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2 tháng 9 2018
a,
\(A\Leftrightarrow\)\(\left(\frac{1}{\sqrt{x}-1}-\frac{\sqrt{x}+1}{\left(\sqrt{x}\right)^2+2\sqrt{x}+1}\right)\)\(\times\frac{x-1}{\sqrt{x}-3}\)
\(\Leftrightarrow\left(\frac{1}{\sqrt{x}-1}-\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)^2}\right)\)\(\times\frac{x-1}{\sqrt{x}-3}\)(1)
Để A xđ <=> \(\hept{\begin{cases}x\ge0\\\sqrt{x}-1\ne0\\\sqrt{x}-3\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge0\\x\ne1\\x\ne9\end{cases}}\)
b , (1) <=> \(\left(\frac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right)\)\(\times\frac{x-1}{\sqrt{x}-3}\)
<=> \(\left(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1-\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right)\)\(\times\frac{x-1}{\sqrt{x}-3}\)
<=> \(\frac{2}{x-1}\times\frac{x-1}{\sqrt{x}-3}\)
<=> \(\frac{2}{\sqrt{x}-3}\)
9 tháng 8 2020
Mình nghĩ đề câu a) là \(\frac{1}{1-\sqrt{x^2-3}}\) khi đó
\(1-\sqrt{x^2-3}\ne0\Rightarrow\sqrt{x^2-3}\ne1\Rightarrow x\ne\pm2\)và \(x^2-3\ge0\Leftrightarrow-\sqrt{3}\le x\le\sqrt{3}\)
b)
\(\sqrt{16-x^2}\ge0;\sqrt{2x+1}\ge0;\sqrt{x^2-8x+14}\ge0\)và \(\sqrt{2x+1}\ne0\)
\(\Leftrightarrow-4\le x\le4;x\ge-\frac{1}{2};4-\sqrt{2}\le x\le4+\sqrt{2};x\ne\frac{1}{2}\)
Như vậy \(-\frac{1}{2}< x\le4+\sqrt{2}\)
18 tháng 10 2020
a) ĐK: \(x\ge1\)
\(\sqrt{x}-\sqrt{x+1}+\frac{1}{\sqrt{x-1}-\sqrt{x}}+\frac{\sqrt{x^3-x}}{\sqrt{x-1}}\)
\(=\sqrt{x}-\sqrt{x-1}+\frac{\sqrt{x-1}+\sqrt{x}}{x-1-x}+\frac{x\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(=\sqrt{x}-\sqrt{x-1}-\sqrt{x-1}-\sqrt{x}+x\)
\(=x-2\sqrt{x-1}\)
\(=\left(x-1\right)-2\sqrt{x-1}+1\)'
\(=\left(\sqrt{x-1}-1\right)^2\)
b) \(P=1\Leftrightarrow\left(\sqrt{x-1}-1\right)^2=1\)\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x-1}-1=1\\\sqrt{x-1}-1=-1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=1\end{cases}}\)
Vậy x=5,x=1
15 tháng 12 2018
P xác định khi \(\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)
\(P=\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\frac{1}{\sqrt{x}+1}+\frac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\left(\frac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\frac{\sqrt{x}-1+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\left(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}}:\frac{1}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}}.\left(\sqrt{x}-1\right)\)
\(=\frac{x-1}{\sqrt{x}}\)
15 tháng 12 2018
P xác định khi \(\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)
\(P=\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\frac{1}{\sqrt{x}+1}+\frac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\left(\frac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\frac{\sqrt{x}-1+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\left(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}}:\frac{1}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}}.\left(\sqrt{x}-1\right)\)
\(=\frac{x-1}{\sqrt{x}}\)
3 tháng 2 2022
a: \(P=\dfrac{x+\sqrt{x}+1+11\sqrt{x}-11+34}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{x+\sqrt{x}+1-x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{x+12\sqrt{x}+24}{\sqrt{x}+2}\)
b: Thay \(x=3-2\sqrt{2}\) vào P, ta được:
\(P=\dfrac{3-2\sqrt{2}+12\left(\sqrt{2}-1\right)+24}{\sqrt{2}-1+2}\)
\(=\dfrac{27-2\sqrt{2}+12\sqrt{2}-12}{\sqrt{2}+1}=5+5\sqrt{2}\)