(Ⅰ)若不等式f(x)≤m的解集为[﹣1,5],求实数a,m的值;
(Ⅱ)当a=2且0≤t<2时,解关于x的不等式f(x)+t≥f(x+2).
∴|x﹣a|≤m,
即a﹣m≤x≤a+m,
∵f(x)≤m的解集为{x|﹣1≤x≤5},
∴ {a−m=−1<
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已知a=3﹣2,b=3+2,试求ba﹣ab的值.
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