PT: \(\sqrt{2x+2\sqrt{2x-1}}-\sqrt{2x-2\sqrt{2x-1}}=\sqrt{2x-1}-10\) (1) (ĐK: \(x\ge\dfrac{1}{2}\))
Đặt: \(y=\sqrt{2x-1}\) (ĐK: \(y\ge0\))
\(\Leftrightarrow x=\dfrac{y^2+1}{2}\)
Thay vào (1) ta có:
\(\sqrt{2\cdot\dfrac{y^2+1}{2}+2y}-\sqrt{2\cdot\dfrac{y^2+1}{2}-2y}=y-10\)
\(\Leftrightarrow\sqrt{y^2+1+2y}-\sqrt{y^2+1-2y}=y-10\)
\(\Leftrightarrow\sqrt{\text{ }y^2+2y+1}-\sqrt{y^2-2y+1}=y-10\)
\(\Leftrightarrow\sqrt{\left(y+1\right)^2}-\sqrt{\left(y-1\right)^2}=y-10\)
\(\Leftrightarrow\left|y+1\right|-\left|y-1\right|=y-10\)
TH1: Với: \(0\le y< 1\)
\(\Leftrightarrow y+1-1+y=y-10\)
\(\Leftrightarrow2y-y=-10\)
\(\Leftrightarrow y=-10\left(ktm\right)\)
TH2: \(y\ge1\)
\(\Leftrightarrow y+1-y+1=y-10\)
\(\Leftrightarrow2=y-10\)
\(\Leftrightarrow y=10+2\)
\(\Leftrightarrow y=12\left(tm\right)\)
Mà: y=12
\(\Rightarrow x=\dfrac{12^2+1}{2}=\dfrac{145}{2}\left(tm\right)\)
Vậy: ...