[(1+x)/2]≤[(2x-1)/3]+1
==>3(x+1)≤2(2x-1)+6
==> 3x+3≤4x-2+6
==> 5-6≤4x-3x
==> -1≤x
不等式的為：{x| -1≤x}
解不等式:(1+x)/2≤(2x-1)/3+1
解不等式:(1+x)/2≤(2x-1)/3+1
“/”為分?jǐn)?shù)線
[(1+x)/2]≤[(2x-1)/3]+1
“/”為分?jǐn)?shù)線
[(1+x)/2]≤[(2x-1)/3]+1
數(shù)學(xué)人氣：803 ℃時(shí)間：2020-07-01 06:57:43
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