Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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, \(P=\left(\frac{\sqrt{x}}{x\sqrt{x}-1}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
\(=\left(\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right):\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
\(=\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{x+\sqrt{x}+1}{\sqrt{x}+1}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
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: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)
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: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
a) ĐKXĐ: \(x\ge0;x\ne1\)
\(A=\left(1+\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right).\left(1-\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right)=\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x\)
b) \(A=1-x\le1\) ( vì \(x\ge0\) )
Vậy max A = 1 khi x = 0