\(\frac{\sqrt{2}-2}{x-4}-\frac{\sqrt{x}5}{x-25}\left(\right)\frac{\sqrt{x}}{7}\)
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5 tháng 10 2018
4) mấy bài kia trình bày dài lắm!! (lười ý mà ahihi)
\(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y+\sqrt{2}\right)^2}+|x+y+z|=0.\)
\(\Leftrightarrow|x-\sqrt{2}|+|y+\sqrt{2}|+|x+y+z|=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-\sqrt{2}=0\\y+\sqrt{2}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{2}\\y=-\sqrt{2}\end{cases}}}\)
Tìm z thì dễ rồi
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 6 2019
ĐKXĐ:...
\(\left(\frac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-1\right):\left(\frac{25-x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}-\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}+\frac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}\right)\)
\(=\left(\frac{\sqrt{x}-\sqrt{x}-5}{\sqrt{x}+5}\right):\left(\frac{25-x-x+9+x-25}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}\right)=\frac{-5}{\left(\sqrt{x}+5\right)}.\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}{\left(9-x\right)}\)
\(=\frac{5\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{5}{\sqrt{x}+3}\)
11 tháng 8 2021
\(A=2-x\sqrt{\frac{x\left(x-2\right)}{\left(x-2\right)^2}+\frac{1}{\left(x-2\right)^2}}=2-x\sqrt{\frac{\left(x-1\right)^2}{\left(x-2\right)^2}}\)
\(=2-x\cdot\frac{x-1}{x-2}=\frac{2x-4}{x-2}-\frac{x^2-x}{x-2}=\frac{-x^2+3x-4}{x-2}\)
\(B=\frac{2\sqrt{5}x}{x-2}\cdot\left|x-2\right|+\frac{3\sqrt{5}x^2}{x}=\frac{2\sqrt{5}x}{x-2}\cdot\left|x-2\right|+3\sqrt{5}x\)
Với 0 < x < 2 \(B=-2\sqrt{5}x+3\sqrt{5}x=\sqrt{5}x\)
Với x > 2 \(B=2\sqrt{5}x+3\sqrt{5}x=5\sqrt{5}x\)
\(C=\frac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\sqrt{x}\left(\sqrt{x}+5\right)}+\sqrt{\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-5\right)^2}}=\frac{\sqrt{x}-5}{\sqrt{x}}+\left|\frac{\sqrt{x}-1}{\sqrt{x}-5}\right|\)
Với 0 < x < 1 \(C=\frac{\sqrt{x}-5}{\sqrt{x}}+\frac{\sqrt{x}-1}{\sqrt{x}-5}=\frac{x-10\sqrt{x}+25}{x\left(\sqrt{x}-5\right)}+\frac{x-\sqrt{x}}{x\left(\sqrt{x}-5\right)}=\frac{2x-11\sqrt{x}+25}{x\left(\sqrt{x}-5\right)}\)
Với 1 < x < 5 \(C=\frac{\sqrt{x}-5}{\sqrt{x}}-\frac{\sqrt{x}-1}{\sqrt{x}-5}=\frac{x-10\sqrt{x}+25}{x\left(\sqrt{x}-5\right)}-\frac{x-\sqrt{x}}{x\left(\sqrt{x}-5\right)}=\frac{-9\sqrt{x}+25}{x\left(\sqrt{x}-5\right)}\)
Với x > 5 \(C=\frac{\sqrt{x}-5}{\sqrt{x}}+\frac{\sqrt{x}-1}{\sqrt{x}-5}=\frac{x-10\sqrt{x}+25}{x\left(\sqrt{x}-5\right)}+\frac{x-\sqrt{x}}{x\left(\sqrt{x}-5\right)}=\frac{2x-11\sqrt{x}+25}{x\left(\sqrt{x}-5\right)}\)
28 tháng 8 2017
1.
= -(13 + 3 căn7 ) / 2 + -(7 + 3 căn7 ) / 2
= -7 + 3 căn7
23 tháng 3 2020
ĐKXĐ :\(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-3\ne0\\\sqrt{x}+5\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}\ne3\\\sqrt{x}\ne-5\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)
- Ta có : \(\left(\frac{x-5\sqrt{x}}{25}-1\right):\left(\frac{25-x}{x+2\sqrt{x}-15}-\frac{\sqrt{x}+3}{\sqrt{x}+5}+\frac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
\(=\left(\frac{x-5\sqrt{x}-25}{25}\right):\left(\frac{25-x}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}+3}{\sqrt{x}+5}+\frac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
\(=\left(\frac{x-5\sqrt{x}-25}{25}\right):\left(\frac{25-x}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}\right)\)
\(=\left(\frac{x-5\sqrt{x}-25}{25}\right):\left(\frac{25-x}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}-\frac{x-9}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}+\frac{x-25}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}\right)\)
\(=\left(\frac{x-5\sqrt{x}-25}{25}\right):\left(\frac{25-x-x+9+x-25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\left(\frac{x-5\sqrt{x}-25}{25}\right):\left(\frac{-x+9}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\left(\frac{x-5\sqrt{x}-25}{25}\right):\left(\frac{-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\left(\frac{x-5\sqrt{x}-25}{25}\right)\left(\frac{\sqrt{x}+5}{-\sqrt{x}-3}\right)\)
\(=\frac{\left(x-5\sqrt{x}-25\right)\left(\sqrt{x}+5\right)}{-25\left(\sqrt{x}+3\right)}=\frac{x\sqrt{x}+5x-5x-25\sqrt{x}-25\sqrt{x}-125}{-25\left(\sqrt{x}+3\right)}\)
\(=\frac{x\sqrt{x}-125-50\sqrt{x}}{-25\left(\sqrt{x}+3\right)}\)
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