\(\sqrt{2x+1}=x-3\)
→ \(\left(\sqrt{2x+1}\right)^2=\left(x-3\right)^2\)
→ \(2x+1=x^2-6x+9\)
→ \(2x+1-x^2+6x-9=0\)
→ \(-x^2+8x-8=0\rightarrow x^2-8x+8=0\)
→ \(x_1=4+2\sqrt{2}\)
\(x_2=4-2\sqrt{2}\)
ĐK: \(2x+1\ge0\Leftrightarrow x\ge-\frac{1}{2}\)
\(pt\Leftrightarrow2x+1=\left(x-3\right)^2\\ \Leftrightarrow2x+1=x^2-6x+9\\ \Leftrightarrow x^2-8x+8=0\\ \Leftrightarrow x^2-2.x.4+4^2-4^2+8=0\\ \Leftrightarrow\left(x-4\right)^2-8=0\\ \Leftrightarrow\left(x-4-2\sqrt{2}\right)\left(x-4+2\sqrt{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-4-2\sqrt{2}=0\\x-4+2\sqrt{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4+2\sqrt{2}\\x=4-2\sqrt{2}\end{matrix}\right.\)
Vậy...............................
\(\sqrt{2x+1}=x-3\) (ĐKXĐ:\(x\ge\frac{1}{2}\))
\(\Leftrightarrow\sqrt{2x+1}^2=\left(x-3\right)^2\)
\(\Leftrightarrow2x+1=x^2-6x+9\)
\(\Leftrightarrow-x^2+8x-8=0\)
\(\Leftrightarrow-\left(x^2-8x+16-8\right)=0\)
\(\Leftrightarrow-\left(\left(x-4\right)^2-\left(2\sqrt{2}\right)^2\right)=0\)
\(\Leftrightarrow-\left(x-4-2\sqrt{2}\right)\left(x-4+2\sqrt{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4-2\sqrt{2}=0\\x-4+2\sqrt{2}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4+2\sqrt{2}\\x=4-2\sqrt{2}\end{matrix}\right.\)
Ta có \(x=4+2\sqrt{2}\) thỏa mãn ĐKXĐ
\(x=4-2\sqrt{2}\) không thỏa mãn ĐKXĐ
nên nghiệm của pt trên là \(x=4+2\sqrt{2}\)
\( \sqrt {2x + 1} = x - 3\\ \Leftrightarrow \left\{ \begin{array}{l} x - 3 \ge 0\\ 2x + 1 = {\left( {x - 3} \right)^2} \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} x \ge 3\\ {x^2} - 8x + 8 = 0 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} x \ge 3\\ \left[ \begin{array}{l} x = 4 + 2\sqrt 2 \\ x = 4 - 2\sqrt 2 \end{array} \right. \end{array} \right. \Leftrightarrow x = 4 + 2\sqrt 2 \)
Điều kiện các bạn trên sai rồi @@