x 1 x 2÷x 3 x 4有意义
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![x 1 x 2÷x 3 x 4有意义](/uploads/image/f/887420-20-0.jpg?t=x+1+x+2%C3%B7x+3+x+4%E6%9C%89%E6%84%8F%E4%B9%89)
n(n+1)(n+2)(n+3)/4
解:令x1=y1+y2,x2=y1-y2,x3=y3,x4=y4f=y1^2-y2^2+y1y3-y2y3+y3y4=(y1+y3/2)^2-(y2+y3/2)^2+y3^2y3y4=z1^2-z2^
一般的,有:(n-1)n(n+1)=n^3-n{n^3}求和公式:Sn=[n(n+1)/2]^2{n}求和公式:Sn=n(n+1)/21x2x3+2x3x4+3x4x5+.+7x8x9=2^3-2+3
解1/(1+2+3)+1/(2+3+4)+1/(3+4+5)……+1/(99+100+101)=1/6+1/9+1/12+1/15.+1/300=1/3*(1/2+1/3.+1/99+1/100)=1
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A(x1,y1),B(x2,y2)x1,x2满足直线y=kx+b与双曲线y=k÷x有交点,所以x1,x2是方程kx+b=k÷x的两根化简得kx²+bx-k=0x1x2=a/c所以x1x2=-
1×2×3+2×3×4+3×4×5+……+8×9×10=(1/4)(1×2×3×4)+(1/4)(2×3×4×5-1×2×3×4)+(1/4)(3×4×5×6-2×3×4×5)+……(1/4)(8×9
nx(n+1)=1/3[n(n+1)(n+2)-(n-1)n(n+1)]1x2+2x3+3x4+...+nx(n+1)=1/3[1x2x3-0x1x2+2x3x4-1x2x3+3x4x5-2x3x4+
读完以上材料,请你计算下列各题:(1)1×2+2×3+3×4+…+10×11(写出过程);1×2+2×3+3×4+…+10×11=1/3(1×2×3-0×1×2)+1/3(2×3×4-1×2×3)+1
3*(1x2+2x3+3x4+...+99x100)=3*1/3*(1x2x3-0x1x2+2x3x4-1x2x3+3x4x5-2x3x4+99x100x101-98x99x100)=99x100x1
1X2+2X3+3X4+、、、、、、+nX(n+1)=(1/3)(1*2*3-0*1*2)+(1/3)(2*3*4-1*2*3)+(1/3)(3*4*5-2*3*4)+.+(1/3)[n*(n+1)(
最简方法:拆项法n(n+1)(n+2)=1/4*[n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2)]...2x3x4=1/4*[2x3x4*5-1*2x3x4]1x2x3=1/4*[
因为1x2x3=(1x2x3×4-0x1x2×3)/42x3x4=(2x3x4×5-1x2x3×4)/4.7x8x9=(7x8x9×10-6x7x8x9)/4所以1x2x3+2x3x4+3x4x5+…
设第n项为anan=n(n+1)(n+2)=n^3+3n^2+2n1×2×3+2×3×4+...+10×11×12=(1^3+2^3+...+10^3)+3(1^2+2^2+...+10^2)+2(1
根据韦达定理x1x2=c>0x3x4=b>0x1+x2=-bx3+x4=-c因为两个方程都有两个正整数根x1,x2,x3,x4都是正整数因此c和b也是正整数c-b=x1x2-x1-x2=(x1-1)(
通分分子=x1x2(x1-x2)-(x1-x2)=(x1-x2)(x1x2-1)
(1)1x2+2x3+…+99x100+100x101==1/3x100x101x102=343400(2)1x2+2x3+3x4+…+n(n+1)(n为正整数)=1/3n(n+1)(n+2)(3)1
(1)1x2+2x3+3x4+…+10x11=1*10*11*12/3=440(2)原式=n(n+1)(n+2)/3(3)1x2x3+2x3x4+3x4x5+…+7x8x9=1x2x3×4/4-0×1
3*(1x2+2x3+3x4+...+99x100)=3*1/3*(1x2x3-0x1x2+2x3x4-1x2x3+3x4x5-2x3x4+99x100x101-98x99x100)=99x100x1
4x(n)x(n+1)+1=(2n+1)^2