a. ĐKXĐ: \(x\ge2\)
\(\sqrt{9\left(x-2\right)}=6\)
\(\Leftrightarrow9\left(x-2\right)=36\)
\(\Leftrightarrow x-2=4\)
\(\Leftrightarrow x=6\) ( thỏa mãn đk )
Vậy \(S=\left\{6\right\}\)
b. ĐKXĐ: mọi \(x\)
\(\sqrt{9\left(x-3\right)^2}=12\)
\(\Leftrightarrow9\left(x-3\right)^2=144\)
\(\Leftrightarrow\left(x-3\right)^2=16\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=4\\x-3=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-1\end{matrix}\right.\)
Vậy \(S=\left\{7;-1\right\}\)
a, ĐK: \(x\ge2\)
\(\sqrt{9\left(x-2\right)}=6\)
\(\Leftrightarrow9x-18=36\)
\(\Leftrightarrow x=6\left(tm\right)\)
b, ĐK: \(x\in R\)
\(\sqrt{9\left(x-3\right)^2}=12\)
\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=4\)
\(\Leftrightarrow\left|x-3\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=4\\x-3=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-1\end{matrix}\right.\)
a) \(\sqrt{9\left(x-2\right)}=6\left(đk:x\ge2\right)\)
\(\Leftrightarrow9\left(x-2\right)=36\Leftrightarrow x-2=4\Leftrightarrow x=6\)(thỏa đk)
b) \(\sqrt{9\left(x-3\right)^2}=12\Leftrightarrow3\left|x-3\right|=12\Leftrightarrow\left|x-3\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=4\\3-x=4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-1\end{matrix}\right.\)
a) ĐKXĐ: x > 2
\(\sqrt{9\left(x-2\right)}=6\)
<=> 9(x - 2) = 36
<=> x = 6 (tm ĐKXĐ)
b) ĐKXĐ: x ∈ R.
\(\sqrt{9\left(x-3\right)^2}=12\)
<=> 9(x - 3)2 = 144.
<=> (x - 3)2 = 16.
<=> \(|x-3|=4\)
Ta có 2 TH:
\(\left[{}\begin{matrix}x-3=4\\x-3=-4\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=7\\x=-1\end{matrix}\right.\)
a: Ta có: \(\sqrt{9\left(x-2\right)}=6\)
\(\Leftrightarrow9\left(x-2\right)=36\)
\(\Leftrightarrow x-2=4\)
hay x=6
b: Ta có: \(\sqrt{9\left(x-3\right)^2}=12\)
\(\Leftrightarrow3\left|x-3\right|=12\)
\(\Leftrightarrow\left|x-3\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=4\\x-3=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-1\end{matrix}\right.\)
