Question

已知x^2+3x-9=0,求代数式(1-3/x)➗x-3/x+3-x+6/x+3？

求大神解答！！

Answer 1

先对后面式子化简，你会发现惊喜的。
(1-3/x)/[(x-3)/(x+3)]-(x+6)/(x+3)
=[(x-3)/x]/[(x-3)/(x+3)]-(x+3+3)/(x+3)
=[(x-3)/x]*[(x+3)/(x-3)]-(x+3+3)/(x+3)
=(x+3)/x -[1+ 3/(x+3)]
=1+ 3/x -1-3/(x+3)
=3/x -3/(x+3) 通分化简得到
=9/(x^2 +3x)
=9/9
=1
由等式x^2+3x-9=0 ,推出x^2+3x=9，代入上面化简式子便可得到。

Answer 2

已知x^2+3x-9=0,求代数式(1-3/x)➗x-3/x+3-x+6/x+3？
(1-3/x)➗(x-3)/(x+3)-(x+6)/(x+3)
=（x-3）/x*(x+3)/(x-3)-(x+6)/(x+3)
=(x+3)/x-(x+6)/(x+3)
={(x+3)*(x+3)-（x+6）*x}/x（x+3）
=（x²+6x+9-x²-6x）/（x²+3x）
=9/（x²+3x）
根据题意可得：x²+3x-9=0
所以x²+3x=9
原式=9/（x²+3x）
=9/9
=1
计算此题关键是先不用解答x，先化简题目中后边的算式。

Answer 3

（1-3/x）÷（x-3）/(x+3)
-(x+6)/(x+3)
=(x-3)/x÷（x-3）/(x+3)
-(x+6)/(x+3)
=(x+3)/x-(x+6)/(x+3)
=[(x+3)^2-x(x+6)]/x(x+3)
=(x^2+6x+9-x^2-6x)/(x^2+3x)
=9/(x^2+3x)
∵x^2+3x-9=0,∴x^2+3x=9代入上式得9/9=1

Answer 4

解答如下