\(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne-1\\x\ne1\end{cases}}\)
\(\left(\frac{a^3-1}{a^2-a}-\frac{a^3+1}{a^2+a}\right):\frac{a+1}{a-1}\)
\(=\left[\frac{\left(a-1\right)\left(a^2+a+1\right)}{a\left(a-1\right)}-\frac{\left(a+1\right)\left(a^2-a+1\right)}{a\left(a+1\right)}\right]:\frac{a+1}{a-1}\)
\(=\left(\frac{a^2+a+1}{a}-\frac{a^2-a+1}{a}\right).\frac{a-1}{a+1}\)
\(=\frac{\left(a^2+a+1\right)-\left(a^2-a+1\right)}{a}.\frac{a-1}{a+1}\)
\(=\frac{a^2+a+1-a^2+a-1}{a}.\frac{a-1}{a+1}\)
\(=\frac{2a}{a}.\frac{a-1}{a+1}=2.\frac{a-1}{a+1}=\frac{2\left(a-1\right)}{a+1}\)
\(\text{ĐK phải là a chứ}\)\(Juventus\)
Còn phần tìm gtri của a để A nguyên nữa mình chưa ghi ạ
ĐKXĐ: a khác -1
a khác 0
a khác 1
\(\left(\frac{a^3-1}{a^2-a}-\frac{a^3+1}{a^2+a}\right):\frac{a+1}{a-1}\)
=\(\left(\frac{\left(a-1\right)\left(a^2+a+1\right)}{a\left(a-1\right)}-\frac{\left(a+1\right)\left(a^2-a+1\right)}{a\left(a+1\right)}\right).\frac{a-1}{a+1}\)
=\(\left(\frac{a^2+a+1}{a}-\frac{a^2-a+1}{a}\right).\frac{a-1}{a+1}\)
=\(\frac{a^2+a+1-a^2+a-1}{a}\)\(.\frac{a-1}{a+1}\)
=\(\frac{2a}{a}.\frac{a-1}{a+1}\)
=\(2.\frac{a-1}{a+1}\)
=\(\frac{2\left(a-1\right)}{a+1}\)
Cảm ơn b nha mình giải đc phần để A nguyên r
