11/【1×3】+11/【3×5】+11/【5×7】+……+11/【2001×2003】计算
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11/【1×3】+11/【3×5】+11/【5×7】+……+11/【2001×2003】计算
(2)已知(x-3y)^2+|3x+y-1|=0,则x^3-3x^2y+3xy^2+y^3-y^2的值是
(3)求1/2(x^2-2xy-y^2)-3(1/6x^2-1/6xy+1/2y^2)的值,其中x=3,y=-2
(2)已知(x-3y)^2+|3x+y-1|=0,则x^3-3x^2y+3xy^2+y^3-y^2的值是
(3)求1/2(x^2-2xy-y^2)-3(1/6x^2-1/6xy+1/2y^2)的值,其中x=3,y=-2
11/【1×3】+11/【3×5】+11/【5×7】+……+11/【2001×2003】
=11/2 ×(2/【1×3】+2/【3×5】+2/【5×7】+……+2/【2001×2003】)
=11/2 ×(【3-1】/【1×3】+【5-3】/【3×5】+【7-5】/【5×7】+……+【2003-2001】/【2001×2003】)
=11/2 ×(1-1/3 + 1/3-1/5 + 1/5-1/7 + …… + 1/2001-1/2003)
=11/2 × (1-1/2003)
=11/2 × (2002/2003)
=11011/2003
(2)已知(x-3y)^2+|3x+y-1|=0,则x^3-3x^2y+3xy^2+y^3-y^2的值是
因(x-3y)^2≥0,|3x+y-1|≥0,
要使(x-3y)^2+|3x+y-1|=0成立,则x-3y=0,3x+y-1=0
解次二元一次方程,得x=3/10,y=1/10
x^3-3x^2y+3xy^2+y^3-y^2
=(3/10)^3-3(3/10)^2×(1/10)+3(3/10)(1/10)^2+(1/10)^3-(1/10)^2
=9/1000+1/1000-1/100
=0
(3)求1/2(x^2-2xy-y^2)-3(1/6x^2-1/6xy+1/2y^2)的值,其中x=3,y=-2
1/2(x^2-2xy-y^2)-3(1/6x^2-1/6xy+1/2y^2)
=1/2(9+12-4)-3(3/2+1+2)
=17/2-27/2
=5
=11/2 ×(2/【1×3】+2/【3×5】+2/【5×7】+……+2/【2001×2003】)
=11/2 ×(【3-1】/【1×3】+【5-3】/【3×5】+【7-5】/【5×7】+……+【2003-2001】/【2001×2003】)
=11/2 ×(1-1/3 + 1/3-1/5 + 1/5-1/7 + …… + 1/2001-1/2003)
=11/2 × (1-1/2003)
=11/2 × (2002/2003)
=11011/2003
(2)已知(x-3y)^2+|3x+y-1|=0,则x^3-3x^2y+3xy^2+y^3-y^2的值是
因(x-3y)^2≥0,|3x+y-1|≥0,
要使(x-3y)^2+|3x+y-1|=0成立,则x-3y=0,3x+y-1=0
解次二元一次方程,得x=3/10,y=1/10
x^3-3x^2y+3xy^2+y^3-y^2
=(3/10)^3-3(3/10)^2×(1/10)+3(3/10)(1/10)^2+(1/10)^3-(1/10)^2
=9/1000+1/1000-1/100
=0
(3)求1/2(x^2-2xy-y^2)-3(1/6x^2-1/6xy+1/2y^2)的值,其中x=3,y=-2
1/2(x^2-2xy-y^2)-3(1/6x^2-1/6xy+1/2y^2)
=1/2(9+12-4)-3(3/2+1+2)
=17/2-27/2
=5
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