+x}=\dfrac{t}{y+x+z}\)
\(\Rightarrow\dfrac{x}{y+z+t}+1=\dfrac{y}{z+t+x}+1=\dfrac{z}{y+t+x}+1=\dfrac{t}{y+x+z}+1\)
\(\Rightarrow\dfrac{x+y+z+t}{y+z+t}=\dfrac{x+y+z+t}{z+t+x}=\dfrac{x+y+z+t}{y+t+x}=\dfrac{x+y+z+t}{y+x+z}\)+) Xét \(x+y+z+t=0\Rightarrow\left\{{}\begin{matrix}x+y=-\left(z+t\right)\\y+z=-\left(x+t\right)\\z+t=-\left(x+y\right)\\x+t=-\left(y+z\right)\end{matrix}\right.\)
\(\Rightarrow A=-1\)
+) Xét \(x+y+z+t\ne0\Rightarrow x=y=z=t\)
\(\Rightarrow A=1\)
Vậy A = -1 hoặc A = 1
Ta có:\(\dfrac{x}{y+z+t}+1=\dfrac{y}{z+t+x}+1=\dfrac{z}{y+t+x}+1=\dfrac{t}{y+x+z}+1\)\(\Rightarrow\dfrac{x+y+z+t}{y+z+t}=\dfrac{x+y+z+t}{z+t+x}=\dfrac{x+y+z+t}{t+x+y}=\dfrac{x+y+z+t}{x+y+z}\)
Nếu x+y+z+t\(\ne\)0 thì y+z+t=z+t+x=t+x+y=x+y+z
=>x=y=z=t nên P=1+1+1+1=4
Nếu X+y+z+t=0 thì P=-4
\(M=\dfrac{x}{x+y+z}+\dfrac{y}{x+y+t}+\dfrac{z}{y+z+t}+\dfrac{t}{x+z+t}\)
\(M+4=\left(\dfrac{x}{x+y+z}+1\right)+\left(\dfrac{y}{x+y+t}+1\right)+\left(\dfrac{z}{y+z+t}+1\right)+\left(\dfrac{t}{x+z+t}+1\right)\)\(M+4=\dfrac{x+t}{x+y+z+t}+\dfrac{y+z}{x+y+z+t}+\dfrac{z+x}{x+y+z+t}+\dfrac{t+y}{x+y+z+t}\)\(M+4=\dfrac{x+t+y+z+z+x+t+y}{x+y+z+t}\)
\(M+4=\dfrac{2\left(x+y+z+t\right)}{x+y+z+t}\)
\(M+4=2\)
\(M=2-4=-2\notin N\)
Ta có đpcm