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$2,\\ 1,ĐK:\left\{{}\begin{matrix}x-1\ge0\\3-x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x\le3\end{matrix}\right.\\ 2,ĐK:x\ne\dfrac{1}{2}\\ BPT\Leftrightarrow2\left(x-1\right)< 2x+1\\ \Leftrightarrow2x-2< 2x+1\\ \Leftrightarrow0< 3\left(luôn.đúng\right)$

Vậy BPT luôn đúng với mọi x

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3 phút trước

Bài 2:

a) $A=\sqrt{x-1}+\sqrt{3-x}$

$ĐK:$ $\left\{{}\begin{matrix}x-1\ge0\\3-x\ge0\end{matrix}\right.$$\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x\le3\end{matrix}\right.$ $\Leftrightarrow3\ge x\ge1$

b) $\dfrac{x-1}{2x+1}< \dfrac{1}{2}$

$\Leftrightarrow2\left(x-1\right)< 1\left(2x+1\right)$

$\Leftrightarrow2x-2< 2x+1$

$\Leftrightarrow-2< 1\left(đúng\forall x\right)$

Vậy $x\in R$

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2 phút trước

$44,\\ 3\sqrt{5}=\sqrt{3^2\cdot5}=\sqrt{45}\\ -5\sqrt{2}=\sqrt{\left(-5\right)^2\cdot2}=\sqrt{50}\\ -\dfrac{2}{3}\sqrt{xy}=\sqrt{\left(-\dfrac{2}{3}\right)^2xy}=\sqrt{\dfrac{4}{9}xy}\\ x\sqrt{\dfrac{2}{x}}=\sqrt{\dfrac{2x^2}{x}}=\sqrt{2x}$

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