Question

cho biểu thức P= \(\left(\dfrac{\sqrt{x}+1}{x-2\sqrt{x}}-\dfrac{1}{\sqrt{x}-2}\right)\times\left(x-3\sqrt{x}+2\right)\)với x>0 và x≠4.

a) Rút gọn P,

b)Tìm x để P< \(\dfrac{1}{2}\)

c, Tìm gt nguyên của x để P có gt nguyên

Answer 1

a) đk: \(x\ne0;4\); \(x>0\)

P = \(\left[\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{1}{\sqrt{x}-2}\right]\times\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)\)

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

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

b) Để P < \(\dfrac{1}{2}\)

<=> \(\dfrac{\sqrt{x}-1}{\sqrt{x}}< \dfrac{1}{2}\)

<=> \(1-\dfrac{1}{\sqrt{x}}< \dfrac{1}{2}\)

<=> \(\dfrac{1}{\sqrt{x}}>\dfrac{1}{2}\)

<=> \(\sqrt{x}< 2\)

<=> x < 4

<=> 0 < x < 4

Answer 2

c, Tìm gt nguyên của x để P có gt nguyên

Answer 3

a) ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)

Answer 4

c) Để P nguyên thì \(\sqrt{x}-1⋮\sqrt{x}\)

\(\Leftrightarrow-1⋮\sqrt{x}\)

\(\Leftrightarrow\sqrt{x}\inƯ\left(-1\right)\)

\(\Leftrightarrow\sqrt{x}\in\left\{1;-1\right\}\)

hay x=1(thỏa ĐK)

Vậy: Khi P nguyên thì x=1

Answer 5

a) Ta có: \(P=\left(\dfrac{\sqrt{x}+1}{x-2\sqrt{x}}-\dfrac{1}{\sqrt{x}-2}\right)\left(x-3\sqrt{x}+2\right)\)

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

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