Question

cho biểu thức A=\(\left(\dfrac{\sqrt{x}+1}{x-2\sqrt{x}}-\dfrac{1}{\sqrt{x}-2}\right)\) x \(\left(x-3\sqrt{x}+2\right)\)

a, rút gọn A

b, tìm x để A<\(\dfrac{1}{2}\)

c, tìm x \(\in\) Z để A \(\in\) Z

Answer 1

a, Mk làm hơi tắt chút bạn thông cảm nha . mk vội ý mà

\(A=\left(\dfrac{\sqrt{x}+1}{x-2\sqrt{x}}-\dfrac{1}{\sqrt{x}-2}\right).\left(x-3\sqrt{x}+2\right)\)

\(A=\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}.\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)\)

\(A=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-2\right)}\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)

Answer 2

Câu c : \(A\in Z\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}}\in Z\Leftrightarrow1-\dfrac{1}{\sqrt{x}}\in Z\)

Để : \(1-\dfrac{1}{\sqrt{x}}\in Z\) thì \(\sqrt{x}\inƯ\left(1\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x}=1\\\sqrt{x}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\x\in\varnothing\end{matrix}\right.\)

Vậy \(x=1\)

Answer 3

Để A <\(\dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}}< \dfrac{1}{2}\left(\text{đ}k\text{x}\text{đ}:x\ne0\right)\)

\(\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}}-\dfrac{1}{2}< 0\)

\(\Leftrightarrow2\sqrt{x}-2-\sqrt{x}< 0\)

\(\Leftrightarrow\sqrt{x}-2< 0\)

\(\Leftrightarrow\sqrt{x}< 2\)

\(\Leftrightarrow x< 4\left(TM\right)\)

Vậy x<4 và x≠0 thì ta có A<\(\dfrac{1}{2}\)