Question

三题初一数学题求解答（过程也要）

1.有一组数 9/5,16/12,25/21,36/32,……(分数）按这种规律写出第7个数是多少？
2.若 x分之1 +x=3,则 x^4+x^2+1分之x^2 =?
3.已知abc=1，求 ab+a+1分之a + bc+b+1分之b + ca+c+1分之c =?

Answer 1

1.81/77,因为第n个数是(n+2)^2/n(n+4)
2.1/x+x=3平方得
1/x^2+x^2=7
x^2/(x^4+x^2+1)=1/(x^2+1/x^2+1)
=1/(7+1)=1/8
3.
a/(ab+a+1)+b/(bc+b+1)+c/ (ca+c+1)
=a/(ab+a+1)+ab/(ab+a+1)+1/(ab+a+1)
=(ab+a+1)/(ab+a+1)
=1

Answer 2

1,分母分子都是+7，+9，+11.....
第七个是81/77

2，1/x+x=3,平方1/x^2+x^2=7,通分(1+x^4)/x^2=7,+1,通分（1+x^2+x^4）/x^2=8,倒数原式=1/8

3，令a=b=c=1。。。。。。正经的，a=1/bc，把a全部这么代换再通分，就出来了，等于1

Answer 3

1`81/77 2`1/8 3`1

Answer 4

1.81/77

2.x+1/x=3
[x+1/x]^2=x^2+1/x^2+2=9
x^2+1/x^2=7
x^2/(x^4+x^2+1)=1/(x^2+1/x^2+1)
=1/(7+1)=1/8

3.因为：a/(ab+a+1)＝a/(ab+a+abc)=1/(b+1+bc)
可知: a/(ab+a+1)=1/b *b/(bc+b+1)
又有:c/(ca+c+1)=c/(ca+c+abc)=1/(a+1+ab)=1／a* a/(ab+a+1)
=1／a* 1/b *b/(bc+b+1)=1/ab* b/(bc+b+1)

所以：a/(ab+a+1) +b/(bc+b+1) +c/(ca+c+1)
＝1/b *b/(bc+b+1)+b/(bc+b+1)+1/ab* b/(bc+b+1)
=(a+ab+abc)/(a+ab+abc)
=1

Answer 5

9/5,16/12,25/21,36/32,49/45,64/60,81/77
X分之4为3，X为4/3，算式自己算。

Answer 6

1.解分子,分母同时+7+(7+2)+(7+2+2).........
所以第七个分数是81/77
2.1/x+x=3平方得 1/x^2+x^2=7 x^2/(x^4+x^2+1)=1/(x^2+1/x^2+1) =1/(7+1)=1/8
3..因为：a/(ab+a+1)＝a/(ab+a+abc)=1/(b+1+bc) 可知: a/(ab+a+1)=1/b *b/(bc+b+1) 又有:c/(ca+c+1)=c/(ca+c+abc)=1/(a+1+ab)=1／a* a/(ab+a+1) =1／a* 1/b *b/(bc+b+1)=1/ab* b/(bc+b+1)

所以：a/(ab+a+1) +b/(bc+b+1) +c/(ca+c+1)＝1/b *b/(bc+b+1)+b/(bc+b+1)+1/ab* b/(bc+b+1)=(a+ab+abc)/(a+ab+abc)=1