Question

已知x=（根号2+1）/2，求代数式4x的平方+4x的三次方-9x的平方-2x+1的值

Answer 1

解：x=(根号2+1)/2
根号2=2x-1
两边平方
(2x-1)^2=2
4x^2-4x+1=2
4x^2-4x=1
4x^4+4x^3-9x^2-2x+1
=4x^4-4x^3+8x^3-9x^2-2x+1=x^2(4x^2-4x)+8x^3-9x^2-2x+1
=x^2*1+8x^3-9x^2-2x+1
=8x^3-8x^2-2x+1
=2x(4x^2-4x)-2x+1
=2x*1-2x+1
=1

Answer 2

x²=(3+2√2)/4=3/4+√2/2=x+1/4
x²-x-1/4=0
4x^2+4x^3-9x^2-2x+1
=4x^3-5x^2-2x+1
=4(x^2-x-1/4)-x^2-2x+2
=-x^2-2x+2
=-(x^2-x-1/4)-3x+7/4
=-3x+7/4
=(-3√2-3)/2+7/4
=(-6√2+1)/4

Answer 3

x²=﹙√2＋1﹚²／4＝﹙2＋1＋2√2﹚／4＝﹙3＋2√2﹚／4
x³＝x²×x＝﹙3＋2√2﹚／4×﹙√2＋1﹚／2＝﹙3√2＋4＋3＋2√2﹚／8＝﹙7＋5√2﹚／8
4x²＋4x³﹣9x²﹣2x＋1＝﹣5x²＋4x³﹣2x＋1
＝﹣5×﹙3＋2√2﹚／4＋4×﹙7＋5√2﹚／8﹣√2-1＋1
=-1／4﹣√2