Question

calculati media   aritmetica   si media  geometrica  a numerelor:
a) (radical din 2 +3) totu la puterea 2 si  ( radical din 2 -3)totu la  puterea 2 
b)a=(3radical din 2 -1)  la puterea 2 si b=19 +6radical 2
c)a=13-4radical din  3  si b = (2rAdical din 3 +1)totu la puterea 2 

Answer 1

a)$\sqrt{ (2+3)^{2} } = \sqrt{ 5^{2} } =5 \\ \sqrt{ (2-3)^{2} } = \sqrt{ (-1)^{2} } = \sqrt{1} =1 \\ m_{a} = \frac{5+1}{2} = \frac{6}{2} =3 \\ m_{g} = \sqrt{5*1} = \sqrt{5}$
b)$a=(3 \sqrt{2-1})^{2} =(3 \sqrt{1}) ^{2} =3 ^{2}=9 \\ b=19 +6\sqrt{2} \\ m_{a} = \frac{a+b}{2} = \frac{9+19+6 \sqrt{2} }{2} = \frac{28+6 \sqrt{2} }{2} =14+3 \sqrt{2} \\ m_{g} = \sqrt{a*b} = \sqrt{9(19 +6\sqrt{2})} =3 \sqrt{19 +6\sqrt{2}} =3 \sqrt{1+6 \sqrt{2}+18 } =$$= 3\sqrt{1+2*3 \sqrt{2}+(3 \sqrt{2}) ^{2}}= 3\sqrt{(1+3 \sqrt{2}) ^{2} } =3(1+3 \sqrt{2} )=3+9 \sqrt{2}$
c)$a=13-4 \sqrt{3}\\b=(2 \sqrt{3}+1)^{2}=(2 \sqrt{3}) ^{2}+2*2 \sqrt{3}*1+1=12+4 \sqrt{3}+1=13+4 \sqrt{3}$$m_{a} = \frac{a+b}{2} = \frac{13-4 \sqrt{3}+13+4 \sqrt{3}}{2} = \frac{26}{2} =13 \\ m_{g}= \sqrt{a*b} \\ a*b=(13-4 \sqrt{3} )*(13+4 \sqrt{3} )=13 ^{2} -(4 \sqrt{3} ) ^{2} =169-48=121 \\ m_{g}= \sqrt{121} =11$