Lời giải:
ĐKXĐ: $x>0$
\(M=\frac{1-\sqrt{x}}{\sqrt{x}(\sqrt{x}+1)}:\frac{x-\sqrt{x}+1}{(\sqrt{x}+1)(x-\sqrt{x}+1)}=\frac{1-\sqrt{x}}{\sqrt{x}(\sqrt{x}+1)}.(\sqrt{x}+1)=\frac{1-\sqrt{x}}{\sqrt{x}}\)
Để $M< 0\Leftrightarrow \frac{1-\sqrt{x}}{\sqrt{x}}< 0$
$\Leftrightarrow 1-\sqrt{x}< 0$
$\Leftrightarrow \sqrt{x}>1$
$\Leftrightarrow x>1$
Kết hợp với đkxđ suy ra $x>1$
\(M=\left(\dfrac{1}{x+\sqrt{x}}-\dfrac{1}{1+\sqrt{x}}\right):\dfrac{x-\sqrt{x}+1}{x\sqrt{x}+1}\)(Đkxđ:x>0)
\(M=\left(\dfrac{1}{\sqrt{x}\left(1+\sqrt{x}\right)}-\dfrac{1}{1+\sqrt{x}}\right):\dfrac{x-\sqrt{x}+1}{\left(\sqrt{x}\right)^3+1^3}\)
\(M=\left(\dfrac{1-\sqrt{x}}{\sqrt{x}\left(1+\sqrt{x}\right)}\right):\dfrac{x-\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(M=\left(\dfrac{1-\sqrt{x}}{\sqrt{x}\left(1+\sqrt{x}\right)}\right)\cdot\left(\sqrt{x}+1\right)\)
\(M=\dfrac{1-\sqrt{x}}{\sqrt{x}}\)
Để M nhận giá trị âm
Thì\(\dfrac{1-\sqrt{x}}{\sqrt{x}}< 0\)
Vì\(x>0\Rightarrow\sqrt{x}>0\)
Nên \(1-\sqrt{x}< 0\Leftrightarrow\sqrt{x}>1\Leftrightarrow x>1\)
Vậy ......
