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\(A=\left(\frac{1}{x-4}-\frac{1}{x+4\sqrt{x}+4}\right):\frac{x+2\sqrt{x}}{\sqrt{x}}.\)
\(A=\left(\frac{1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\frac{1}{\left(\sqrt{x}+2\right)^2}\right):\frac{x+2\sqrt{x}}{\sqrt{x}}.\)
\(A=\left(\frac{\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}-\frac{\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}\right):\frac{x+2\sqrt{x}}{\sqrt{x}}.\)
\(A=\left(\frac{\sqrt{x}+2-\sqrt{x}+2}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}\right):\frac{x+2\sqrt{x}}{\sqrt{x}}.\)
\(A=\frac{4}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}:\frac{x+2\sqrt{x}}{\sqrt{x}}.\)
\(A=\frac{4}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}\cdot\frac{\sqrt{x}}{x+2\sqrt{x}}\)
\(A=\frac{4}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}\cdot\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x+2}\right)}\)
\(A=\frac{4}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}\cdot\frac{1}{\left(\sqrt{x+2}\right)}\)
\(A=\frac{4}{\left(\sqrt{x}+2\right)^3\left(\sqrt{x}-2\right)}\)
a) Ta có: \(A=\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{2+5\sqrt{x}}{4-x}\)
\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2+5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{3\sqrt{x}}{\sqrt{x}+2}\)
a) Ta có: \(x=9\)thỏa mãn đk
\(\Rightarrow\)Thay \(x=9\)vào biểu thức ta được:
\(A=\frac{3\sqrt{9}}{1-\sqrt{9}}=\frac{9}{-2}=\frac{-9}{2}\)
b) Với x thỏa mãn ĐKXĐ thì ta có:
\(B=\frac{1}{\sqrt{x}+2}-\frac{x+12}{4-x}-\frac{4}{\sqrt{x}-2}\)
\(=\frac{1}{\sqrt{x}+2}+\frac{x+14}{x-4}-\frac{4}{\sqrt{x}-2}\)
\(=\frac{\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\frac{x+12}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{4\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\left(\sqrt{x}-2\right)+\left(x+12\right)-4\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}-2+x+12-4\sqrt{x}-8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x-3\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}-1}{\sqrt{x}+2}\)
1. x = 9 => A = \(\frac{3\sqrt{9}}{1-\sqrt{9}}=\frac{9}{-2}=-\frac{9}{2}\)
2. \(B=\frac{1}{\sqrt{x}+2}-\frac{x+12}{4-x}-\frac{4}{\sqrt{x}-2}=\frac{\sqrt{x}-2+x+12-4\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x-3\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{x-\sqrt{x}-2\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}-1}{\sqrt{x}+2}\)
3. \(AB>-\frac{3}{4}\) <=> \(\frac{3\sqrt{x}}{1-\sqrt{x}}\cdot\frac{\sqrt{x}-1}{\sqrt{x}+2}>-\frac{3}{4}\)
<=> \(-\frac{3\sqrt{x}}{\sqrt{x}+2}+\frac{3}{4}>0\)
<=> \(\frac{12\sqrt{x}-3\sqrt{x}-4}{4\left(\sqrt{x}+2\right)}< 0\)
<=> \(\frac{9\sqrt{x}-4}{4\sqrt{x}+8}< 0\)
Do \(4\sqrt{x}+8>0\)với mọi x => \(9\sqrt{x}-4< 0\) <=> \(x< \frac{16}{81}\)
ĐKXĐ\(\left\{{}\begin{matrix}x\ge0\\4-x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\-x\ne-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
Cậu nhầm ở chỗ chuyển từ dương 4 sang thì phải thành -4