Giải:
a) \(x+\left(-\dfrac{31}{12}\right)^2=\left(\dfrac{49}{12}\right)^2-x=y\)
\(\Leftrightarrow x+\left(-\dfrac{31}{12}\right)^2=\left(\dfrac{49}{12}\right)^2-x\)
\(\Leftrightarrow x+\left(-\dfrac{31}{12}\right)^2-\left(\dfrac{49}{12}\right)^2+x=0\)
\(\Leftrightarrow2x+\left(-\dfrac{31}{12}\right)^2-\left(\dfrac{49}{12}\right)^2=0\)
\(\Leftrightarrow2x+\dfrac{\left(-31\right)^2}{12^2}-\dfrac{49^2}{12^2}=0\)
\(\Leftrightarrow2x+\dfrac{\left(-31\right)^2-49^2}{144}=0\)
\(\Leftrightarrow2x+\dfrac{961-2401}{144}=0\)
\(\Leftrightarrow2x+\dfrac{-1440}{144}=0\)
\(\Leftrightarrow2x+\left(-10\right)=0\)
\(\Leftrightarrow2x=10\)
\(\Leftrightarrow x=5\)
Mà \(x+\left(-\dfrac{31}{12}\right)^2=y^2\)
\(\Leftrightarrow5+\dfrac{961}{144}=y^2\)
\(\Leftrightarrow y^2=\dfrac{1681}{144}\)
\(\Leftrightarrow y=\pm\dfrac{41}{12}\)
Vậy ...
b) \(\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\le0\)
Vì \(\left(\dfrac{1}{2}x-5\right)^{20}\ge0;\forall x\)
và \(\left(y^2-\dfrac{1}{4}\right)^{10}\ge0;\forall y\)
\(\Rightarrow\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\le0\)
\(\Leftrightarrow\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x-5=0\\y^2-\dfrac{1}{4}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x=5\\y^2=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=\pm\dfrac{1}{2}\end{matrix}\right.\)
Vậy ...
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