Question

M=\(\left(\dfrac{1}{x+\sqrt{x}}-\dfrac{1}{1+\sqrt{x}}\right):\dfrac{x-\sqrt{x}+1}{x\sqrt{x}+1}\)

tìm các gt của x để M nhận giá trị âm

Answer 1

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 $x>1$

Answer 2

\(M=\left(\dfrac{1}{x+\sqrt{x}}-\dfrac{1}{1+\sqrt{x}}\right):\dfrac{x-\sqrt{x}+1}{x\sqrt{x}+1}\)(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 ......