你好令f（x）=（3/2）ax^2-（abx-x^2+2cx-c^2）=（3/2a-1）x^2-（ab+2c）x+c^2f′（x）=0f′（x）=2（3/2a-1）x-（ab+2c）=（3a-2）x-（ab+2c）=0当x=（ab+2c）/（3a-2）时,函数有最值.很高兴为您解答,...

　　特别抱歉，我把题写错了，已更正

　　你好，（3/2）ay^2=abx-x^2+2cx-c^2是一个椭圆方程，要移项化为y^2/u^2+（x-t)^2/v^2=1的形式,u就是y的最值这里（3/2）ay^2+x^2-(ab+2c)x+c^2=0（3/2）ay^2+[x-(ab+2c)/2]^2-[(ab+2c)/2]^2+c^2=0（3/2）ay^2+[x-(ab+2c)/2]^2=a²b²/4+2abc（3/2）ay^2/(a²b²/4+2abc)+[x-(ab+2c)/2]^2/(a²b²/4+2abc)=1y^2/(ab²/6+4/3abc)+[x-(ab+2c)/2]^2/(a²b²/4+2abc)=1y的最大值=√(ab²/6+4/3abc)y的最小值=-√(ab²/6+4/3abc)