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\(a,\sqrt{x^2-9}=\sqrt{\left(x-3\right)\left(x+3\right)}\) xác định \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3\ge0\\x+3\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-3\le0\\x+3\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge3\\x\ge-3\end{matrix}\right.\\\left\{{}\begin{matrix}x\le3\\x\le-3\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\le-3\\x\ge3\end{matrix}\right.\)
\(b,\sqrt{49x^2-24x+4}=\sqrt{\left(7x-2\right)^2}\ge0\forall x\)
\(\Rightarrow\) Căn thức có nghĩa \(\forall x\)
`a,` Điều kiện: `x^2 - 9 >=0 <=>` \(\left[{}\begin{matrix}x\ge3\\x\le-3\end{matrix}\right.\)
`b,` Điều kiện: `49x^2-24x+4 = (7x-2)^2 >=0`.
`-> x in RR`.
6) ĐKXĐ: \(x\le-6\)
\(\sqrt{\left(x+6\right)^2}=-x-6\Leftrightarrow\left|x+6\right|=-x-6\)
\(\Leftrightarrow x+6=x+6\left(đúng\forall x\right)\)
Vậy \(x\le-6\)
7) ĐKXĐ: \(x\ge\dfrac{2}{3}\)
\(pt\Leftrightarrow\sqrt{\left(3x-2\right)^2}=3x-2\Leftrightarrow\left|3x-2\right|=3x-2\)
\(\Leftrightarrow3x-2=3x-2\left(đúng\forall x\right)\)
Vậy \(x\ge\dfrac{2}{3}\)
8) ĐKXĐ: \(x\ge5\)
\(pt\Leftrightarrow\sqrt{\left(4-3x\right)^2}=2x-10\)\(\Leftrightarrow\left|4-3x\right|=2x-10\)
\(\Leftrightarrow4-3x=10-2x\Leftrightarrow x=-6\left(ktm\right)\Leftrightarrow S=\varnothing\)
9) ĐKXĐ: \(x\ge\dfrac{3}{2}\)
\(pt\Leftrightarrow\sqrt{\left(x-3\right)^2}=2x-3\Leftrightarrow\left|x-3\right|=2x-3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=2x-3\left(x\ge3\right)\\x-3=3-2x\left(\dfrac{3}{2}\le x< 3\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=2\left(tm\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(4x-3\right)^2-\left(x-2\right)^2=0\\x>=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(4x-3-x+2\right)\left(4x-3+x-2\right)=0\\x>=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(3x-1\right)\left(5x-5\right)=0\\x>=2\end{matrix}\right.\Leftrightarrow x\in\varnothing\)