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x=[√19-√3]/8 . Let a=2x > a+1/a= (√19)/2. Again, a^4+1/a^4 = (a+1/a)^4 -4(a+1/a)^2 + 2 = [(a+1/a)2 -2]^2 -2. So, E=[(a+1/a)^4]/[{(a+1/a)2 -2}^2 -2] = 361/89.
E=361/89
? = 361/89
2x=(√19-√3) /4;(1/2x) =(√19+√3) /4(2x+1/2x) ^4=361/16;(2x) ^4+(1/2x) ^4=89/16;result:361/89👍
I have got the rjght answer.
369/89
{44x+44x ➖ } 88x^2{19x+19x ➖} =38x^2 +{3x+3x ➖ }= 6x^2 .{88x^2+38x^2+6x^2}= 122x^8/:{19x+19x ➖}{11x+11x ➖}+(11x+11x ➖}{3x +3x ➖} 122x^8/{38x^2+22x^2}+{22x^2+6x^2} = 122x^8/{60x^4+28x^4} = 122x^8/88x^8 =1.34x^1 2^17 2^17^1 2^1^1 2^1 (x ➖ 2x+1). {2x+2x+2x+2x2x ➖ 2x ➖ 2x 2x} {1x+1x+1x+1x+1x ➖ 1x ➖ 1x ➖ 1x}/{2x+2x+2x+2x2x ➖ 2x ➖ 2x ➖ 2x}= {8x^4+4x^4}=12x^8/8x^4 {2x+2x+2x+2x2x ➖ 2x ➖ 2x ➖ 2x} {1x+1x+1x+1x1x ➖ 1x ➖ 1x ➖ 1x }/{2x+2x+2x+2x2x ➖ 2x ➖ 2x ➖ 2x } = {8x^4+4x^4}/8x^4 = 12x^8/8x^4 {12x^8/8x^4+12x^8/8x^4}= 24x^16/16x^8= 1.8x^2 1^1.2^3x^2^1 2^1x^2^1 2^1x1^1 2x^1(x ➖ 2x+1).
x=[√19-√3]/8 . Let a=2x > a+1/a= (√19)/2. Again, a^4+1/a^4 = (a+1/a)^4 -4(a+1/a)^2 + 2 = [(a+1/a)2 -2]^2 -2. So, E=[(a+1/a)^4]/[{(a+1/a)2 -2}^2 -2] = 361/89.
E=361/89
? = 361/89
2x=(√19-√3) /4;(1/2x) =(√19+√3) /4(2x+1/2x) ^4=361/16;(2x) ^4+(1/2x) ^4=89/16;result:361/89👍
I have got the rjght answer.
369/89
{44x+44x ➖ } 88x^2{19x+19x ➖} =38x^2 +{3x+3x ➖ }= 6x^2 .{88x^2+38x^2+6x^2}= 122x^8/:{19x+19x ➖}{11x+11x ➖}+(11x+11x ➖}{3x +3x ➖} 122x^8/{38x^2+22x^2}+{22x^2+6x^2} = 122x^8/{60x^4+28x^4} = 122x^8/88x^8 =1.34x^1 2^17 2^17^1 2^1^1 2^1 (x ➖ 2x+1). {2x+2x+2x+2x2x ➖ 2x ➖ 2x 2x} {1x+1x+1x+1x+1x ➖ 1x ➖ 1x ➖ 1x}/{2x+2x+2x+2x2x ➖ 2x ➖ 2x ➖ 2x}= {8x^4+4x^4}=12x^8/8x^4 {2x+2x+2x+2x2x ➖ 2x ➖ 2x ➖ 2x} {1x+1x+1x+1x1x ➖ 1x ➖ 1x ➖ 1x }/{2x+2x+2x+2x2x ➖ 2x ➖ 2x ➖ 2x } = {8x^4+4x^4}/8x^4 = 12x^8/8x^4 {12x^8/8x^4+12x^8/8x^4}= 24x^16/16x^8= 1.8x^2 1^1.2^3x^2^1 2^1x^2^1 2^1x1^1 2x^1(x ➖ 2x+1).