Câu hỏi của hêllu the world

Question

Chứng minh rằng $M=\frac{x}{x+y+z}+\frac{y}{x+y+t}+\frac{z}{y+z+t}+\frac{t}{x+z+t}$ có giá trị không phải là số tự nhiên

Phùng Minh Quân · Accepted Answer

Ta có :$\frac{x}{x+y+z}>\frac{x}{x+y+z+t}$$;$$\frac{y}{x+y+t}>\frac{y}{x+y+z+t}$$;$$\frac{z}{y+z+t}>\frac{z}{x+y+z+t}$$;$$\frac{t}{x+z+t}>\frac{t}{x+y+z+t}$$\Rightarrow$$M=\frac{x}{x+y+z}+\frac{y}{x+y+t}+\frac{z}{y+z+t}+\frac{t}{x+z+t}>\frac{x}{x+y+z+t}+\frac{y}{x+y+z+t}+\frac{z}{x+y+z+t}$$+\frac{t}{x+y+z+t}=\frac{x+y+z+t}{x+y+z+t}=1$Suy ra $M>1$$\left(1\right)$Lại có :$\frac{x}{x+y+z}< \frac{x+t}{x+y+z+t}$$;$$\frac{y}{x+y+t}< \frac{y+z}{x+y+z+t}$$;$$\frac{z}{y+z+t}< \frac{z+x}{x+y+z+t}$$;$$\frac{t}{x+z+t}< \frac{t+y}{x+y+z+t}$$\Rightarrow$$M=\frac{x}{x+y+z}+\frac{y}{x+y+t}+\frac{z}{y+z+t}+\frac{t}{x+z+t}< \frac{x+t}{x+y+z+t}+\frac{y+z}{x+y+z+t}+\frac{z+x}{x+y+z+t}$$+\frac{t+y}{x+y+z+t}=\frac{x+t+y+z+z+x+t+y}{x+y+z+t}=\frac{2x+2y+2z+2t}{x+y+z+t}=\frac{2\left(x+y+z+t\right)}{x+y+z+t}=2$Suy ra $M< 2$$\left(2\right)$Từ $\left(1\right)$ và $\left(2\right)$ suy ra $1< M< 2$ Vậy $M$ không là số tự nhiên Câu trả lời: Ta có :$\frac{x}{x+y+z}>\frac{x}{x+y+z+t}$$;$$\frac{y}{x+y+t}>\frac{y}{x+y+z+t}$$;$$\frac{z}{y+z+t}>\frac{z}{x+y+z+t}$$;$$\frac{t}{x+z+t}>\frac{t}{x+y+z+t}$$\Rightarrow$$M=\frac{x}{x+y+z}+\frac{y}{x+y+t}+\frac{z}{y+z+t}+\frac{t}{x+z+t}>\frac{x}{x+y+z+t}+\frac{y}{x+y+z+t}+\frac{z}{x+y+z+t}$$+\frac{t}{x+y+z+t}=\frac{x+y+z+t}{x+y+z+t}=1$Suy ra $M>1$$\left(1\right)$Lại có :$\frac{x}{x+y+z}< \frac{x+t}{x+y+z+t}$$;$$\frac{y}{x+y+t}< \frac{y+z}{x+y+z+t}$$;$$\frac{z}{y+z+t}< \frac{z+x}{x+y+z+t}$$;$$\frac{t}{x+z+t}< \frac{t+y}{x+y+z+t}$$\Rightarrow$$M=\frac{x}{x+y+z}+\frac{y}{x+y+t}+\frac{z}{y+z+t}+\frac{t}{x+z+t}< \frac{x+t}{x+y+z+t}+\frac{y+z}{x+y+z+t}+\frac{z+x}{x+y+z+t}$$+\frac{t+y}{x+y+z+t}=\frac{x+t+y+z+z+x+t+y}{x+y+z+t}=\frac{2x+2y+2z+2t}{x+y+z+t}=\frac{2\left(x+y+z+t\right)}{x+y+z+t}=2$Suy ra $M< 2$$\left(2\right)$Từ $\left(1\right)$ và $\left(2\right)$ suy ra $1< M< 2$ Vậy $M$ không là số tự nhiên

hêllu the world · Answer

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Thanh Hà · Answer

$\frac{x}{x+y+z}>\frac{x}{x+y+z+t}$$\frac{y}{x+y+t}>\frac{y}{x+y+z+t}$$\frac{z}{y+z+t}>\frac{z}{x+y+z+t}$$\frac{t}{x+z+t}>\frac{t}{x+y+z+t}$$\Rightarrow$$M>\frac{x+y+z+t}{x+y+z+t}=1$$\frac{x}{x+y+z}< \frac{x+t}{x+y+z+t}$$\frac{y}{x+y+t}< \frac{y+z}{x+y+z+t}$$\frac{z}{y+z+t}< \frac{z+x}{x+y+z+t}$$\frac{t}{x+z+t}< \frac{t+y}{x+y+z+t}$$\Rightarrow M< \frac{2\left(x+y+z+t\right)}{x+y+z+t}=2$Vif 1\frac{x}{x+y+z+t}$$\frac{y}{x+y+t}>\frac{y}{x+y+z+t}$$\frac{z}{y+z+t}>\frac{z}{x+y+z+t}$$\frac{t}{x+z+t}>\frac{t}{x+y+z+t}$$\Rightarrow$$M>\frac{x+y+z+t}{x+y+z+t}=1$$\frac{x}{x+y+z}< \frac{x+t}{x+y+z+t}$$\frac{y}{x+y+t}< \frac{y+z}{x+y+z+t}$$\frac{z}{y+z+t}< \frac{z+x}{x+y+z+t}$$\frac{t}{x+z+t}< \frac{t+y}{x+y+z+t}$$\Rightarrow M< \frac{2\left(x+y+z+t\right)}{x+y+z+t}=2$Vif 1

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