Question

Giải hệ PT: \(\left\{{}\begin{matrix}a^3+15ab^2=2\\3a^2b+5b^3=1\end{matrix}\right.\)

Answer 1

\(\Leftrightarrow\left\{{}\begin{matrix}a^3+15ab^2=2\\6a^2b+10b^3=2\end{matrix}\right.\)

\(\Rightarrow a^3+15ab^2-6a^2b-10b^3=0\)

\(\Leftrightarrow\left(a-b\right)\left(a^2-5ab+10b^2\right)=0\)

\(\Leftrightarrow a=b\)

Thay vào pt đầu:

\(a^3+15a^3=2\Rightarrow a^3=\frac{1}{8}\Rightarrow a=b=\frac{1}{2}\)