Question

高一数学 第二题 速度 写下拍下 速度！！！谢谢！！！！！

Answer 1

由立方差公式a^3-b^3=(a-b)(a^2+ab+b^2)得到：
第一项(x-1)/[x^(2/3)+x^(1/3)+1]
={[x^(1/3)-1]*[x^(2/3)+x^(1/3)+1]}/[x^(2/3)+x^(1/3)+1]
=x^(1/3)-1

由立方和公式a^3+b^3=(a+b)(a^2-ab+b^2)得到：
第二项(x+1)/[x^(1/3)+1]
={[x^(1/3)+1]*[x^(2/3)-x^(1/3)+1]}/[x^(1/3)+1]
=x^(2/3)-x^(1/3)+1

第三项[x-x^(1/3)]/[x^(1/3)-1]
=x^(1/3)*[x^(2/3)-1]/[x^(1/3)-1]
=x^(1/3)*[x^(1/3)+1]*[x^(1/3)-1]/[x^(1/3)-1]【平方差公式】
=x^(1/3)*[x^(1/3)+1]
=x^(2/3)+x^(1/3)

所以，原式=x^(1/3)-1-[x^(2/3)-x^(1/3)+1]-[x^(2/3)-x^(1/3)]
=x^(1/3)-1-x^(2/3)+x^(1/3)-1-x^(2/3)+x^(1/3)
=x^(1/3)-2x^(2/3)-2

Answer 2

。。。化简么？

追问

蒽