1,3,5,7,9,m,13,15, m=

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1,3,5,7,9,m,13,15, m=
m-2m-3m+4m-5m-6m+7m-8m-9m+.+2002m-2003m-2004m

等差数列求和,每三项合并,有668项,就可以用一加二一直加到一百的方法算了

先化简 再求值(3+m)(3-m)+m(m-6)-7,其中m=1/2

原式=9-m2+m2-6m-7=-6m+2=-6*1/2+2=-1

1.m^3+2m^2-9m-18 2.(x+1)(x+3)(x+5)(x+7)-9

1.m^3+2m^2-9m-18=m^2(m+2)-9(m+2)=(m+2)(m^2-3^2)=(m+2)(m+3)(m-3)2.(x+1)(x+3)(x+5)(x+7)-9=[x^2+8x+7][(

已知m>4,化简:|m-4|+|7-2m|+|m^2-2m+1|-|m^2-2m-3|

∵m>4∴m-4>0,7-2m0,m+1>0,m-3>0∴(m+1)(m-3)>0∴原式=|m-4|+|7-2m|+|(m-1)²|-|(m+1)(m-3)|=m-4+2m-7+(m-1)&

有9条线段分别长1m,2m,3m,4m,5m,6m,7m,8m,9m;从中取出若干条,顺次连接,围成一个正方形,问有多少

共有三种情况:>边长为9:L1=1+8.L2=2+7.L3=3+6.L4=4+5.L5=9.(5A4)/4=30种(分子表示五个元素取四个进行排列,除以4是因为有每四个取法可围成同一正方形)>边长为1

先化简再求值(3m²+5)(-3m²+5)-m²(7m+8)(7m+8)+(2m+1)(-

(3m²+5)(-3m²+5)-m²(7m+8)(7m-8)+(7m-8)+(m+1)(-2m-1)=25-9m^4-49m^4+64m²+7m-8-2m

(m+3m+5m+7m+.+2009m)-(2m-4m-6m-.-2008m)=?

(m+3m+5m+7m+.+2009m)-(2m+4m+6m+.+2008m)=(1+3+5+7+.+2009)m-(2+4+6+.+2008)m=【(1+2009)×1005÷2-(2+2008)×

m-2m-3m+4m-5m-6m+7m.2004m=?

m-2m-3m+4m-5m-6m+7m.2004m=(1-2-3+4-5-6+7-8-9+...2002-2003-2004)m=(-4-7-10-..-2005)m=-(4+7+10+..+2005

(1m+3m+5m+...+2009m)-(2m+4m+6m+...=2008m)

(1m+3m+5m+...+2009m)-(2m+4m+6m+...+2008m)=1m+3m+5m+...+2009m-2m-4m-6m-...-2008m=1m+(3m-2m)+(5m-4m)+.

(m+3m+5m+7m+.+2011m)-(2m+4m+6m+.+2010m) 要说理由!

(1+3+5+...+2011)m-(2+4+6+...+2010)m=(1+(3-2)+(5-4)+...+(2011-2010))m=1+1+1+...+1从2到2010是2010/2=1005项

若m为正整数,且m不能被4整除,试说明1^m+2^m+3^m+4^m+…+9^m一定是5的倍数

1^m+2^m+3^m+4^m+…+9^m|5=(1^m+2^m+3^m+4^m)*2+0^m|5=2+(2^m+3^m+4^m)*2|5当M被4除余1时2+(2^m+3^m+4^m)*2|5=2+(

2m+3m+5m+7m+9m+…+2003m+2005m+2007m

2m+m(3+2007)*1003/2=2m+1008015m=1005017m

已知m是有理数,/m-1/+/m-3/+/m-5/+/m-7/的最小值是————

8画个数轴,这道题就是在数轴上找一个点到1357这几个点的距离和最小,所以在3到5之间找一个数就行了

(3m+5)(-3m的平方+5)-m的平方(7m+8)(7m-8)+(2m+1)(-2m-1),其中m使(m-1)的0次

∵m^4(m-1)º无意义∴m=1(3m+5)(-3m的平方+5)-m的平方(7m+8)(7m-8)+(2m+1)(-2m-1),=25-9m²-m²(49m²

3m^2-m-[7m-2(4m-5)++9m^2]

3m^2-m-[7m-2(4m-5)+9m^2]=3m^2-m-(7m-8m+10+9m^2)=3m^2-m-7m+8m-10-9m^2=-6m^2-10

(1)已知9(m+3)*27(m+1)/3(4m+7)=81,求m的值(2)若5m=32,5n=11,求5(3m-2n)

(1)即(m+3)*(m+1)/(4m+7)=1化简得m2+4m+3=4m+7即m^2=4解得m=±2检验的当m=±2时4m+7≠0,故都为实根(2)原式=3*5m-2*5n=3*32-2*11=96