本x2 2 1/5X2.7+5.3X2 1/5+2 1/5

2 1/5X2.7+5.3X2 1/5+2 1/5  

2 1/5X2.7+5.3X2 1/5+2 1/5

2 1/5X2.7+5.3X2 1/5+2 1/5
=2又1/5x（2.7+5.3+1）
=2又1/5×9
=5分之99
=19又5分之4
祝开心

1/x2+3x+2 +1/x2+5x+6 +1/x2+7x+12=?

通过因式分解可将原式化为1/（x+1）（x+2）+1/（x+2）（x+3）+1/（x+3）（x+4）
利用公式1/n（n+1）=1/n-1/（n+1）可将此式化为1/（x+1）-1/（x+2）+1/（x+2）-1/（x+3）+1/（x+3）-1/（x+4），合并同类项得1/（x+1）-1/（x+4）所以结果是3/（x+1）（x+4）

x+(1/x)=5,x2+(1/x2)=?

答：
x+1/x=5
两边平方得：
x^2+2+1/x^2=25
x^2+1/x^2=23

已知:1/3X5=1/2X(1/3-1/5),1/5X7=1/2X(1/5-1/7),求1/19X21+1/21X23+.+1/97X99

由前面规律 原式=1/2（1/19-1/21)+1/2(1/21-1/23)....+1/2(1/97-1/99)=1/2(1/19-1/21+1/21-1/23+1/23.....+1/97-1/99)=1/2(1/19-1/99)=40/1881

先化解，再求值：【1/(x2+3x+2)+1/(x2+5x+6)+1/(x2+7x+12)】/1/(x2+9x+20),其中x=√2

原式=[1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)] ×(x+4)(x+5)
=[1/(x+1)-1/(x+2) +1/(x+2)-1/(x+3) +1/(x+3)-1/(x+4)]×(x+4)(x+5)
= [1/(x+1)-1/(x+4)]×(x+4)(x+5)
=3/(x+1)(x+4)×(x+4)(x+5)
=3(x+5)/(x+1)
当x=√2时
=3(√2+5)/(√2+1)
=3(√2+5)(√2-1)
=3(2-√2+5√2-5)
=12√2 -9

1/(1x2x3)+1/(2x3x4)+1/(3x4x5)+.1/(20x21x22)=?

1/(1*2*3)+1/(2*3*4)+1/(3*4*5)+.........1/(20*21*22)
=1/2[1/1*2-1/2*3]+1/2[1/2*3-1/3*4]+1/2[1/3*4-1/4*5]+...+1/2[1/20*21-1/21*22]
=1/2[1/1*2-1/2*3+1/2*3-1/3*4+1/3*4-1/4*5+...+1/20*21-1/21*22]
=1/2[1/2-1/21*22]
=1/2*115/231
=115/462

1/3x2-（3x2+3xy-3/5y2)+(5/3x2+3xy+2/5y2)其中x=-1/2,y=2 化简求值 快！

1/3x2-（3x2+3xy-3/5y2)+(5/3x2+3xy+2/5y2)=1/3x2-3x2-3xy+3/5y2+5/3x2+3xy+2/5y2=-x2+y2把x=-1/2,y=2 代入-x2+y2=-1/4+4=15/4=3.75

奥数题: 1/2X5 + 1/3X5 + 1/8X11 + .+1/20X23 = ?

1/2*5=(1/2-1/5)1/3
1/3*5=(1/2-1/5)1/3
所以原式=1/3[1/2-1/5+1/5-1/8……+1/20-1/23]
=1/3(1/2-1/23)
=1/3(23/46-2/46)
=1/3(21/46)
=7/46

已知x/x2-3x+1=1/5,求x2;/x4-9x2+1

x/x2-3x+1=1/5,
5x=x²-3x+1
x²-8x+1=0
全部都除以x得到
x-8+1/x=0
x+1/x=8 两边平方
x²+2+1/x²=64
x²+1/x²=62

x²/(x^4-9x²+1）
=1/(x^2-9+1/x^2)
=1/(62-9)
=1/53
如有不明白，可以追问
如有帮助，记得采纳，谢谢
祝学习进步！

我们知道1/1x2=1-1/2,1/2x3=1/2-1/3,1/3x4=1/3-1/4,并依此计算1/1x3+1/3x5+1/5x7+.1/2003x2005

1/1x3+1/3x5+1/5x7+....1/2003x2005
=(1/2)(1-1/3)+(1/2)(1/3-1/5)+………………+(1/2)(1/2003-1/2005)
=(1/2)*[1-1/3+1/3-1/5+…………+1/2003-1/2005]
=(1/2)*(1-1/2005)
=(1/2)*(2004/2005)
=1002/2005