\(\frac{\left(3\sqrt{x}-\sqrt{x+8}\right)\left(3\sqrt{x}+\sqrt{x+8}\right)}{3\sqrt{x}+\sqrt{x+8}}\)(4+\(\sqrt{x^2+8x}\)) =16(x-1)
\(\frac{8\left(x-1\right)}{3\sqrt{x}+\sqrt{x+8}}\)(4+\(\sqrt{x^2+8x}\)) =16(x-1)
Xét x=1 => 0=0 (TM)
Xét x khác 1
Chia 2 vế cho 8(x-1) ya có pt
\(\frac{4+\sqrt{x^2+8x}}{3\sqrt{x}+\sqrt{x+8}}\) =2
=> 4+\(\sqrt{x^2+8x}\) = 2(3\(\sqrt{x}\) +\(\sqrt{x+8}\) )
bình phương 2 vế ta có
x2 +8x +16 +8\(\sqrt{x^2+8x}\) =32 +40x +24\(\sqrt{x^2+8x}\)
=>x2 - 32x -16 =16\(\sqrt{x^2+8x}\)
(x2 - 32x -16)2 = (16\(\sqrt{x^2+8x}\) )2
x4 - 64x3 + 992x2 + 1024x +256 = 256x2 +2048x
x4 - 64x3 +736x2 - 1024x +256 = 0
(x2 - 32x +16 )2 + 704x2 = 0 (vô lý)
Vậy x=1
