Question

根式计算

化简1、(1/x^2-3x+2) +(1/x^2-x)+(1/x^2+x)+(1/x^2+3X+2)
2、(2a^2+3a+2/a+1)-(a^2-a-5/a+2)-(3a^2-4a-5/a-2)+(2a^2-8a+5/a-3)
3。1/x-1/y=3，求2x+3xy-2y/x-2xy-y的值
4.设x-1/x=3,求x^10+X^8+x^2+1/x^10+X^6+x^4+1的值
请大家帮忙( ⊙ o ⊙ )！谢谢大家,感激不尽(⊙o⊙)哦 ~

Answer 1

1、(1/x^2-3x+2) +(1/x^2-x)+(1/x^2+x)+(1/x^2+3X+2)

有个公式：1/n(n+1)=1/n-1/(n+1)
1/x^2-3x+2=1/(x-1)(x-2)=1/(x-2)-1/(x-1)

1/x^2-x=1/x(x-1)=1/(x-1)-1/x

1/x^2+x=1/x(x+1)=1/x-1/(x+1)

1/x^2+3x+2=1/(x+1)(x+2)=1/(x+1)-1/(x+2)
所以 原式
=【1/(x-2)-1/(x-1)】+【1/(x-1)-1/x】+【1/x-1/(x+1) 】+【1/(x+1)-1/(x+2)】
=1/(x-2)-1/(x+2)
=[(x+2)-(x-2)]/(x+2)(x-2)
=4/(x²-4)

2）(2a^2+3a+2/a+1)-(a^2-a-5/a+2)-(3a^2-4a-5/a-2)+(2a^2-8a+5/a-3)

2a^2+3a+2/a+1=[(2a²+3a+1)+1]/(a+1)=[(a+1)(2a+1)+1]/(a+1)=2a+1+[1/(a+1)]

a^2-a-5/a+2=[(a²-a-6)+1]/(a+2)=[(a+2)(a-3)+1]/(a+2)=a-3+[1/(a+2)]

3a^2-4a-5/a-2=[(3a²-4a-4)-1]/(a-2)=[(3a+2)(a-2)-1]/(a-2)=3a+2-[1/(a-2)]

2a^2-8a+5/a-3=[(2a²-8a+6)-1]/(a-3)=[(2a-2)(a-3)-1]/(a-3)=2a-2-[1/(a-3)]

所以原式
=2a+1+[1/(a+1)]- 【a-3+[1/(a+2)]】-【3a+2-[1/(a-2)]】+ 2a-2-[1/(a-3)]

=(2a+1-a+3-3a-2+2a-2)+[1/(a+1)-1/(a+2)]+[1/(a-3)-1/(a-2)]

=[2a]+1/(a+1)(a+2)+1/(a-2)(a-3)

=[2a]+[(a-2)(a-3)+(a+1)(a+2)]/(a+1)(a+2)(a-2)(a-3)

=[2a]+(a²-5a+6+a²+3a+2) /(a+1)(a+2)(a-2)(a-3)

=[2a]+(2a²-2a+8)/ (a+1)(a+2)(a-2)(a-3)

=[2a]+2(a²-a+4)/ (a+1)(a+2)(a-2)(a-3)

3) 1/x-1/y=3，求2x+3xy-2y/x-2xy-y
1/x-1/y=3
(y-x)/xy=3
x-y=-3xy
所以2x+3xy-2y/x-2xy-y
=2×（-3xy）+3xy/-3xy-2xy
=3/5

4）x-1/x=3,求x^10+X^8+x^2+1/x^10+X^6+x^4+1
x-1/x=3
两边平方 x²-2+1/x²=9，x²+1/x²=11
再两边平方，x^4+2+1/x^4=121，x^4+1/x^4=109

x^10+X^8+x^2+1/x^10+X^6+x^4+1
=(x²+1)(x^8+1)/(x^4+1)(x^6+1)

=(x²+1)(x^8+1)/(x^4+1)(x²+1)(x^4-x²+1)

=(x^8+1)/(x^4+1) (x^4-x²+1)

=x^4(x^4+1/x^4)/x²(x²+1/x²)×x²(x²+1/x²-1)

=109x^4/11x²×10x²

=109/110

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Answer 2