Question

[minhh minh]

giup minh cau nay voi

Answer 1

[Nguyễn Việt Lâm]

a. Em tự tính

b.

\(B=\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{x-9-2\sqrt{x}+9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{\sqrt{x}}{\sqrt{x}+3}\)

c.

\(P=\dfrac{B}{A}=\dfrac{\sqrt{x}}{\sqrt{x}-3}\)

\(\left|P\right|>P\Leftrightarrow P< 0\)

\(\Rightarrow\dfrac{\sqrt{x}}{\sqrt{x}-3}< 0\) \(\Rightarrow\left\{{}\begin{matrix}x\ne0\\\sqrt{x}-3< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x< 9\end{matrix}\right.\)

Kết hợp điều kiện đề bài \(\Rightarrow\left\{{}\begin{matrix}0< x< 9\\x\ne4\end{matrix}\right.\)

Answer 2

[Nguyễn Lê Phước Thịnh]

a: Khi x=4 thì \(A=\dfrac{2-3}{2+3}=\dfrac{-1}{5}\)

b: \(=\dfrac{x-9-2\sqrt{x}+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\dfrac{x-2\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}}{\sqrt{x}+3}\)

c: P=B:A

\(=\dfrac{\sqrt{x}}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}+3}{\sqrt{x}-3}=\dfrac{\sqrt{x}}{\sqrt{x}-3}\)

Để |P|>P thì P<=0

=>căn x-3<0

=>0<x<9