rut gon bieu thuc
A=\(\frac{2}{5}.\sqrt{50x}-\frac{3}{4}.\sqrt{8}\left(x\ge0\right)\)
K
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5 tháng 7 2020
a) Ta có: \(A=\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{2+5\sqrt{x}}{4-x}\)
\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2+5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{3\sqrt{x}}{\sqrt{x}+2}\)
30 tháng 5 2016
\(A=\left(\frac{\sqrt{x}}{x-4}+\frac{2}{2-\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\frac{10-x}{\sqrt{x}+2}\right)\)(DK : \(x\ge0;x\ne4\))
\(=\frac{\sqrt{x}-2\left(\sqrt{x}+2\right)+\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\frac{x-4+10-x}{\sqrt{x}+2}\)
\(=\frac{-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\frac{\sqrt{x}+2}{6}=\frac{1}{2-\sqrt{x}}\)
Để A > 0 thì \(2-\sqrt{x}>0\Rightarrow\sqrt{x}< 2\Rightarrow x< 4\)
Vậy để A < 0 thì x < 4
CH
Cô Hoàng Huyền
Admin
VIP
31 tháng 5 2016
Bảo Ngọc kết luận hơi sai một chút nhé. Để A > 0 thì x < 4 nhé :)
29 tháng 7 2019
\(B=\left(\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-3}-\frac{\sqrt{x}+3}{\sqrt{x}-1}\right)\)
\(=\frac{\sqrt{x}-\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)\(:\left(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\frac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\)\(\frac{x-1-x+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{2}\)
\(=\frac{\sqrt{x}+3}{2\sqrt{x}}\)
24 tháng 11 2019
a. A có nghĩa khi \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-1\ne\\\frac{x+\sqrt{x}}{\sqrt{x}+1}\ne0\end{matrix}\right.0\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
A\(=\frac{x-\sqrt{x}+\sqrt{x}-1}{\sqrt{x}-1}.\frac{\sqrt{x}+1}{x+\sqrt{x}}\)\(=\frac{x-1}{\sqrt{x}-1}.\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}.\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}}\)
b. \(x=7+4\sqrt{3}\Rightarrow\)A = \(\frac{\sqrt{7+4\sqrt{3}}+1}{\sqrt{7+4\sqrt{3}}}=\frac{\sqrt{\left(2+\sqrt{3}\right)^2}+1}{\sqrt{\left(2+\sqrt{3}\right)^2}}=\frac{3+\sqrt{3}}{2+\sqrt{3}}\)
6 tháng 12 2015
\(=\left(\frac{x+2-\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right):\left(\frac{\left(\sqrt{x}-4\right)\left(x+1\right)-\sqrt{x}\left(1-x\right)}{1-x^2}\right)\)
\(=\left(\frac{x+2-x-\sqrt{x}}{\sqrt{x}+1}\right):\left(\frac{x\sqrt{x}+\sqrt{x}-4x-4-\sqrt{x}+x\sqrt{x}}{1-x^2}\right)\)
\(=\frac{2-\sqrt{x}}{\sqrt{x}+1}:\frac{2x\sqrt{x}-3x-4}{\left(1-x\right)\left(1+x\right)}\)
\(=\frac{2-\sqrt{x}}{\sqrt{x}+1}.\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(1+x\right)}{2x\sqrt{x}-3x-4}\)
\(=\frac{\left(2-\sqrt{x}\right)\left(\sqrt{x}+x\sqrt{x}-1+x\right)}{2x\sqrt{x}-3x-4}\)
\(=\frac{2\sqrt{x}+2x\sqrt{x}-2+2x-x-x^2+\sqrt{x}-x\sqrt{x}}{2x\sqrt{x}-3x-4}\)
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