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a/ đkxđ \(\hept{\begin{cases}\sqrt{1+x}-\sqrt{1-x}\ne0\\\sqrt{1-x^2}-1+x\ne0\\x\ne0\end{cases}}va\hept{\begin{cases}1+x>0\\1-x>0\\1-x^2>0\end{cases}va}\sqrt{\frac{1}{x^2}-1}>0\)
\(\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne1\\-1< x< 1\end{cases}}vax>0\)
b =/\(\left[\frac{\sqrt{1+x}}{\sqrt{1+x}-\sqrt{1-x}}+\frac{1-x}{\sqrt{1-x^2}-1+x}\right].\left[\frac{\sqrt{1-x^2}}{x}-\frac{1}{x}\right]\)=
\(\left[\frac{\sqrt{1+x}}{\sqrt{1+x}-\sqrt{1-x}}+\frac{1-x}{\sqrt{1-x}\left[\sqrt{1+x}-\sqrt{1-x}\right]}\right].\frac{\sqrt{1-x^2}-1}{x}\)=\(\left[\frac{\sqrt{1+x}}{\sqrt{1+x}-\sqrt{1-x}}+\frac{\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}\right].\frac{\sqrt{1-x^2}-1}{x}\)=\(\frac{\left[\sqrt{1+x}+\sqrt{1-x}\right]\left[\sqrt{1-x^2}-1\right]}{\left[\sqrt{1+x}-\sqrt{1-x}\right].x}\)
c/ khi x=1/2 thi A=\(\frac{\left[\sqrt{1+\frac{1}{2}}+\sqrt{1-\frac{1}{2}}\right]\left[\sqrt{1-\frac{1}{4}}-1\right]}{\left[\sqrt{1+\frac{1}{2}}-\sqrt{1-\frac{1}{2}}\right].\frac{1}{2}}=-1\)
a/ đkxđ
| √1+x−√1−x≠0 |
| √1−x2−1+x≠0 |
| x≠0 |
va{
| 1+x>0 |
| 1−x>0 |
| 1−x2>0 |
va√1x2 −1>0
| x≠0 |
| x≠1 |
| −1<x<1 |
vax>0
b =/[√1+x√1+x−√1−x +1−x√1−x2−1+x ].[√1−x2x −1x ]=
[√1+x√1+x−√1−x +1−x√1−x[√1+x−√1−x] ].√1−x2−1x =[√1+x√1+x−√1−x +√1−x√1+x−√1−x ].√1−x2−1x =[√1+x+√1−x][√1−x2−1][√1+x−√1−x].x
c/ khi x=1/2 thi A=[√1+12 +√1−12 ][√1−14 −1][√1+12 −√1−12 ].12 =−1
1/
a/ ĐKXĐ: \(x\ge0\) và \(x\ne\frac{1}{9}\)
b/ \(P=\left[\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}-1\right)+8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\right]:\left(\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right)\)
\(=\frac{3x-2\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}.\frac{3\sqrt{x}+1}{3}\)
\(=\frac{3x+3\sqrt{x}}{3\sqrt{x}-1}.\frac{1}{3}=\frac{x+\sqrt{x}}{3\sqrt{x}-1}\)
c/ \(P=\frac{6}{5}\Rightarrow\frac{x+\sqrt{x}}{3\sqrt{x}-1}=\frac{6}{5}\Rightarrow6\left(3\sqrt{x}-1\right)=5\left(x+\sqrt{x}\right)\)
\(\Rightarrow5x-13\sqrt{x}+6=0\Rightarrow\left(5\sqrt{x}-3\right)\left(\sqrt{x}-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{x}=\frac{3}{5}\\\sqrt{x}=2\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{9}{25}\\x=4\end{cases}}}\)
Vậy x = 9/25 , x = 4
1) a) ĐKXĐ : \(0\le x\ne\frac{1}{9}\)
b) \(P=\left(\frac{\sqrt{x}-1}{3\sqrt{x}-1}-\frac{1}{3\sqrt{x}+1}+\frac{8\sqrt{x}}{9x-1}\right):\left(1-\frac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)\)
\(=\left[\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}-\frac{3\sqrt{x}-1}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}+\frac{8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right]:\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\)
\(=\frac{3x-2\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\frac{3\sqrt{x}+1}{3}=\frac{3x+3\sqrt{x}}{3\left(3\sqrt{x}-1\right)}=\frac{x+\sqrt{x}}{3\sqrt{x}-1}\)
c) \(P=\frac{6}{5}\Leftrightarrow18\sqrt{x}-6=5x+5\sqrt{x}\Leftrightarrow5x-13\sqrt{x}+6=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{9}{25}\\x=4\end{cases}}\)