解x²-(2m-3)x+m²-3m≤0
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解x²-(2m-3)x+m²-3m≤0解x²-(2m-3)x+m²-3m≤0解x²-(2m-3)x+m²-3m≤0x²-(2m-3
解x²-(2m-3)x+m²-3m≤0
解x²-(2m-3)x+m²-3m≤0
解x²-(2m-3)x+m²-3m≤0
x²-(2m-3)x+m²-3m≤0
[x-m][x-(m-3)]<=0
由于m>m-3
解得 m-3 ≤ x ≤m
答：
x²-(2m-3)x+m²-3m<=0
利用十字相乘法：
1 -m
X
1 -(m-3)
所以原不等式化为：
(x-m)[x-(m-3)]<=0
x1=m，x2=m-3