Question

[Phạm Mạnh Cường]

a^2+3a=b^2+3b=2 tính a^5+b^5

Answer 1

[Xyz OLM]

Ta có : a² + 3a = b² + 3b

<=> (a² - b²) + (3a - 3b) = 0

<=> (a - b)(a + b + 3) = 0

<=> \(\left[{}\begin{matrix}a=b\\a=-b-3\end{matrix}\right.\)(1)

Khi đó \(\left\{{}\begin{matrix}a^2+3a=2\\b^2+3b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(a+\dfrac{3}{2}\right)^2=\dfrac{17}{4}\\\left(b+\dfrac{3}{2}\right)^2=\dfrac{17}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{\pm\sqrt{17}-3}{2}\\b=\dfrac{\pm\sqrt{17}-3}{2}\end{matrix}\right.\) (2)

Với a = b

Từ (1) và (2) => \(P=a^5+b^5=2a^5=2.\left(\dfrac{\pm\sqrt{17}-3}{2}\right)^5\)

Với a = -b - 3

=> P = \(b^5-\left(b+3\right)^5=\left[{}\begin{matrix}\left(\dfrac{\sqrt{17}-3}{2}\right)^5-\left(\dfrac{\sqrt{17}+3}{2}\right)^5\\\left(\dfrac{-\sqrt{17}-3}{2}\right)^5-\left(\dfrac{-\sqrt{17}+3}{2}\right)^5\end{matrix}\right.\)