Question

三道数学问题``在线急等

(1)化简S=(1/1+x)+(2/1+x^2)+(2^2/1+x^4)+...+(2^(n-1)/1+x^(2n)

(2)比较5678901234/6789012345与5678901235/6789012347的大小

(3)如果11x/3x^2+14x-24=(A/x+6 )+(B/4-3x),求常数A,B的值

谢谢哦``要过程

Answer 1

(1)化简S=(1/1+x)+(2/1+x^2)+(2^2/1+x^4)+...+(2^(n-1)/1+x^(2n)

S=1/(1-x)+1/(1+x)+2/(1+x^2)+2^2/(1+x^4)+...+2^n-1/(1+x^2n)-1/(1-x)

=2/(1-x^2)+2/(1+x^2)+2^2/(1+x^4)+...+2^n-1/(1+x^2n)-1/(1-x)

=2^2/(1-x^4)+2^2/(1+x^4)+...+2^n-1/(1+x^2n)-1/(1-x)

=2^3/(1-x^8)+....+2^n-1/(1+x^2n)-1/(1-x)

=.....

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

=2^n/(1-x^4n)-1/(1-x)

(2)比较5678901234/6789012345与5678901235/6789012347的大小

设x=5678901234
y=6789012345
2y>x

则：A=x/y
B=(x+1)/(y+2)

A-B=(2y-x)/[y(y+2)]>0

即：A>B，5678901234/6789012345>5678901235/6789012347

(3)如果11x/3x^2+14x-24=(A/x+6 )+(B/4-3x),求常数A,B的值

11x/(3x^2+14x-24)=[A(4-3x)+B(x+6)]/(x+6)(4-3x)

11x/(3x^2+14x-24)=[4A-3Ax+Bx+6B]/(4x-3x^2+24-18x)

11x/(3x^2+14x-24)=[(3A-B)x-(4A+6B)]/(3x^2+14x-24)

3A-B=11

4A+6B=0

A=3

B=-2