a) \(\sqrt{2x+1}-\sqrt{x-3}=2\) ĐKXĐ: \(x\ge3\)
Ta có: \(x\ge3\Rightarrow x< 2x\Rightarrow x-3< 2x+1\Rightarrow\sqrt{x-3}< \sqrt{2x+1}\Rightarrow\sqrt{2x+1}-\sqrt{x-3}>0\)
\(\Leftrightarrow\left(\sqrt{2x+1}-\sqrt{x-3}\right)^2=4\) (Vì \(\sqrt{2x+1}-\sqrt{x-3}>0\))
\(\Leftrightarrow2x+1-2\sqrt{\left(2x+1\right)\left(x-3\right)}+x-3=4\)
\(\Leftrightarrow-2\sqrt{\left(2x+1\right)\left(x-3\right)}=6-3x\)
\(\Leftrightarrow\sqrt{\left(2x+1\right)\left(x-3\right)}=\frac{3x-6}{2}\)
Ta có: \(x\ge3\Rightarrow3x\ge9\Rightarrow3x-6\ge3\Rightarrow\frac{3x-6}{2}>0\)
\(\Leftrightarrow\left(2x+1\right)\left(x-3\right)=\frac{\left(3x-6\right)^2}{4}\) (Vì \(\frac{3x-6}{2}>0\))
\(\Leftrightarrow2x^2-6x+x-3=\frac{9x^2-36x+36}{4}\)
\(\Leftrightarrow4\left(2x^2-5x-3\right)=9x^2-36x+36\)
\(\Leftrightarrow9x^2-36x+36-8x^2+20x+12=0\)
\(\Leftrightarrow x^2-16x+48=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\text{ }\left(tmđkxđ\right)\\x=12\text{ }\left(tmđkxđ\right)\end{matrix}\right.\)
Vậy \(S=\left\{4;12\right\}\)
b) \(x^2+4x-7=\left(x+4\right)\sqrt{x^2-7}\) ĐKXĐ: \(\left[{}\begin{matrix}x\ge\sqrt{7}\\x\le-\sqrt{7}\end{matrix}\right.\)
\(\Leftrightarrow\frac{x^2+4x-7}{x+4}=\sqrt{x^2-7}\)
Vì \(\sqrt{x^2-7}\ge0\) nên \(\frac{x^2+4x-7}{x+4}\ge0\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2+4x-7\ge0\\x+4>0\end{matrix}\right.\\\left\{{}\begin{matrix}x^2+4x-7\le0\\x+4< 0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x\ge-2+\sqrt{11}\\-2-\sqrt{11}\le x< -4\end{matrix}\right.\) (Kết hợp ĐKXĐ)
\(\Rightarrow\left[{}\begin{matrix}x\ge\sqrt{7}\\-2-\sqrt{11}\le x< -4\end{matrix}\right.\)
Ta có: \(\frac{x^2+4x-7}{x+4}=\sqrt{x^2-7}\)
\(\Leftrightarrow\frac{x\left(x+4\right)}{x+4}-\frac{7}{x+4}=\sqrt{x^2-7}\)
\(\Leftrightarrow x-\frac{7}{x+4}=\sqrt{x^2-7}\)
\(\Leftrightarrow\left(x-\frac{7}{x+4}\right)^2=x^2-7\) (vì \(x-\frac{7}{x+4}>0\))
\(\Leftrightarrow x^2-\frac{14x}{x+4}+\frac{49}{\left(x+4\right)^2}=x^2-7\)
\(\Leftrightarrow-\frac{14x\left(x+4\right)}{\left(x+4\right)^2}+\frac{49}{\left(x+4\right)^2}=-7\)
\(\Leftrightarrow\frac{-14x^2-56x+49}{x^2+8x+16}=-7\)
\(\Leftrightarrow2x^2+8x-7=x^2+8x+16\)
\(\Leftrightarrow x^2=23\)\(\Leftrightarrow x=\pm\sqrt{23}\) (tmđkxđ)
Vậy \(S=\left\{\pm\sqrt{23}\right\}\)
Có vẻ khá dài dòng vậy, nhưng mình chưa có cách khác.
