a) Ta có: \(A=\left(\frac{\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}\right)\cdot\frac{\sqrt{x}-2}{2}\)
\(=\left(\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\cdot\frac{\sqrt{x}-2}{2}\)
\(=\frac{2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\frac{\sqrt{x}-2}{2}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}+2}\)
b) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
Ta có: \(x=7-4\sqrt{3}\)
\(=4-2\cdot2\cdot\sqrt{3}+3\)
\(=\left(2-\sqrt{3}\right)^2\)(nhận)
Thay \(x=\left(2-\sqrt{3}\right)^2\) vào biểu thức \(A=\frac{\sqrt{x}+1}{\sqrt{x}+2}\), ta được:
\(A=\frac{\sqrt{\left(2-\sqrt{3}\right)^2}+1}{\sqrt{\left(2-\sqrt{3}\right)^2}+2}\)
\(=\frac{\left|2-\sqrt{3}\right|+1}{\left|2-\sqrt{3}\right|+2}\)
\(=\frac{2-\sqrt{3}+1}{2-\sqrt{3}+2}\)(Vì \(2>\sqrt{3}\))
\(=\frac{3-\sqrt{3}}{4-\sqrt{3}}\)
Vậy: Khi \(x=7-4\sqrt{3}\) thì \(A=\frac{3-\sqrt{3}}{4-\sqrt{3}}\)
c) Để \(A=\frac{3}{4}\) thì \(\frac{\sqrt{x}+1}{\sqrt{x}+2}=\frac{3}{4}\)
\(\Leftrightarrow4\cdot\left(\sqrt{x}+1\right)=3\left(\sqrt{x}+2\right)\)
\(\Leftrightarrow4\sqrt{x}+4=3\sqrt{x}+6\)
\(\Leftrightarrow4\sqrt{x}-3\sqrt{x}=6-4=2\)
\(\Leftrightarrow\sqrt{x}=2\)
hay x=4(loại)
Vậy: Không có giá trị nào của x để \(A=\frac{3}{4}\)
d) Để \(A=-\sqrt{x}+3\) thì \(\frac{\sqrt{x}+1}{\sqrt{x}+2}=-\sqrt{x}+3\)
\(\Leftrightarrow\sqrt{x}+1=\left(3-\sqrt{x}\right)\left(\sqrt{x}+2\right)\)
\(\Leftrightarrow\sqrt{x}+1=3\sqrt{x}+6-x-2\sqrt{x}\)
\(\Leftrightarrow\sqrt{x}+1=-x+\sqrt{x}+6\)
\(\Leftrightarrow\sqrt{x}+1+x-\sqrt{x}-6=0\)
\(\Leftrightarrow x-5=0\)
\(\Leftrightarrow x=5\)(nhận)
Vậy: Để \(A=-\sqrt{x}+3\) thì x=5
e) Để \(A< \frac{2}{3}\) thì \(A-\frac{2}{3}< 0\)
\(\Leftrightarrow\frac{\sqrt{x}+1}{\sqrt{x}+2}-\frac{2}{3}< 0\)
\(\Leftrightarrow\frac{3\left(\sqrt{x}+1\right)}{3\left(\sqrt{x}+2\right)}-\frac{2\left(\sqrt{x}+2\right)}{3\left(\sqrt{x}+2\right)}< 0\)
\(\Leftrightarrow\frac{3\sqrt{x}+3-2\sqrt{x}-4}{3\left(\sqrt{x}+2\right)}< 0\)
\(\Leftrightarrow\frac{\sqrt{x}-1}{3\left(\sqrt{x}+2\right)}< 0\)
\(\Leftrightarrow\sqrt{x}-1\) và \(3\left(\sqrt{x}+2\right)\) khác dấu
mà \(3\left(\sqrt{x}+2\right)>0\forall x\) thỏa mãn ĐKXĐ
nên \(\sqrt{x}-1< 0\)
\(\Leftrightarrow\sqrt{x}< 1\)
\(\Leftrightarrow\left|x\right|< 1\)
\(\Leftrightarrow-1< x< 1\)
Kết hợp ĐKXĐ,ta được: \(0\le x< 1\)
Vậy: Để \(A< \frac{2}{3}\) thì \(0\le x< 1\)
