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tresset_1 [31]
3 years ago
8

A^3+a+a^2+1 2a^2+2ab+ab^2+b^3

Mathematics
1 answer:
Vikentia [17]3 years ago
6 0
\frac{a^3+a+a^2+1}{a^3+a^2+ab^2+b^2} \cdot  \frac{2a^2+2ab+ab^2+b^3}{a^3+a+a^2b+b} =\frac{a(a^2+1)+1(a^2+1)}{a^2(a+1)+b^2(a+1)} \cdot  \frac{2a(a+b)+b^2(a+b)}{a(a^2+1)+b(a^2+1)} =\\\\=\frac{(a^2+1)(a+1)}{(a+1)(a^2+b^2)} \cdot  \frac{(a+b)(2a+b^2)}{(a^2+1)(a+b)} = \frac{2a+b^2}{a^2+b^2}
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The solution of the absolute inequality |y + 8| ≤ 2 is -10 ≤ y ≤ -6

<h3>What is an equation?</h3>

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1 year ago
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