KN

Giải giúp mình với ạ :3

a / \(\sqrt{2x^2+6x+1}=x+2\)

b /  \(\left\{{}\begin{matrix}x+\dfrac{1}{y}=2\\\dfrac{x}{y}+\dfrac{1}{y}=2\end{matrix}\right.\)

KH
14 tháng 5 2021 lúc 20:40

\(a.\sqrt{2x^2+6x+1}=x+2\Leftrightarrow\left\{{}\begin{matrix}x+2\ge0\\2x^2+6x+1=x^2+4x+4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\x^2+2x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\end{matrix}\right.\Leftrightarrow x=1\\ \Rightarrow S=\left\{1\right\}\)

\(b.\) ĐKXĐ: \(y\ne0\)\(\left(I\right)\Rightarrow x+\dfrac{1}{y}=\dfrac{x}{y}+\dfrac{1}{y}\Leftrightarrow x=\dfrac{x}{y}\Leftrightarrow x\left(1-\dfrac{1}{y}\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\y=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=\dfrac{1}{2}\\x=1\end{matrix}\right.\left(TM\right)\Rightarrow S=\left\{\left(0;\dfrac{1}{2}\right);\left(1;1\right)\right\}\)

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