Question

a= radical din (6+radical din 11)
b= radical din (6-radical din 11)
Aflati media aritmetica a celor 2

Answer 1

Răspuns:

$a=\sqrt{6+\sqrt{11}}=\sqrt{\dfrac{12+2\sqrt{11}}{2}}=\sqrt{\dfrac{11+2\sqrt{11}+1}{2}}=\\\sqrt{\dfrac{(\sqrt{11}+1)^2}{2}}=\dfrac{\sqrt{11}+1}{\sqrt 2}=\dfrac{\sqrt{22}+\sqrt 2}{2}\\\texttt{Analog gasim ca }b=\dfrac{\sqrt{22}-\sqrt 2}{2}\\m_a=\dfrac{a+b}{2}=\dfrac{\frac{\sqrt{22}-\sqrt2}{2}+\frac{\sqrt{22}+\sqrt2}{2}}{2}=\dfrac{\frac{2\sqrt{22}}{2}}{2}=\boxed{\dfrac{\sqrt{22}}{2}}$

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