B=\(\left(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{4}{x-2\sqrt{x}}\right).\left(\frac{1}{\sqrt{x}+2}+\frac{4}{x-4}\right)\)
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29 tháng 7 2020
\(B=\left(\frac{1}{\sqrt{x}-2}-\frac{2}{\sqrt{x}+2}+\frac{x}{x\sqrt{x}-4\sqrt{x}}\right):\left(\frac{6-x}{\sqrt{x}+2}+2+\sqrt{x}\right)\)
\(B=\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x+2}\right)}\right):\left(\frac{6-x+2\left(\sqrt{x}+2\right)+\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}+2}\right)\)
\(B=\left(\frac{\sqrt{x}+2-2\sqrt{x}+4+\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right):\left(\frac{6-x+2\sqrt{x}+4+x+2\sqrt{x}}{\sqrt{x}+2}\right)\)
\(B=\frac{6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\frac{\sqrt{x}+2}{10+4\sqrt{x}}\)
\(B=\frac{6}{\sqrt{x}-2}\cdot\frac{1}{2\left(5+2\sqrt{x}\right)}\)
B = \(\frac{3}{\left(\sqrt{x}-2\right)\left(5+2\sqrt{x}\right)}\)
NC
21 tháng 8 2020
đkxđ: \(x\ge0;x\ne4\)
\(Q=\left[\frac{x-\sqrt{x}+7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{1}{\sqrt{x}-2}\right]\div\left[\frac{\sqrt{x}+2}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+2}-\frac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right]\)
\(Q=\left[\frac{x-\sqrt{x}+7+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right]\div\left[\frac{\left(\sqrt{x}+2\right)^2-\left(\sqrt{x}-2\right)^2-2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right]\)
\(Q=\frac{x+9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\div\frac{x+4\sqrt{x}+4-x+4\sqrt{x}-4-2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(Q=\frac{x+9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{6\sqrt{x}}\)
\(Q=\frac{\left(x+9\right)\sqrt{x}}{6x}\)
\(Q=\frac{x\sqrt{x}+9\sqrt{x}}{6x}\)
NC
21 tháng 8 2020
đkxđ sửa tí thành \(\hept{\begin{cases}x>0\\x\ne4\end{cases}}\)
30 tháng 1 2022
a: \(Q=\dfrac{4\sqrt{x}\left(\sqrt{x}-2\right)-8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{\sqrt{x}-1-2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{-8\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-1-2\sqrt{x}+4}\)
\(=\dfrac{-8x}{\sqrt{x}+2}\cdot\dfrac{1}{3-\sqrt{x}}=\dfrac{8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
b: Để Q=-1 thì \(8x=-\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)\)
\(\Leftrightarrow8x+x-\sqrt{x}-6=0\)
\(\Leftrightarrow9x-\sqrt{x}-6=0\)
Bạn xem lại đề, nghiệm này rất xấu
B= \(\left(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{4}{x-2\sqrt{2}}\right):\left(\frac{1}{\sqrt{x}+2}+\frac{4}{x-4}\right)\)
= \(\left(\frac{\sqrt{x}}{\sqrt{x}-2}+\frac{4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right):\left(\frac{\sqrt{x}-2+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right)\)
= \(\frac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\):\(\frac{1}{\sqrt{x}-2}\)
=\(\frac{\sqrt{x}+2}{\sqrt{x}}.\sqrt{x}-2\)
Nhớ tick cho mk nhé
=\(\frac{x-4}{\sqrt{x}}\)
Mk đang cần gấp nhé