Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(P=\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}}{x-1}\right):\left(\frac{2}{x}-\frac{2-x}{x\sqrt{x}+x}\right)\left(ĐK:x>0;x\ne1\right)\)
\(=\left[\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]:\left[\frac{2}{x}-\frac{2-x}{x\left(\sqrt{x}+1\right)}\right]\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\frac{2\left(\sqrt{x}+1\right)-2+x}{x\left(\sqrt{x}+1\right)}\)
\(=\frac{x+\sqrt{x}+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\frac{x\left(\sqrt{x}+1\right)}{2\sqrt{x}+2-2+x}\)
\(=\frac{x+2\sqrt{x}}{\sqrt{x}-1}\cdot\frac{x}{2\sqrt{x}+x}=\frac{x}{\sqrt{x}-1}\)
b)Để P>2
\(\Leftrightarrow\frac{x}{\sqrt{x}-1}>2\)
\(\Leftrightarrow\frac{x}{\sqrt{x}-1}-2>0\)
\(\Leftrightarrow\frac{x-2\sqrt{x}+2}{\sqrt{x}-1}>0\)
\(\Leftrightarrow\frac{\left(\sqrt{x}-1\right)^2+1}{\sqrt{x}-1}>0\)
\(\Leftrightarrow\sqrt{x}-1>0\Leftrightarrow x>1\left(tm\right)\)
Vậy x>1 thì P>2
a) đk: \(x\ge0;x\ne\left\{\frac{1}{4};1\right\}\)
\(P=\left(\frac{2x\sqrt{x}+x-\sqrt{x}}{x\sqrt{x}-1}-\frac{x+\sqrt{x}}{x-1}\right)\cdot\frac{x-1}{2x+\sqrt{x}-1}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)
\(P=\left[\frac{\left(2x+\sqrt{x}-1\right)\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}+1\right)\sqrt{x}}{x-1}\right]\cdot\frac{x-1}{2x+\sqrt{x}-1}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)
\(P=\frac{\left(x-1\right)\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}+1\right)\sqrt{x}}{2x+\sqrt{x}-1}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)
\(P=\frac{x+\sqrt{x}}{x+\sqrt{x}+1}-\frac{\sqrt{x}}{2\sqrt{x}-1}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)
\(P=\frac{x+\sqrt{x}}{x+\sqrt{x}+1}\)
b) Ta có:
\(P=\frac{x+\sqrt{x}}{x+\sqrt{x}+1}=\frac{\left(x+\sqrt{x}+1\right)-1}{x+\sqrt{x}+1}=1-\frac{1}{x+\sqrt{x}+1}\)
Mà \(x+\sqrt{x}\ge0\left(\forall x\right)\)
\(\Leftrightarrow x+\sqrt{x}+1\ge1\left(\forall x\right)\)
\(\Leftrightarrow\frac{1}{x+\sqrt{x}+1}\le1\left(\forall x\right)\)
\(\Leftrightarrow P=1-\frac{1}{x+\sqrt{x}+1}\ge0\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(x+\sqrt{x}=0\Leftrightarrow x=0\)
Vậy Min(P) = 0 khi x = 0
điều kiện \(\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)
a) A= (\(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{x-1}\)\(+\frac{\sqrt{x}}{x-1}\)) : \(\frac{2\left(\sqrt{x}+1\right)}{x\left(\sqrt{x}+1\right)}-\frac{2-x}{x\left(1+\sqrt{x}\right)}\))
=\(\frac{x+2\sqrt{x}}{x-1}:\frac{x+2\sqrt{x}}{x\left(1+\sqrt{x}\right)}\)=\(\frac{x\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{x}{\sqrt{x}-1}\)
b) A<1 <=> \(\frac{x}{\sqrt{x}-1}< 1< =>\frac{x-\sqrt{x}+1}{\sqrt{x}-1}< 0\)<=> \(\frac{\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}}{\sqrt{x}-1}< 0\)<=> \(\sqrt{x}-1< 0< =>x< 1\)kết hợp với điều kiện x>0 ta được 0<x<1
c) Min \(\sqrt{A}\)
Điều kiện A \(\ge0< =>\frac{x}{\sqrt{x}-1}\ge0< =>\hept{\begin{cases}x\ge0\\\sqrt{x}-1>0\end{cases}}< =>x>1;\)
(\(\sqrt{x}-2\))2 = x-4\(\sqrt{x}+4\)\(\ge0\)<=>x\(\ge4\left(\sqrt{x}-1\right)\) <=> \(\frac{x}{\sqrt{x}-1}\ge4\) (vì \(\sqrt{x}-1>0\))
hay A \(\ge4=>\sqrt{A}\ge2\)
\(\sqrt{A}=2\) khi \(\sqrt{x}-2=0< =>x=4\)
\(C=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{2x-\sqrt{x}}{\sqrt{x}}+\frac{2\left(x-1\right)}{\sqrt{x}-1}\) (tự tìm ĐKXĐ)
\(=\frac{\sqrt{x}\left(\sqrt{x}^3-1\right)}{x+\sqrt{x}+1}-\frac{\sqrt{x}\left(2\sqrt{x}-1\right)}{\sqrt{x}}+\frac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\left(2\sqrt{x}-1\right)+2\left(\sqrt{x}+1\right)\)
\(=\sqrt{x}\left(\sqrt{x}-1\right)-2\sqrt{x}+1+2\sqrt{x}+2\)
\(=x-\sqrt{x}+3\)
GTNN:\(x-\sqrt{x}+3=\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{11}{4}\ge\frac{11}{4}\)
\(\Rightarrow Min\left(C\right)=\frac{11}{4}khi..\sqrt{x}-\frac{1}{2}=0\Leftrightarrow\sqrt{x}=\frac{1}{2}\Leftrightarrow x=\frac{1}{4}\)