Question

Giải hệ phương trình:

a) \(\left\{{}\begin{matrix}a\left(a+b\right)=3\\b\left(b+c\right)=30\\c\left(c+a\right)=12\end{matrix}\right.\)                       b) \(\left\{{}\begin{matrix}x^3-y^3+3y^2-3x=2\\x^2+\sqrt{1-x^2}-3\sqrt{2-y-y^2}=-2\end{matrix}\right.\)

Answer 1

\(\left\{{}\begin{matrix}x^3-y^3+3y^2-3x=2\left(1\right)\\x^2+\sqrt{1-x^2}-3\sqrt{2-y-y^2}=-2\left(2\right)\end{matrix}\right.\)

\(đk:\left\{{}\begin{matrix}1-x^2\ge0\\2-y-y^2\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-1\le x\le1\\-2\le y\le1\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow x^3-y^3+3y^2-3x-2=0\)

\(\Leftrightarrow\left(x-y+1\right)\left(x^2+xy-x+y^2-2y-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-y+1=0\Leftrightarrow x=y-1\\x^2+xy-x+y^2-2y-2=0\left(3\right)\end{matrix}\right.\)

\(\left(2\right)\Rightarrow\left(y-1\right)^2+\sqrt{-y^2+2y-1}-3\sqrt{2-y-y^2}=-2\left(4\right)\)

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