已知x的平方=x+1求代数式x的五次方-5x+2的值
4个回答
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x^2 = x+1
x^5 - 5x +2 = (x^2)^2 * x -5x +2
= [(x+1)^2] *x -5x +2 ( x^2 用x+1代替)
=x^3 + 2x^2 +x -5x +2
=(x^2)*x +2x^2-4x +2
=(x+1)*x +2x^2 -4x +2
=3x^2 - 3x +2
= 3(x+1) -3x +2
=5
x^5 - 5x +2 = (x^2)^2 * x -5x +2
= [(x+1)^2] *x -5x +2 ( x^2 用x+1代替)
=x^3 + 2x^2 +x -5x +2
=(x^2)*x +2x^2-4x +2
=(x+1)*x +2x^2 -4x +2
=3x^2 - 3x +2
= 3(x+1) -3x +2
=5
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展开全部
x^5-5x+2
=x(x^2)^2-5x+2
=x(x+1)^2-5x+2
=x(x^2+2x+1)-5x+2
=x(x+1+2x+1)-5x+2
=x(3x+2)-5x+2
=3x^2-3x+2
=3(x+1)-3x+2=5
=x(x^2)^2-5x+2
=x(x+1)^2-5x+2
=x(x^2+2x+1)-5x+2
=x(x+1+2x+1)-5x+2
=x(3x+2)-5x+2
=3x^2-3x+2
=3(x+1)-3x+2=5
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展开全部
x5-5x+2
=x(x4-4)-x+2
=x(x2-2)(x2+2)-x+2 代入x2=x+1
=x(x+1-2)(x+1+2)-x+2
=x(x-1)(x+3)-x+2
=(x2-x)(x+3)-x+2 代入x2=x+1
=(x+1-x)(x+3)-x+2
=x+3-x+2
=3+2
=5
=x(x4-4)-x+2
=x(x2-2)(x2+2)-x+2 代入x2=x+1
=x(x+1-2)(x+1+2)-x+2
=x(x-1)(x+3)-x+2
=(x2-x)(x+3)-x+2 代入x2=x+1
=(x+1-x)(x+3)-x+2
=x+3-x+2
=3+2
=5
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