Ta có:
\(\dfrac{x}{2}=\dfrac{y}{4}\Rightarrow\dfrac{x^2}{2^2}=\dfrac{y^2}{4^2}=\dfrac{x^2}{4}=\dfrac{y^2}{16}=k\)
\(\Rightarrow x^2=4k\)
\(\Rightarrow y^2=16k\)
\(\Leftrightarrow4k.16k=4\)
\(\Leftrightarrow64k^2=4\)
\(\Leftrightarrow k^2=\dfrac{1}{16}\)
\(\Leftrightarrow k=\dfrac{1}{4}\)
\(x^2=4k\Rightarrow x^2=4.\dfrac{1}{4}=1\Rightarrow x=1\)
\(y^2=16k\Rightarrow y^2=16.\dfrac{1}{4}=4\Rightarrow y=2\)
Vậy x = 1 ; = 2.
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