a:\(A=\dfrac{x\sqrt{x}+1-\left(x-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x\sqrt{x}+1-x\sqrt{x}+x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
b: Khi x=1/4 thì \(A=\dfrac{1}{2}:\left(\dfrac{1}{2}-1\right)=\dfrac{-1}{2}:\dfrac{1}{2}=-1\)
c: Để A<0 thì \(\sqrt{x}-1< 0\)
=>0<=x<1
Lời giải:
ĐKXĐ: $x\geq 0; x\neq 1$
a.
\(A=\frac{(\sqrt{x}+1)(x-\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)}-\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}+1}\)
\(=\frac{x-\sqrt{x}+1}{\sqrt{x}-1}-(\sqrt{x}-1)=\frac{x-\sqrt{x}+1-(\sqrt{x}-1)^2}{\sqrt{x}-1}=\frac{\sqrt{x}}{\sqrt{x}-1}\)
b.
Tại $x=\frac{1}{4}$ thì $\sqrt{x}=\frac{1}{2}$
$A=\frac{\frac{1}{2}}{\frac{1}{2}-1}=-1$
c.
$A=\frac{\sqrt{x}}{\sqrt{x}-1}<0$ \(\Leftrightarrow \left\{\begin{matrix} x>0\\ \sqrt{x}-1<0\end{matrix}\right.\Leftrightarrow 0< x< 1\)
Kết hợp với ĐKXĐ suy ra $0< x< 1$ thì $A<0$
