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\(đkxđ\Leftrightarrow\hept{\begin{cases}x\ge0\\\sqrt{x}-1\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ge0\\\sqrt{x}\ne1\end{cases}\Rightarrow}\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}}\)
\(M=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}-\frac{6\sqrt{x}-4}{x-1}.\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{x-1}+\frac{3\left(\sqrt{x}-1\right)}{x-1}-\frac{6\sqrt{x}-4}{x-1}\)
\(=\frac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
\(b,M< \frac{1}{2}\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}+1}< \frac{1}{2}\)
\(\Rightarrow\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{1}{2}< 0\)\(\Rightarrow\frac{2\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}+1}{2\left(\sqrt{x}+1\right)}< 0\)
\(\Rightarrow\frac{2\sqrt{x}-1-\sqrt{x}-1}{2\left(\sqrt{x}+1\right)}< 0\)\(\Rightarrow\frac{\sqrt{x}-2}{2\left(\sqrt{x}+1\right)}< 0\)
Vì \(2\left(\sqrt{x}+1\right)>0\Rightarrow\sqrt{x}-2>0\Rightarrow\sqrt{x}>2\)
\(\Rightarrow\sqrt{x}>\sqrt{4}\Leftrightarrow x>4\)
\(M=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}-\frac{6\sqrt{x}-4}{x-1}\left(x\ge0;x\ne1\right)\)
\(M=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{6\sqrt{x}-4}{x-1}\)
\(M=\frac{x+\sqrt{x}+3\sqrt{x}-3}{\left(\sqrt{x}\right)^2-1^2}-\frac{6\sqrt{x}-4}{x-1}\)
\(M=\frac{x-2\sqrt{x}+1}{x-1}\)
\(M=\frac{\left(\sqrt{x}-1\right)^2}{x-1}\)
\(1,A=\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\sqrt{x}-1}{x+\sqrt{x}+1}\)
2, Với x>1 ta có \(\frac{1}{A}=\frac{x+\sqrt{x}+1}{\sqrt{x}-1}=\frac{\sqrt{x}\left(\sqrt{x}-1\right)+2\left(\sqrt{x}-1\right)+3}{\sqrt{x}-1}\)
\(=\sqrt{x}-1+\frac{3}{\sqrt{x}-1}+3\)
Áp dụng bđt AM-GM ta có
\(\frac{1}{A}\ge2\sqrt{\left(\sqrt{x}-1\right).\frac{3}{\sqrt{x}-1}}+3=2\sqrt{3}+3\)
Dấu "=" xảy ra khi \(\left(\sqrt{x}-1\right)^2=3\Rightarrow\sqrt{x}=\pm\sqrt{3}+1\)
\(\Rightarrow x=\left(\pm\sqrt{3}+1\right)^2=4\pm2\sqrt{3}\)
@Mai.T.Loan câu a pha cuối hơi tắt đó nhìn khó hiểu lắm
còn câu b kl sai r nha
\(1,P=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{\sqrt{x}+1}{x-1}\)
\(=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{\left(x+\sqrt{x}+1\right)}-\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{\left(x+\sqrt{x}+1\right)}-\frac{1}{\sqrt{x}-1}\)
\(=\frac{x+2}{x\sqrt{x}-1}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{x+2+x-1-x-\sqrt{x}-1}{x\sqrt{x}-1}\)
\(=\frac{x-\sqrt{x}}{x\sqrt{x}-1}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
a) \(Q=\left(\frac{\sqrt{x}}{1-\sqrt{x}}+\frac{\sqrt{x}}{1+\sqrt{x}}\right)+\frac{3-\sqrt{x}}{x-1}\left(x\ge0;x\ne1\right)\)
\(=-\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{3-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{-\sqrt{x}\left(\sqrt{x}+1\right)+\sqrt{x}\left(\sqrt{x}-1\right)+3-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{-x-\sqrt{x}+x-\sqrt{x}+3-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{-3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=-\frac{3}{\sqrt{x}+1}\)
b) Để \(Q=-1\)
\(\Leftrightarrow-\frac{3}{\sqrt{x}+1}=-1\)
\(\Leftrightarrow\sqrt{x}+1=3\)
\(\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)
Bài 1:
a) \(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{2}-\frac{2\left(\sqrt{3}-1\right)}{2}\)
\(=\sqrt{3}+1-\left(\sqrt{3}-1\right)=2\)
b) \(\frac{2}{5-\sqrt{3}}+\frac{3}{\sqrt{6}+\sqrt{3}}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{\left(5-\sqrt{3}\right)\left(5+\sqrt{3}\right)}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{\left(\sqrt{6}+\sqrt{3}\right)\left(\sqrt{6}-\sqrt{3}\right)}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{2}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{3}\)
\(=5+\sqrt{3}+\sqrt{6}-\sqrt{3}=5+\sqrt{6}\)
c) ĐK: \(a\ge0;a\ne1\)
\(\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{1+\sqrt{a}}\right).\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)+a\)
\(=1-a+a=1\)
Ta có
\(1P=\left(\frac{2x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\frac{x\sqrt{x}-1}{\sqrt{x}+1}-\sqrt{x}\right)\)
\(=\frac{1}{\sqrt{x}-1}.\frac{x\sqrt{X}-x-\sqrt{x}-1}{\sqrt{x}+1}\)
\(=1\frac{x\sqrt{x}-x-\sqrt{x}-1}{x-1}\)
Ta có thao câu b thì 1 - x > 0
<=> x < 1
=> \(0\le x< 1\)
Ta có \(P\sqrt{1-x}=\frac{x\sqrt{x}-x-\sqrt{x}-1}{-\sqrt{1-x}}< 0\)
\(\Leftrightarrow x\sqrt{x}-x-\sqrt{x}-1>0\)
Ta thấy \(0\le x< 1\Rightarrow x\sqrt{x}< x+\sqrt{x}+1\)
Vậy không có giá trị nào của x để cái trên xảy ra