\(\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)\left(\frac{a-\sqrt{a}}{\sqrt{a}+1}-\frac{a+\sqrt{a}}{\sqrt{a}-1}\right)\)
giúp mk vs ạ
K
Khách
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12 tháng 6 2017
\(A=\left(\frac{\left(\sqrt{a}+1\right)^2-\left(\sqrt{a}-1\right)^2}{a-1}+4\sqrt{a}\right)\left(\frac{a+1}{\sqrt{a}}\right)\)
\(A=\left(\frac{4\sqrt{a}}{a-1}+\frac{4\sqrt{a}\left(a-1\right)}{a-1}\right)\left(\frac{a+1}{\sqrt{a}}\right)\)
\(A=\frac{4a\sqrt{a}}{a-1}.\frac{a+1}{\sqrt{a}}=\frac{4a\left(a+1\right)}{a-1}\)
....... Tới đây được chưa bạn?
5 tháng 1 2020
\(\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{a-\sqrt{a}}\right):\left(\frac{1}{\sqrt{a}+1}+\frac{2}{a-1}\right)\) (\(a>0;a\ne1\))
=\(\left(\frac{a}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right)\):\(\left(\frac{\sqrt{a}-1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\frac{2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)
=\(\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
=\(\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}.\left(\sqrt{a}-1\right)\)
\(=\frac{a-1}{\sqrt{a}}\)
vậy...
B
30 tháng 7 2019
\(a,\left(1+\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)\left(1-\frac{a+\sqrt{a}}{1+\sqrt{a}}\right)=\left(1+\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\left(1-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1^2-\sqrt{a}^2=1-a\)
\(b,\left(2-\frac{a-3\sqrt{a}}{\sqrt{a}-3}\right)\left(2-\frac{5\sqrt{a}-\sqrt{ab}}{\sqrt{b}-5}\right)=\left(2-\frac{\sqrt{a}\left(\sqrt{a}-3\right)}{\sqrt{a}-3}\right)\left(2-\frac{-\sqrt{a}\left(\sqrt{b}-5\right)}{\sqrt{b}-5}\right)\)
\(=\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)=2^2-\sqrt{a}^2=2-a\)
\(c,\left(3+\frac{a-2\sqrt{a}}{\sqrt{a}-2}\right)\left(3-\frac{3a+\sqrt{a}}{3\sqrt{a}+1}\right)=\left(3+\frac{\sqrt{a}\left(\sqrt{a}-2\right)}{\sqrt{a}-2}\right)\left(3-\frac{\sqrt{a}\left(3\sqrt{a}+1\right)}{3\sqrt{a}+1}\right)\)
\(=\left(3+\sqrt{a}\right)\left(3-\sqrt{a}\right)=3^2-\sqrt{a}^2=3-a\)
\(d,\left(\frac{a-\sqrt{a}}{\sqrt{a}-1}+2\right)\left(2-\frac{\sqrt{a}+a}{1+\sqrt{a}}\right)=\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}+2\right)\left(2-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\)
\(=\left(\sqrt{a}+2\right)\left(2-\sqrt{a}\right)=2^2-\sqrt{a}^2=2-a\)
NV
Nguyễn Việt Lâm
Giáo viên
9 tháng 10 2019
ĐKXĐ: \(y\ge0;y\ne4;9\)
\(A=\left(\frac{8\sqrt{y}-4y+8y}{\left(2+\sqrt{y}\right)\left(2-\sqrt{y}\right)}\right):\left(\frac{\sqrt{y}-1}{\sqrt{y}\left(\sqrt{y}-2\right)}-\frac{2\left(\sqrt{y}-2\right)}{\sqrt{y}\left(\sqrt{y}-2\right)}\right)\)
\(=\left(\frac{4\sqrt{y}\left(2+\sqrt{y}\right)}{\left(2+\sqrt{y}\right)\left(2-\sqrt{y}\right)}\right):\left(\frac{-\sqrt{y}+3}{\sqrt{y}\left(\sqrt{y}-2\right)}\right)\)
\(=\left(\frac{4\sqrt{y}}{2-\sqrt{y}}\right):\left(\frac{\sqrt{y}-3}{\sqrt{y}\left(2-\sqrt{y}\right)}\right)\)
\(=\frac{4\sqrt{y}}{\left(2-\sqrt{y}\right)}.\frac{\sqrt{y}\left(2-\sqrt{y}\right)}{\left(\sqrt{y}-3\right)}=\frac{4y}{\sqrt{y}-3}\)
\(A=-2\Leftrightarrow\frac{4y}{\sqrt{y}-3}=-2\)
\(\Rightarrow2y=-\sqrt{y}+3\Rightarrow2y+\sqrt{y}-3=0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{y}=1\\\sqrt{y}=-\frac{3}{2}< 0\left(l\right)\end{matrix}\right.\) \(\Rightarrow y=1\)
- ĐKXĐ : \(\left\{{}\begin{matrix}a\ge0\\\sqrt{a}-1\ne0\\\sqrt{a}+1\ne0\\2\sqrt{a}\ne0\end{matrix}\right.\)=> \(\left\{{}\begin{matrix}a\ne0\\a\ge0\\a\ne1\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}a>0\\a\ne1\end{matrix}\right.\)
- Ta có phương trình : \(\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)\left(\frac{a-\sqrt{a}}{\sqrt{a}+1}-\frac{a+\sqrt{a}}{\sqrt{a}-1}\right)\)
=\(\left(\frac{a}{2\sqrt{a}}-\frac{1}{2\sqrt{a}}\right)\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}+1}-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}-1}\right)\)
= \(\left(\frac{a-1}{2\sqrt{a}}\right)\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)^2}{a-1}-\frac{\sqrt{a}\left(\sqrt{a}+1\right)^2}{a-1}\right)\)
= \(\left(\frac{a-1}{2\sqrt{a}}\right)\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)^2-\sqrt{a}\left(\sqrt{a}+1\right)^2}{a-1}\right)\)
= \(\left(\frac{a-1}{2\sqrt{a}}\right)\left(\frac{\sqrt{a}\left(\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2\right)}{a-1}\right)\)
= \(\left(\frac{a-1}{2\sqrt{a}}\right)\left(\frac{\sqrt{a}\left(a-2\sqrt{a}+1-a-2\sqrt{a}-1\right)}{a-1}\right)\)
= \(\left(\frac{a-1}{2\sqrt{a}}\right)\left(\frac{\sqrt{a}\left(-4\sqrt{a}\right)}{a-1}\right)\)
= \(\left(\frac{a-1}{2\sqrt{a}}\right)\left(\frac{-4a}{a-1}\right)\)= \(\frac{-4a\left(a-1\right)}{2\sqrt{a}\left(a-1\right)}\) = \(\frac{-4a}{2\sqrt{a}}\)
= \(\frac{-4\sqrt{a}\sqrt{a}}{2\sqrt{a}}\) = \(-2\sqrt{a}\)