\(=a^5+a^3b^2+b^3a^2+b^5-\left(a+b\right)\)

\(=a^5+b^5+\left(a^3b^2+b^3a^2\right)-\left(a+b\right)\)

\(=a^5+b^5+a^2b^2\left(a+b\right)-\left(a+b\right)\)

\(=a^5+b^5+\left[\left(ab\right)^2-1\right]\left(a+b\right)\)

Mà \(ab=1\Rightarrow\left(ab\right)^2-1=1^2-1=0\)

\(\Rightarrow\left(a^3+b^3\right)\left(a^2+b^2\right)-\left(a+b\right)=a^5+b^5+0=a^5+b^5\)

Vậy ...

\(a^2+b^2\le1+ab\)

\(\Leftrightarrow a^2-ab+b^2\le1\)

\(\Leftrightarrow\left(a+b\right)\left(a^2-ab+b^2\right)\le a+b\)

\(\Leftrightarrow a^3+b^3\le a+b\)

\(\Leftrightarrow\left(a^3+b^3\right)\left(a^3+b^3\right)\le\left(a+b\right)\left(a^5+b^5\right)\) ( \(a^3+b^3=a^5+b^5\))

\(\Leftrightarrow a^6+2a^3b^3+b^6\le a^6+ab^5+a^5b+b^6\)

\(\Leftrightarrow a^5b+ab^5\ge2a^3b^3\)

\(\Leftrightarrow a^5b+ab^5-2a^3b^3\ge0\)

\(\Leftrightarrow ab\left(a^4-2a^2b^2+b^4\right)\ge0\)

\(\Leftrightarrow ab\left(a^2-b^2\right)^2\ge0\) (luôn đúng \(\forall a;b>0\))

Vậy \(a^2+b^2\le1+ab\)

Biến đổi VP:

\(\left(a^3+b^3\right)\left(a^2+b^2\right)-\left(a+b\right)\)

\(=a^5+b^5+a^3b^2+a^2b^3-\left(a+b\right)\)

\(=a^5+b^5+a^2b^2\left(a+b\right)-\left(a+b\right)\)

\(=a^5+b^5+\left(a+b\right)-\left(a+b\right)\)

\(=a^5+b^5\left(ĐPCM\right)\)

(a³+b³)(a²+b²)-(a+b)

=a⁵+a³b²+ b³a²+b⁵-(a+b)

=a⁵+b⁵+a²b²(a+b)-(a+b)

=a⁵+b⁵+(a+b)-(a+b)(vì ab=1 nên a²b²=1)

=a⁵+b⁵(điều phải chứng minh)

\(\left(a^3+b^3\right)\left(a^2+b^2\right)-\left(a+b\right)\)

\(=a^5+a^3b^2+b^3a^2+b^5-\left(a+b\right)\)

\(=a^5+b^5+a^2b^2\left(a+b\right)-\left(a+b\right)\)

\(=a^5+b^5+\left(a+b\right)\)

\(=a^5+b^5\)

Cho A=x²+2x+24, với x thuộc N

Tìm x để A là số chính phương

Lời giải:
Ta có:

$(a^3+b^3)(a^2-b^2)-(a+b)=a^5+a^3b^2+a^2b^3+b^5-(a+b)$

$=(a^5+b^5)+(a^3b^2+a^2b^3)-(a+b)$

$=(a^5+b^5)+a^2b^2(a+b)-(a+b)=a^5+b^5+(a+b)-(a+b)=a^5+b^5$

(đpcm)

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