Toán 8

D

demon311

$a,x,y,z \in N^* ; x > y > z $

$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=a \\
a \le \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4} < 2 \\
\rightarrow a=1 \\
\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1 \\
\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z} \le \dfrac{3}{z} \rightarrow z \le 3 \\
z=1 \rightarrow x=-y \;\; \text{(loại)} \\
z=2 \rightarrow y \le 4 \\
y=1 \rightarrow x<0 \\
y=2 \rightarrow x \; \text{không tồn tại} \\
y=3 \rightarrow x=6 \\
y=4 \rightarrow x=4 \;\; \text{(loại)} \\
z=3 \rightarrow y=3 \;\; \text{(loại)} \\
x=6 \; ; \; y=3 \; ; \; z=2$
 
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