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若（ x-a）(x-b)-k 中含有因式 x+b, 则 k=_______

1.(12+13+14+.....12008)x(1+12+13+.....+12007)x(1+12+13+....12008)x(12+13+....+12007)=?. 答案. 36. 2. 若（ x-a）(x-b)-k 中含有因式 x+b, 则 k=_______. 答案： 解：设（ x-a ） (x-b)-k =0 ， x+b=0 则 x=﹣b 把 x=﹣ b 代入可得： （ ﹣b-a ） (﹣b-b)=k 则 k= 2b(a+b).

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若（ x-a）(x-b)-k 中含有因式 x+b, 则 k=_______

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  1. 1.(1\2+1\3+1\4+.....1\2008)x(1+1\2+1\3+.....+1\2007)x(1+1\2+1\3+....1\2008)x(1\2+1\3+....+1\2007)=?1.(1\2+1\3+1\4+.....1\2008)x(1+1\2+1\3+.....+1\2007)x(1+1\2+1\3+....1\2008)x(1\2+1\3+....+1\2007)=?

  2. 答案 36

  3. 2. 若（x-a）(x-b)-k 中含有因式x+b,则k=_______

  4. 答案：解：设（x-a）(x-b)-k =0 ，x+b=0则x=﹣b把x=﹣ b代入可得：（﹣b-a）(﹣b-b)=k则 k= 2b(a+b)

  5. 3.设a，b，c为实数，且｜a｜+a=0，｜ab｜=ab，｜c｜-c=0，求代数式｜b｜-｜a＋b｜-｜c-b｜＋｜a-c｜的值．

  6. ．因为｜a｜=-a，所以a≤0，又因为｜ab｜=ab，所以b≤0，因为｜c｜=c，所以c≥0．所以a＋b≤0，c-b≥0，a-c≤0．所以 原式=-b＋(a＋b)-(c-b)-(a-c)=b．

  7. 4．若m＜0，n＞0，｜m｜＜｜n｜，且｜x＋m｜＋｜x-n｜=m＋n， 求x的取值范围

  8. ．因为m＜0，n＞0，所以｜m｜=-m，｜n｜=n．所以｜m｜＜｜n｜可变为m＋n＞0．当x+m≥0时，｜x+m｜=x＋m；当x-n≤0时，｜x-n｜=n-x．故当-m≤x≤n时， ｜x＋m｜＋｜x-n｜=x＋m-x＋n=m＋n．

  9. 5.设(3x-1)7=a7x7＋a6x6+…+a1x＋a0，试求a0+a2＋a4＋a6的值．

  10. 分别令x=1，x=-1，代入已知等式中，得 a0+a2＋a4＋a6=-8128．

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