Question

giúp mình va

Answer 1

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

\(=\dfrac{x\sqrt{x}+1-x\sqrt{x}+x+\sqrt{x}-1}{x-1}=\dfrac{x+\sqrt{x}}{x-1}=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)

b: Thay x=1/4 vào A, ta được:

\(A=\dfrac{\dfrac{1}{2}}{\dfrac{1}{2}-1}=\dfrac{1}{2}:\dfrac{-1}{2}=-1\)

c: Để A<0 thì \(\sqrt{x}-1< 0\)

hay 0<=x<1

d: Để |A|=A thì A>=0

=>\(\sqrt{x}-1>=0\)

hay x>1

Answer 2

Bài 4:

a) ĐKXĐ: \(x\ne1;x\ge0\)

\(A=\dfrac{x\sqrt{x}+1}{x-1}-\dfrac{x-1}{\sqrt{x}+1}\)

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

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

\(=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)

b) \(\left|x\right|=\dfrac{1}{4}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)

Thay  \(x=\dfrac{1}{4}\) vào \(A=\dfrac{\sqrt{x}}{\sqrt{x}-1}\) ta có:

\(A=\dfrac{\sqrt{\dfrac{1}{4}}}{\sqrt{\dfrac{1}{4}}-1}=-1\)

Thay \(x=-\dfrac{1}{4}\) vào \(A=\dfrac{\sqrt{x}}{\sqrt{x}-1}\) ta có:

\(A=\dfrac{\sqrt{-\dfrac{1}{3}}}{\sqrt{-\dfrac{1}{4}}-1}=-1\) (vô lí, vì \(\sqrt{x}\ge0\forall x\))

Vậy \(A=-1\) khi \(\left|x\right|=\dfrac{1}{4}\)

c) Để \(A< 0\) thì \(\dfrac{\sqrt{x}}{\sqrt{x}-1}< 0\)

Mà \(\sqrt{x}>0\) (luôn dương \(\forall x\))

\(\Rightarrow\left(\sqrt{x}-1\right)< 0\)

\(\Leftrightarrow0\le x< 1\)

Vậy \(0\le x< 1\) thì \(A< 0\)

d) Ta có: \(\left|A\right|=A\) (ĐK: \(A>0\))

\(\Rightarrow\dfrac{\sqrt{x}}{\sqrt{x}-1}>0\)

\(\Rightarrow\sqrt{x}-1>0\)

\(\Leftrightarrow x>1\)

Vậy \(x>1\) thì \(\left|A\right|=A\)