【急!】几道初中有关的因式分解的填空,1.已知m^2+m-1=0,则m^3+2m^2+2004=______2.(-2)^2002+(-2)2001=_______3.[(5^3)*15^(-2)]/3^(-2)=_______4.若a-b=1,则(1/2)(a^2+b^2)-ab=_______5.若单项式-6x^2y^m与(1/3)x^(n-1)·y^3是同类项,
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![【急!】几道初中有关的因式分解的填空,1.已知m^2+m-1=0,则m^3+2m^2+2004=______2.(-2)^2002+(-2)2001=_______3.[(5^3)*15^(-2)]/3^(-2)=_______4.若a-b=1,则(1/2)(a^2+b^2)-ab=_______5.若单项式-6x^2y^m与(1/3)x^(n-1)·y^3是同类项,](/uploads/image/z/1851296-32-6.jpg?t=%E3%80%90%E6%80%A5%21%E3%80%91%E5%87%A0%E9%81%93%E5%88%9D%E4%B8%AD%E6%9C%89%E5%85%B3%E7%9A%84%E5%9B%A0%E5%BC%8F%E5%88%86%E8%A7%A3%E7%9A%84%E5%A1%AB%E7%A9%BA%2C1.%E5%B7%B2%E7%9F%A5m%5E2%2Bm-1%3D0%2C%E5%88%99m%5E3%2B2m%5E2%2B2004%3D______2.%28-2%29%5E2002%2B%28-2%292001%3D_______3.%5B%285%5E3%29%2A15%5E%28-2%29%5D%2F3%5E%28-2%29%3D_______4.%E8%8B%A5a-b%3D1%2C%E5%88%99%EF%BC%881%2F2%29%28a%5E2%2Bb%5E2%29-ab%3D_______5.%E8%8B%A5%E5%8D%95%E9%A1%B9%E5%BC%8F-6x%5E2y%5Em%E4%B8%8E%281%2F3%29x%5E%28n-1%29%C2%B7y%5E3%E6%98%AF%E5%90%8C%E7%B1%BB%E9%A1%B9%2C)
【急!】几道初中有关的因式分解的填空,1.已知m^2+m-1=0,则m^3+2m^2+2004=______2.(-2)^2002+(-2)2001=_______3.[(5^3)*15^(-2)]/3^(-2)=_______4.若a-b=1,则(1/2)(a^2+b^2)-ab=_______5.若单项式-6x^2y^m与(1/3)x^(n-1)·y^3是同类项,
【急!】几道初中有关的因式分解的填空,
1.已知m^2+m-1=0,则m^3+2m^2+2004=______
2.(-2)^2002+(-2)2001=_______
3.[(5^3)*15^(-2)]/3^(-2)=_______
4.若a-b=1,则(1/2)(a^2+b^2)-ab=_______
5.若单项式-6x^2y^m与(1/3)x^(n-1)·y^3是同类项,那么这两个单项式的积是________
【急!】几道初中有关的因式分解的填空,1.已知m^2+m-1=0,则m^3+2m^2+2004=______2.(-2)^2002+(-2)2001=_______3.[(5^3)*15^(-2)]/3^(-2)=_______4.若a-b=1,则(1/2)(a^2+b^2)-ab=_______5.若单项式-6x^2y^m与(1/3)x^(n-1)·y^3是同类项,
1.已知m^2+m-1=0,则m^3+2m^2+2004=__2005____
m²+m=1
m³+2m²+2004
=m(m²+m)+m²+2004
=m+m²+2004
=1+2004
=2005
2.(-2)^2002+(-2)2001=___2^2001____
=(-2)^2001[(-2)+1]
=2^2001
3.[(5^3)*15^(-2)]/3^(-2)=__5_____
=5^3*5^(-2)
=5
4.若a-b=1,则(1/2)(a^2+b^2)-ab=__1/2_____
=(1/2)(a²+b²-2ab)
=(1/2)(a-b)²
=(1/2)×1²
=1/2
5.若单项式-6x^2y^m与(1/3)x^(n-1)·y^3是同类项,那么这两个单项式的积是__-2x^4y^6______
根据同类项的概念可得到
2=n-1,n=3
m=3
所以他们的积是-2x^4y^6
【写了半天了,要多多给分哦】
1. 由m^2+m-1=0,得m^2+m=1,所以
m^3+2m^2+2004 = m^3+m^2+m^2+2004 =m(m^2+m) +m^2 +2004 =m^2+m+2004 = 2005
2. .(-2)^2002+(-2)^2001 = 2^2002 - 2^2001 = 2^2001*(2-1) = 2^2001
3. .[(5^3)*15^(-2)]/3^(...
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1. 由m^2+m-1=0,得m^2+m=1,所以
m^3+2m^2+2004 = m^3+m^2+m^2+2004 =m(m^2+m) +m^2 +2004 =m^2+m+2004 = 2005
2. .(-2)^2002+(-2)^2001 = 2^2002 - 2^2001 = 2^2001*(2-1) = 2^2001
3. .[(5^3)*15^(-2)]/3^(-2) = (5^3) / (3*5)^2 *3^2 = 5^3 / (3^2 *5^2) * 3^2 = 5
4. (1/2)(a^2+b^2)-ab = 1/2 (a^2+b^2-2ab) =1/2 (a-b)^2 = 1/2
5. 单项式-6x^2y^m与(1/3)x^(n-1)·y^3是同类项,
则 n-1 = 2, m= 3,即 m=3,n=3
-6x^2y^m = -6 x^2 y^3,
(1/3)x^(n-1)·y^3 =1/3 x^2 y^3
-6 x^2 y^3 * 1/3 x^2 y^3 = -2 x^4 y^6
收起
1.2005
2.2^2001
3.5(我没看错符号的话……)
4.1/2
5.-2x^4·y^6