Question #104862
If 3x=3y=3z and x+y+z=27√29

Then √x²+y²+z²=???
1
Expert's answer
2020-03-09T14:14:28-0400

Solution. From 3x=3y=3z get x=y=z. As result


x=y=z=x+y+z3=929x=y=z=\frac{x+y+z}{3}=9\sqrt{29}

Find the value x²+y²+z²


x2+y2+z2=3×81×29x^2+y^2+z^2=3\times81\times 29

Therefore


x2+y2+z2=987\sqrt {x^2+y^2+z^2}=9\sqrt{87}

Answer.


x2+y2+z2=987\sqrt {x^2+y^2+z^2}=9\sqrt{87}



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