Question

3) cho bt P= \(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{3x+9}{9-x}\)

a) rút gọn bt P

b) tìm điều kiện của x để P > 0

c) tìm x nguyên để P nhận giá trị nguyên

giúp mk vs ạ mk cần gấp

Answer 1

a, ĐK: \(x\ge0;x\ne9\)

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

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

\(=\dfrac{2x-6\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{x+3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{3x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

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

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

Answer 2

b, \(P>0\Leftrightarrow-\dfrac{3}{\sqrt{x}-3}>0\)

\(\Leftrightarrow\sqrt{x}-3>0\)

\(\Leftrightarrow x>9\)

c, \(P=-\dfrac{3}{\sqrt{x}-3}\in Z\)

\(\Leftrightarrow\sqrt{x}-3\inƯ_3=\left\{\pm1;\pm3\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{0;2;4;6\right\}\)

\(\Leftrightarrow x\in\left\{0;4;16;36\right\}\)

Answer 3

a: Ta có: \(P=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{3x+9}{9-x}\)

\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

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

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

b: Để P<0 thì \(\sqrt{x}-3< 0\)

hay x<9

Kết hợp ĐKXĐ, ta được: \(0\le x< 9\)