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Note that [7/3+4/(√3)]^1/2= (2+√3)/(√3). Thus, E= [44√3 +76]^1/5. Let x^5=44√3 +76 and y^5 = 44√3 -76. Then, xy=2 and x^5-y^5=152. Let x-y=a. So, a satisfies a^5+10a^3+20a-152=0 which implies a=x-y=2. So, x-2/x=2 which gives x =E=√3 +1.
χ=1+(3)^(1/2)
1+√3
x^5+{10+10 ➖ }+{66+66 ➖ }{7+7 ➖ }/{3+3 ➖ }+{4+4 ➖ }/{3+3 ➖ }=x^5{20+132}{14/6+8/6}=x^5 {152+22}/6={x^5*174}/6=174x^5/6=29x^5 29^1x^5 1^1x^2^3 x^2^3 (x ➖ 3x+2).
1+ √3
Note that [7/3+4/(√3)]^1/2= (2+√3)/(√3). Thus, E= [44√3 +76]^1/5. Let x^5=44√3 +76 and y^5 = 44√3 -76. Then, xy=2 and x^5-y^5=152. Let x-y=a. So, a satisfies a^5+10a^3+20a-152=0 which implies a=x-y=2. So, x-2/x=2 which gives x =E=√3 +1.
χ=1+(3)^(1/2)
1+√3
x^5+{10+10 ➖ }+{66+66 ➖ }{7+7 ➖ }/{3+3 ➖ }+{4+4 ➖ }/{3+3 ➖ }=x^5{20+132}{14/6+8/6}=x^5 {152+22}/6={x^5*174}/6=174x^5/6=29x^5 29^1x^5 1^1x^2^3 x^2^3 (x ➖ 3x+2).
1+ √3