ĐKXĐ : \(x\ge-1\)
P/t \(\Leftrightarrow\left(2x^2-16x\right)-5\left(x-8\right)-5\left(x-2\right)\left[\sqrt{x+1}-3\right]=0\)
\(\Leftrightarrow2x\left(x-8\right)-5\left(x-8\right)-5\left(x-2\right)\dfrac{x-8}{\sqrt{x+1}+3}=0\)
\(\Leftrightarrow\left(x-8\right)\left[2x-5-\dfrac{5\left(x-2\right)}{\sqrt{x+1}+3}\right]=0\)
x = 8 là no của p/t
Xét [...] \(\Leftrightarrow\left(2x-5\right)\left[\sqrt{x+1}+3\right]-5\left(x-2\right)=0\)
\(\Leftrightarrow\left(2x+2\right)\sqrt{x+1}+\left(\sqrt{x+1}-\dfrac{7}{2}\right)^2=\dfrac{73}{4}\)
Đặt a = \(\sqrt{x+1}\ge0\) ; p/t \(\Leftrightarrow2a^3+\left(a-\dfrac{7}{2}\right)^2=\dfrac{73}{4}\)
\(\Leftrightarrow2a^3+a^2-7a-6=0\) \(\Leftrightarrow\left(a-2\right)\left(a+1\right)\left(a+\dfrac{3}{2}\right)=0\)
\(\Leftrightarrow a=2\) hay \(\sqrt{x+1}=2\Leftrightarrow x=3\)
Vậy p/t có 2 no x = 3 và x = 8
