Question

[Nguyễn Mạnh Hưng]

Nhờ các bạn giải giúp.

Q=(\(\frac{2\sqrt{x}}{\sqrt{x}+3}\)+\(\frac{\sqrt{x}+1}{\sqrt{x}-3}\)-\(\frac{3-11\sqrt{x}}{x-9}\)

a. Tìm ĐXĐ.

b. Rút gọn Q.

c. Tìm x để Q<1

Câu:

A=(\(\frac{\sqrt{a}}{\sqrt{a}-1}\)-\(\frac{1}{a-\sqrt{a}}\)):(\(\frac{1}{\sqrt{a}+1}\)-\(\frac{2}{a-1}\))

a. Rút gọn A.

Answer 1

[Hoàng Thị Lan Hương]

a. ĐK \(\hept{\begin{cases}x\ge0\\x\ne9\end{cases}}\)

b. \(Q=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}-\frac{3-11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

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

\(=\frac{2x-6\sqrt{x}+x+4\sqrt{x}+3-3+11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

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

c. Để \(Q< 1\Rightarrow Q-1< 0\Leftrightarrow\frac{3\sqrt{x}-\sqrt{x}+3}{\sqrt{x}-3}< 0\Leftrightarrow\frac{2\sqrt{x}+3}{\sqrt{x}-3}< 0\)

\(\Rightarrow\sqrt{x}-3< 0\Rightarrow0\le x< 9\)

Vậy \(0\le x< 9\)thì \(Q< 1\)