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a, Với \(x\ge0;x\ne1\)
\(Q=\left(\frac{x-1}{\sqrt{x}-1}-\frac{x\sqrt{x}-1}{x-1}\right):\left(\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}+\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\left(\sqrt{x}+1-\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x-1}\right):\left(\frac{x-\sqrt{x}+1}{\sqrt{x}+1}\right)\)
\(=\left(\sqrt{x}+1-\frac{x+\sqrt{x}+1}{\sqrt{x}+1}\right):\left(\frac{x-\sqrt{x}+1}{\sqrt{x}+1}\right)\)
\(=\left(\frac{x+2\sqrt{x}+1-x-\sqrt{x}-1}{\sqrt{x}+1}\right):\left(\frac{x-\sqrt{x}+1}{\sqrt{x}+1}\right)\)
\(=\frac{\sqrt{x}}{x-\sqrt{x}+1}\)
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
\(P=\dfrac{x\sqrt{x}-x-\sqrt{x}-2}{\left(x-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{\left(1-x^2\right)^2}{2}\)
\(P=\dfrac{\left(\sqrt{x}-2\right)\left(x-1\right)}{\left(x+\sqrt{x}+1\right)}.\dfrac{\left(1-x^2\right)\left(x-1\right)}{2}\)
\(P=\dfrac{\left(\sqrt{x}-2\right)\left(x-1\right)\left(1-x^2\right)}{2\left(x+\sqrt{x}+1\right)}\)
\(đkxđ\Leftrightarrow\hept{\begin{cases}x\ge0\\\sqrt{x}-1\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ge0\\\sqrt{x}\ne1\end{cases}\Rightarrow}\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}}\)
\(M=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}-\frac{6\sqrt{x}-4}{x-1}.\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{x-1}+\frac{3\left(\sqrt{x}-1\right)}{x-1}-\frac{6\sqrt{x}-4}{x-1}\)
\(=\frac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
\(b,M< \frac{1}{2}\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}+1}< \frac{1}{2}\)
\(\Rightarrow\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{1}{2}< 0\)\(\Rightarrow\frac{2\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}+1}{2\left(\sqrt{x}+1\right)}< 0\)
\(\Rightarrow\frac{2\sqrt{x}-1-\sqrt{x}-1}{2\left(\sqrt{x}+1\right)}< 0\)\(\Rightarrow\frac{\sqrt{x}-2}{2\left(\sqrt{x}+1\right)}< 0\)
Vì \(2\left(\sqrt{x}+1\right)>0\Rightarrow\sqrt{x}-2>0\Rightarrow\sqrt{x}>2\)
\(\Rightarrow\sqrt{x}>\sqrt{4}\Leftrightarrow x>4\)
\(M=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}-\frac{6\sqrt{x}-4}{x-1}\left(x\ge0;x\ne1\right)\)
\(M=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{6\sqrt{x}-4}{x-1}\)
\(M=\frac{x+\sqrt{x}+3\sqrt{x}-3}{\left(\sqrt{x}\right)^2-1^2}-\frac{6\sqrt{x}-4}{x-1}\)
\(M=\frac{x-2\sqrt{x}+1}{x-1}\)
\(M=\frac{\left(\sqrt{x}-1\right)^2}{x-1}\)
\(\(b)\frac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}\left(a,b\ge0;a,b\ne1\right)\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\left(a\sqrt{b}-b\sqrt{a}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab+1}\right)}\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)
\(\(=\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{ab}-1\right)}\left(a,b\ge0.a,b\ne1\right)\)\)
_Minh ngụy_
\(\(c)\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)\)( tự ghi điều kiện )
\(\(=\frac{x\sqrt{x}+y\sqrt{y}-\left(\sqrt{x}-\sqrt{y}\right)^2.\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)\)
\(\(=\frac{x\sqrt{x}+y\sqrt{y}-\left(x\sqrt{x}+x\sqrt{y}-2x\sqrt{y}-2y\sqrt{x}+y\sqrt{x}+y\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)\)
\(\(=\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)\)( phá ngoặc và tính )
\(\(=\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}=\sqrt{xy}\)\)
_Minh ngụy_
\(A=\frac{\left(\sqrt{x}+1\right)^2+\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\frac{3\sqrt{x}+1}{x-1}\)
\(A=\frac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{2x-3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(A=\frac{2\sqrt{x}-1}{\sqrt{x}+1}\)