Question

1. Cho \(Q=\left(\dfrac{2\sqrt{a}}{\sqrt{a}+3}+\dfrac{\sqrt{a}}{\sqrt{a}-3}-\dfrac{3a+3}{a-9}\right):\left(\dfrac{2\sqrt{a}-2}{\sqrt{a}-3}-1\right)\)

a) Rút gọn Q.

b) Tìm x để \(Q< \dfrac{-1}{2}\)

c) Tìm Q_min

2. \(P=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}-\dfrac{\sqrt{x}+3}{5-\sqrt{x}}-\dfrac{3x+4\sqrt{x}-5}{x-4\sqrt{x}-5}\)

a) Rút gọn P.

b) Tìm x để P > -2

c) Tìm số chính phương x để P có giá trị nguyên.

Answer 1

1.

a. \(Q=\left(\dfrac{2\sqrt{a}}{\sqrt{a}+3}+\dfrac{\sqrt{a}}{\sqrt{a}-3}-\dfrac{3a+3}{a-9}\right):\left(\dfrac{2\sqrt{a}-2}{\sqrt{a}-3}-1\right)=\left(\dfrac{2\sqrt{a}}{\sqrt{a}+3}+\dfrac{\sqrt{a}}{\sqrt{a}-3}-\dfrac{3a+3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\right):\left(\dfrac{2\sqrt{a}-2-\sqrt{a}+3}{\sqrt{a}-3}\right)=\)

\(\dfrac{2\sqrt{a}\left(\sqrt{a}-3\right)+\sqrt{a}\left(\sqrt{a}+3\right)-3a-3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}:\dfrac{\sqrt{a}+1}{\sqrt{a}-3}=\dfrac{2a-6\sqrt{a}+a+3\sqrt{a}-3a-3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}.\dfrac{\sqrt{a}-3}{\sqrt{a}+1}=\dfrac{-3\sqrt{a}-3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+1\right)}.\dfrac{\sqrt{a}-3}{\sqrt{a}+1}=\dfrac{-3\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}.\dfrac{\sqrt{a}-3}{\sqrt{a}+1}=-\dfrac{3}{\sqrt{a}+3}\)

Answer 2

@Ace Legona @Nguyễn Huy Tú @Trần Thiên Kim

Giúp với ԅ[ •́ ﹏├┬┴┬┴

Answer 3

1b. Để \(Q< -\dfrac{1}{2}\) thì \(\dfrac{-3}{\sqrt{a}+3}< -\dfrac{1}{2}\) (ĐKXĐ: \(a\ge0\))

\(\Leftrightarrow-\dfrac{3}{\sqrt{a}+3}+\dfrac{1}{2}< 0\Leftrightarrow\dfrac{-6+\sqrt{a}+3}{2\left(\sqrt{a}+3\right)}< 0\)

\(\Rightarrow\sqrt{a}-3< 0\Leftrightarrow\sqrt{a}< 3\Leftrightarrow a< 9\)

Vậy để \(Q< -\dfrac{1}{2}\)thì \(0\le a< 9\)

Answer 4

2a. \(\dfrac{\sqrt{x}+2}{\sqrt{x}+1}-\dfrac{\sqrt{x}+3}{5-\sqrt{x}}-\dfrac{3x+4\sqrt{x}-5}{x-4\sqrt{x}-5}=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}+\dfrac{\sqrt{x}+3}{\sqrt{x}-5}-\dfrac{3x+4\sqrt{x}-5}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-5\right)+\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)-3x-4\sqrt{x}+5}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+1\right)}=\dfrac{x-5\sqrt{x}+2\sqrt{x}-10+x+\sqrt{x}+3\sqrt{x}+3-3x-4\sqrt{x}+5}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{-x-3\sqrt{x}-2}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}-5}\) b. Để \(P>-2\) thì \(\dfrac{\sqrt{x}+2}{\sqrt{x}-5}>-2\) (ĐKXĐ: \(x\ge0\))

\(\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}-5}+2>0\Leftrightarrow\dfrac{\sqrt{x}+2+2\left(\sqrt{x}-5\right)}{\sqrt{x}-5}>0\Rightarrow\sqrt{x}+2+2\sqrt{x}-10>0\Leftrightarrow3\sqrt{x}-8>0\Leftrightarrow3\sqrt{x}>8\Leftrightarrow\sqrt{x}>\dfrac{8}{3}\Leftrightarrow x>\dfrac{64}{9}\)

Vậy để \(P>-2\) thì \(x>\dfrac{64}{9}\)

P/s: Làm xong câu b thấy sai sai nhể =="

1c và 2c đành để lại vậy =(((