Question

[Thục Quyên]

giải bài 1 nhé mn

Answer 1

[An Thy]

a) \(P=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\left(x\ge0,x\ne1\right)\)

\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(=\dfrac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(2-5\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}\)

b) \(P=\dfrac{1}{2}\Rightarrow\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}=\dfrac{1}{2}\Rightarrow4-10\sqrt{x}=\sqrt{x}+3\Rightarrow11\sqrt{x}=1\)

\(\Rightarrow x=\dfrac{1}{121}\)

c) \(P\le\dfrac{2}{3}\Rightarrow\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}\le\dfrac{2}{3}\Rightarrow\dfrac{2}{3}-\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}\ge0\)

\(\Rightarrow\dfrac{2\left(\sqrt{x}+3\right)-3\left(2-5\sqrt{x}\right)}{3\left(\sqrt{x}+3\right)}\ge0\Rightarrow\dfrac{17\sqrt{x}}{3\left(\sqrt{x}+3\right)}\ge0\) (luôn đúng)

Answer 2

[Nguyễn Lê Phước Thịnh]

Bài 1: 

a) Ta có: \(P=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(=\dfrac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{-\left(\sqrt{x}-1\right)\left(5\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}\)

b) Để \(P=\dfrac{1}{2}\) thì \(4-10\sqrt{x}-\sqrt{x}-3=0\)

\(\Leftrightarrow-11\sqrt{x}=-1\)

\(\Leftrightarrow x=\dfrac{1}{121}\)