Question

giải hệ phương trình: \(\left\{{}\begin{matrix}2\left(x+y\right)=3\left(\sqrt[3]{x^2y}+\sqrt[3]{xy^2}\right)\\\sqrt[3]{x}+\sqrt[3]{y}=6\end{matrix}\right.\)

Answer 1

https://i.imgur.com/Xdu2t93.jpg

Answer 2

đặt \(\left\{{}\begin{matrix}\sqrt[3]{x}=a\\\sqrt[3]{y=b}\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}4\left(a^3+b^3\right)=\left(a+b\right)^3\\a+b=6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a^3+b^3=\frac{6^3}{4}\left(1\right)\\a+b=6\end{matrix}\right.\)

(1)\(\Leftrightarrow\left(a+b\right)\left(a^2-ab+b^2\right)=54\)

\(\Leftrightarrow\left(a+b\right)\left(\left(a+b\right)^2-3ab\right)=54\)\(\Rightarrow ab=9\)

Lại có a+b=6

áp dụng định lí Vi-ét đảo, tìm được a, b

Answer 3

Đặt \(a=\sqrt[3]{x},b=\sqrt[3]{y}\). Khi đó hệ trở thành:

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

Đặt \(S=a+b,P=ab\), ta được:

\(\left\{{}\begin{matrix}2\left(S^3-3SP\right)=3SP\\S=6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2\left(36-3P\right)=3P\\S=6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}S=6\\P=8\end{matrix}\right.\)

\(\Rightarrow\) a, b là nghiệm của PT: \(X^2-6X+8=0\Leftrightarrow X_1=2;X_2=4\)

\(\Rightarrow\left\{{}\begin{matrix}a=2\Leftrightarrow x=8\\b=4\Leftrightarrow y=64\end{matrix}\right.\)

hoặc \(\left\{{}\begin{matrix}a=4\Rightarrow x=64\\b=2\Rightarrow y=8\end{matrix}\right.\)

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