阿Q吧 > 高一数学解方程，急!

高一数学解方程，急!

2025-04-04 02:59:50
推荐回答（2个）
回答1：

2x²+3x-5√（2x²+3x+9）+3=0
设√(2X²+3X+9)=Y

Y²-9-5Y+3=0
Y²-5Y-6=0
Y=-1（舍去） Y=6

2X²+3X+9=36
2X²+3X-27=0
(X-3)(2X+9)=0
X=3或X=-9/2

祝学习进步

回答2：

令2x^2+3x=t,
原式为
(t+3)^2=25(t+9)
t^2-19t-216=0
t=-8或者t=27

第一种情况：t=-8
2x^2+3x+8=0
△=9-4*2*8=-55
该方程无解
第二种情况：t=27
2x^2+3x-27=0
△=9+4*2*27=225

解得：
x=3或者x=-4.5
综上：
x=3或者x=-4.5均满足该方程

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