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兩式相加得 2c=10+4b^2=》c=2b^2+5 相减得 2a=2+8b+2b^2=》a=b^2+4b+1 計算c-a 則 2b^2+5-(b^2+4b+1)=> (b-2)^2=0恒成立 所以c>=a 其他的也是一樣的辦法,算一下相减以後的值再與0比較就行了
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