\(\left(3x-5y\right)^2+\left(xy-135\right)^2=0\)
Vì \(\left\{{}\begin{matrix}\left(3x-5y\right)^2\ge0\forall x,y\\\left(xy-135\right)^2\ge0\forall x,y\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}3x-5y=0\\xy-135=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{3}y\\xy=135\end{matrix}\right.\)
\(\Rightarrow\dfrac{5}{3}y.y=135\)\(\Rightarrow y^2=81\)
\(\Leftrightarrow\left[{}\begin{matrix}y=9\Rightarrow x=15\\y=-9\Rightarrow x=-15\end{matrix}\right.\)
Ta có: \(\left(3x-5y\right)^2\ge0\forall x;y\)
\(\left(xy-135\right)^2\ge0\forall x;y\)
\(\Rightarrow\left(3x-5y\right)^2+\left(xy-135\right)^2\ge0\forall x\)
Mặt khác: \(\left(3x-5y\right)^2+\left(xy-135\right) ^2=0\)
nên: \(\left\{{}\begin{matrix}\left(3x-5y\right)^2=0\\\left(xy-135\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x-5y=0\\xy-135=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x=5y\\xy=135\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{3}y\\\dfrac{5}{3}y^2=135\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{3}y\\y^2=81\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{3}y\\\left[{}\begin{matrix}y=9\\y=-9\end{matrix}\right.\end{matrix}\right.\)
\(+,TH1:y=9\Leftrightarrow x=\dfrac{5}{3}\cdot9=15\left(tm\right)\)
\(+,TH2:y=-9\Leftrightarrow x=\dfrac{5}{3}\cdot\left(-9\right)=-15\left(tm\right)\)
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#\(Toru\)
