((√x)+1)^4*(x-1)^5 展开式中x^4的系数等于?
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((√x)+1)^4*(x-1)^5 展开式中x^4的系数等于?
((√x)+1)^4*(x-1)^5 展开式中x^4的系数等于?
((√x)+1)^4*(x-1)^5 展开式中x^4的系数等于?
((√x)+1)^4中x的整数系数只有 0 1 2
所以要(x-1)^5中x的4,3和2次幂
所以x^4的系数是4C5+2C4*2C5+4C4*3C5*(-1)^3=5+60-10=55
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已知1/√(x)+√(x)=√5,求√(x/x2+x+1)+√(x/x2-x+1)的值定义域 x>0,两边平方,x+2+1/x=5,x+1/x=3,x^2-3x+1=0,x^2+x+1-4x=0,x^2+x+1=4x,x^2-x+1-2x=0.x^2-x+1=2x,√[x/(x^2+x+1)]-√[x/(x^2-x+1)]=√[x/(4x)]-√[x/2x]=√(1/4)-√(1/2)
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