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Giúp mk nhanh nha.mk cần gấp...........ai nhanh mak đúng mk k cho
a, \(P=\left(\frac{\sqrt{x}}{x\sqrt{x}-1}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
\(=\left(\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right):\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
\(=\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{x+\sqrt{x}+1}{\sqrt{x}+1}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
b, Vì x > 1, g/s : Thay x = 4 vào P ta được :
\(\frac{\sqrt{4}+1}{\sqrt{4}-1}=\frac{3}{1}=3\)
Thay x = 4 vào căn P ta được : \(\sqrt{\frac{\sqrt{4}+1}{\sqrt{4}-1}}=\sqrt{3}\)
mà \(3>\sqrt{3}\Rightarrow P>\sqrt{P}\)với x > 1
a. ĐKXĐ: \(x>0,x\ne1\)
A=Đề\(=\left[\frac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]:\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)
\(=\frac{-1}{\sqrt{x}}\cdot\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)}\)\(=\frac{-\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
Đề sai hả bạn ?
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
a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)
b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)
c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)