1.
\(\sqrt{x^2-6x+9}=2x+1\)
\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=2x+1\)
\(\Leftrightarrow\left|x-3\right|=2x+1\)
TH1: \(\left\{{}\begin{matrix}x-3=2x+1\\x-3\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-4\\x\ge3\end{matrix}\right.\left(l\right)\)
TH2: \(\left\{{}\begin{matrix}3-x=2x+1\\x-3< 0\end{matrix}\right.\Leftrightarrow x=\dfrac{2}{3}\)
2.
ĐK: \(x\ge3\)
\(\sqrt{x+2\sqrt{x-3}-2}+\sqrt{x-2\sqrt{x-3}-2}=2\)
\(\Leftrightarrow\sqrt{x-3+2\sqrt{x-3}+1}+\sqrt{x-3-2\sqrt{x-3}+1}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-3}+1\right)^2}+\sqrt{\left(\sqrt{x-3}-1\right)^2}=2\)
\(\Leftrightarrow\left|\sqrt{x-3}+1\right|+\left|\sqrt{x-3}-1\right|=2\)
\(\Leftrightarrow\left|\sqrt{x-3}+1\right|+\left|1-\sqrt{x-3}\right|=2\)
Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\).
\(\left|\sqrt{x-3}+1\right|+\left|1-\sqrt{x-3}\right|\ge\left|\sqrt{x-3}+1+1-\sqrt{x-3}\right|=2\)
Đẳng thức xảy ra khi:
\(\left(\sqrt{x-3}+1\right)\left(1-\sqrt{x-3}\right)\ge0\)
\(\Leftrightarrow1-x+3\ge0\)
\(\Leftrightarrow x\le4\)
Vậy \(3\le x\le4\).
1: Ta có: \(\sqrt{x^2-6x+9}=2x+1\)
\(\Leftrightarrow\left|x-3\right|=2x+1\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=2x+1\left(x\ge3\right)\\3-x=2x+1\left(x< 3\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=4\\-3x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\left(loại\right)\\x=\dfrac{2}{3}\left(nhận\right)\end{matrix}\right.\)
