Ta có:\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)và(x+y)2-4(z-x)2=9 =>\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{\left(x+y\right)^2-4\left(z-x\right)^2}{\left(2+3\right)^2-4\left(4-3\right)^2}=\dfrac{9}{21}=\dfrac{3}{7}\) =>\(\dfrac{x}{2}=\dfrac{3}{7};x=\) \(\dfrac{6}{7}\)
\(\dfrac{y}{3}=\dfrac{3}{7};y=\)\(\dfrac{9}{7}\)
\(\dfrac{z}{4}=\dfrac{3}{7};z=\)\(\dfrac{12}{7}\)
Vậy x=\(\dfrac{6}{7}\)
y=\(\dfrac{9}{7}\)
z=\(\dfrac{12}{7}\)
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