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≤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所以不等式的解为:m-3<=x<=m