Question

a ^ 3 + b ^ 3 = 2sqrt(5) a ^ 2 * b + a * b ^ 2 = sqrt(5)
Знайти a та b​

Answer 1

Ответ:

$\begin{cases}a=\frac{\sqrt{5}+1}{2}\\b=\frac{\sqrt{5}-1}{2} \end{cases}$

$\begin{cases}a=\frac{\sqrt{5}-1}{2}\\b=\frac{\sqrt{5}+1}{2} \end{cases}$

Пошаговое объяснение:

$a^3+b^3=2\sqrt{5}\\\\a^2b+ab^2=\sqrt5\\\\(a+b)^3=a^3+3a^2b+3ab^2+b^3\\\\(a+b)^3=(a^3+b^3)+3(a^2b+ab^2)\\\\(a+b)^3=2\sqrt{5}+3\sqrt{5}\\\\(a+b)^3=5\sqrt{5}\\\\(a+b)^3=5\cdot5^{\frac{1}{2}}(a+b)^3=5^{\frac{3}{2}}\ \ \ |()^{\frac{1}{3}}\\\\a+b=5^{\frac{1}{2}}\\\\a+b=\sqrt{5}$

$a^2b+ab^2=ab(a+b)\\\\ab(a+b)=\sqrt{5}\\\\ab\sqrt{5}=\sqrt{5}\ \ \ |:\sqrt{5}\\\\ab=1$

$\begin{cases}a+b=\sqrt{5}\\ab=1\end{cases}$

$\begin{cases}a=\sqrt{5}-b\\ab=1\end{cases}$

$\begin{cases}a=\sqrt{5}-b\\(\sqrt{5}-b)b=1\end{cases}$

$(\sqrt{5}-b)b=1$

$\sqrt{5}b-b^2-1=0\ \ \ |:(-1)$

$b^2-\sqrt{5}b+1=0$

$D=(-\sqrt{5})^2-4\cdot1\cdot1=5-4=1$

$\sqrt{D}=\sqrt{1}=1$

$b_1=\frac{\sqrt{5}-1}{2\cdot 1}=\frac{\sqrt{5}-1}{2}$

$b_2=\frac{\sqrt{5}+1}{2\cdot 1}=\frac{\sqrt{5}+1}{2}$

$a_1=\sqrt{5}-b_1=\sqrt{5}-\frac{\sqrt{5}-1}{2}=\frac{2\sqrt{5}-\sqrt{5}+1}{2}=\frac{\sqrt{5}+1}{2}$

$a_2=\sqrt{5}-b_1=\sqrt{5}-\frac{\sqrt{5}+1}{2}=\frac{2\sqrt{5}-\sqrt{5}-1}{2}=\frac{\sqrt{5}-1}{2}$

$\begin{cases}a=\frac{\sqrt{5}+1}{2}\\b=\frac{\sqrt{5}-1}{2} \end{cases}$

$\begin{cases}a=\frac{\sqrt{5}-1}{2}\\b=\frac{\sqrt{5}+1}{2} \end{cases}$