ĐKXĐ: x>=5/4
\(\dfrac{\sqrt{3x^2-9x+21}}{\sqrt{3}}+10\sqrt{\dfrac{16x-20}{25}}=5\sqrt{4x-5}\)
=>\(\sqrt{\dfrac{3\left(x^2-3x+7\right)}{3}}+10\cdot\dfrac{2\sqrt{4x-5}}{5}=5\sqrt{4x-5}\)
=>\(\sqrt{x^2-3x+7}=5\sqrt{4x-5}-4\sqrt{4x-5}=\sqrt{4x-5}\)
=>\(x^2-3x+7=4x-5\)
=>x^2-7x+12=0
=>(x-3)(x-4)=0
=>\(\left[{}\begin{matrix}x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(nhận\right)\\x=4\left(nhận\right)\end{matrix}\right.\)
=>Phương trình có 2 nghiệm nguyên dương