Đk: x >/ 4
\(3x+7\sqrt{x-4}=14\sqrt{x+4}-20\)
\(\Leftrightarrow3x-15+15+7\sqrt{x-4}-7+7=14\sqrt{x+4}-42+42-20\)
\(\Leftrightarrow3\left(x-5\right)+15+7\cdot\dfrac{x-5}{\sqrt{x-4}+1}+7=14\cdot\dfrac{x-5}{\sqrt{x+4}+3}+42-20\)
\(\Leftrightarrow3\left(x-5\right)+7\cdot\dfrac{x-5}{\sqrt{x-4}+1}-14\cdot\dfrac{x-5}{\sqrt{x+4}+3}=0\)
\(\Leftrightarrow\left(x-5\right)\left(3+\dfrac{7}{\sqrt{x-4}+1}-\dfrac{14}{\sqrt{x+4}+3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\left(N\right)\\3+\dfrac{7}{\sqrt{x-4}+1}-\dfrac{14}{\sqrt{x+4}+3}=0\left(1\right)\end{matrix}\right.\)
Xét pt (1), ta có: \(\dfrac{7}{\sqrt{x-4}+1}>0\)
\(\sqrt{x+4}>2\) (vì x > 4)
\(\Leftrightarrow\sqrt{x+4}+3>5\Leftrightarrow\dfrac{1}{\sqrt{x+4}+3}< \dfrac{1}{5}\Leftrightarrow\dfrac{14}{\sqrt{x+4}+3}< \dfrac{14}{5}\Leftrightarrow-\dfrac{14}{\sqrt{x+4}+3}>-\dfrac{14}{5}\Leftrightarrow3-\dfrac{14}{\sqrt{x+4}+3}>3-\dfrac{14}{5}=\dfrac{1}{5}\)
=> VT > 1/5
Vậy pt (1) vô nghiệm
Kl: x=5
