题目

解方程根号（9-x^2）+根号(4-x^2)=4

提问时间：2021-01-02

答案

如果(9-x^2)^0.5-(4-x^2)^0.5≠0的话
方程两边同乘(9-x^2)^0.5-(4-x^2)^0.5
先证明(9-x^2)^0.5-(4-x^2)^0.5≠0
9≠4
9-x^2≠4-x^2
(9-x^2)^0.5≠(4-x^2)^0.5
(9-x^2)^0.5-(4-x^2)^0.5≠0
方程两边同乘(9-x^2)^0.5-(4-x^2)^0.5
[(9-x^2)^0.5+(4-x^2)^0.5]×[(9-x^2)^0.5-(4-x^2)^0.5]=4[(9-x^2)^0.5-(4-x^2)^0.5]
[(9-x^2)-(4-x^2)]=4[(9-x^2)^0.5-(4-x^2)^0.5]
(9-x^2)^0.5-(4-x^2)^0.5=5/4
和原方程相加
2(9-x^2)^=4+5/4=21/4
9-x^2=441/64
x=±3/8×15^0.5
(9-x^2)^0.5-(4-x^2)^0.5=5/4和(9-x^2)^0.5+(4-x^2)^0.5=4两式相减求出x也=±3/8×15^0.5