a) \(\sqrt{\left(2x+3\right)^2}=4< =>\left|2x+3\right|=4\)
* TH 1 :
Với : 2x + 3 \(\ge0< =>x\ge-\dfrac{3}{2}\)
Ta có PT : 2x +3 = 4 <=> 2x = 1 => x = \(\dfrac{1}{2}\left(TM\right)\)
*TH 2 :
Với 2x + 3 < 0 <=> x < \(-\dfrac{3}{2}\)
Ta có PT : - 2x - 3 = 4 <=> -2x = 7 => x = \(-\dfrac{7}{2}\left(TM\right)\)
Vậy S = { \(-\dfrac{7}{2};\dfrac{1}{2}\) }
b) \(\sqrt{9x}-5\sqrt{x}=6-4\sqrt{x}< =>3\sqrt{x}-5\sqrt{x}=6-4\sqrt{x}< =>\) \(2\sqrt{x}=6< =>\sqrt{x}=3=>x=9\)
Vậy S = { 9 }
a,\(\sqrt{\left(2x+3\right)^2}=4\)
\(\Rightarrow\left|2x+3\right|=4\)
\(\Rightarrow\left\{{}\begin{matrix}2x+3=-4\\2x+3=4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-3,5\\x=0,5\end{matrix}\right.\)
b, \(\sqrt{9x}-5\sqrt{x}=6-4\sqrt{x}\)
\(\Rightarrow3\sqrt{x}-5\sqrt{x}=6-4\sqrt{x}\)
\(\Rightarrow2\sqrt{x}=6\Rightarrow\sqrt{x}=3\Rightarrow x=9\)
Chúc bạn học tốt!!!
a ) \(\sqrt{\left(2x-3\right)^2}=4\)
\(\Leftrightarrow\left|2x-3\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=4\\2x-3=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=7\\2x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy .............
b ) \(\sqrt{9x}-5\sqrt{x}=6-4\sqrt{x}\)
\(\Leftrightarrow\sqrt{9x}-\sqrt{x}=6\)
\(\Leftrightarrow\sqrt{x}\left(3-1\right)=6\)
\(\Leftrightarrow2\sqrt{x}=6\)
\(\Leftrightarrow\sqrt{x}=3\Leftrightarrow x=9\)
