\(e)ĐK:x\ge\dfrac{1}{2}\)
\(pt\Leftrightarrow\sqrt{2x-1-2\sqrt{2x-1}+1}+2\sqrt{2x-1-2.2\sqrt{2x-1}+4}+3\sqrt{2x-1-2.3\sqrt{2x-1}+9}=4\)
\(\Leftrightarrow\sqrt{\left(\sqrt{2x-1}-1\right)^2}+2\sqrt{\left(\sqrt{2x-1}-2\right)^2}+3\sqrt{\left(\sqrt{2x-1}-3\right)^2}=4\)
\(\Leftrightarrow\left|\sqrt{2x-1}-1\right|+2\left|\sqrt{2x-1}-2\right|+3\left|\sqrt{2x-1}-3\right|=4\left(1\right)\)
Xét 3 TH:
TH1: \(\sqrt{2x-1}\ge1\)
\(pt\left(1\right)\Leftrightarrow\sqrt{2x-1}-1+2\left(\sqrt{2x-1}-2\right)+3\left(\sqrt{2x-1}-3\right)=4\)
\(\Leftrightarrow6\sqrt{2x-1}-14=4\)
\(\Leftrightarrow6\sqrt{2x-1}=18\Leftrightarrow\sqrt{2x-1}=3\)
\(\Leftrightarrow2x-1=9\Leftrightarrow x=5\left(tm\right)\)
TH2: \(\sqrt{2x-1}\le1\Leftrightarrow x\le1\)
\(pt\left(1\right)\Leftrightarrow1-\sqrt{2x-1}+2\left(2-\sqrt{2x-1}\right)+3\left(3-\sqrt{2x-1}\right)=4\)
\(\Leftrightarrow14-6\sqrt{2x-1}=4\)
\(\Leftrightarrow6\sqrt{2x-1}=10\Leftrightarrow\sqrt{2x-1}=\dfrac{5}{3}\)
\(\Leftrightarrow2x-1=\dfrac{25}{9}\Leftrightarrow x=\dfrac{17}{9}>1\left(l\right)\)
TH3: \(1\le\sqrt{2x-1}\le3\Leftrightarrow1\le x\le5\)
\(pt\Leftrightarrow\sqrt{2x-1}-1+2\left(\sqrt{2x-1}-2\right)+3\left(3-\sqrt{2x-1}\right)=4\)
\(\Leftrightarrow4+0\sqrt{2x-1}=4\Leftrightarrow0\sqrt{2x-1}=0\)
\(\Leftrightarrow\sqrt{2x-1}=0\Leftrightarrow x=\dfrac{1}{2}\left(l\right)\)
Vậy, pt có nghiệm duy nhất là x=5
