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a) ĐK: x > 1
\(P=\left(\frac{\sqrt{x-1}}{3+\sqrt{x-1}}+\frac{x+8}{9-\left(x-1\right)}\right):\left(\frac{3\sqrt{x-1}+1}{\left(x-1\right)-3\sqrt{x-1}}-\frac{1}{\sqrt{x-1}}\right)\)
\(P=\frac{\sqrt{x-1}\left(3-\sqrt{x-1}\right)+x+8}{9-\left(x-1\right)}:\frac{3\sqrt{x-1}+1-\left(\sqrt{x-1}-3\right)}{\sqrt{x-1}\left(\sqrt{x-1}-3\right)}\)
\(P=\frac{3\sqrt{x-1}-x+1+x+8}{10-x}:\frac{2\sqrt{x-1}+4}{\sqrt{x-1}\left(\sqrt{x-1}-3\right)}\)
\(P=\frac{3\left(\sqrt{x-1}+3\right)}{10-x}.\frac{\sqrt{x-1}\left(\sqrt{x-1}-3\right)}{2\sqrt{x-1}+4}\)
\(P=\frac{-3\sqrt{x-1}}{2\sqrt{x-1}+4}\)
b) \(x=\sqrt[4]{\frac{17+12\sqrt{2}}{1}}-\sqrt[4]{\frac{17-12\sqrt{2}}{1}}=1+\sqrt{2}-\left(\sqrt{2}-1\right)=2\)
Vậy \(P=\frac{-3\sqrt{2-1}}{2\sqrt{2-1}+4}=-\frac{1}{2}\)
cô Hoàng Thị Thu Huyền làm rõ cho em ý b đc ko ạ chỗ biến đổi x
\(A=\left(\frac{x-1}{\sqrt{x}-1}+\frac{x+2\sqrt{x}+1}{\sqrt{x}+1}\right).\frac{1}{2\sqrt{x}}=\left[\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x-1}}+\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}+1}\right].\frac{1}{2\sqrt{x}}\)
\(A=2\left(\sqrt{x}+1\right).\frac{1}{2\sqrt{x}}=\frac{\sqrt{x}+1}{\sqrt{x}}>1=\sqrt{\frac{2019}{2019}}>\sqrt{\frac{2018}{2019}}\) ( đpcm )
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a) Đkxđ : \(\left\{{}\begin{matrix}a\ge0\\a\ne9\end{matrix}\right.\)
A = \(\left(\frac{\sqrt{a}+3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}+\frac{\sqrt{a}-3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\right)\left(1-\frac{3}{\sqrt{a}}\right)\)
= \(\frac{2\sqrt{a}}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}.\frac{\sqrt{a}-3}{\sqrt{a}}\)
= \(\frac{2}{\sqrt{a}+3}\)
b) Để A > \(\frac{1}{2}\)
<=> \(\frac{2}{\sqrt{a}+3}>\frac{1}{2}\Leftrightarrow\frac{2}{\sqrt{a}+3}-\frac{1}{2}>0\)
<=> \(4-\sqrt{a}-3>0\Leftrightarrow1-\sqrt{a}>0\Leftrightarrow a< 1\)
Vậy để A >1/2 thì a <1