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Ta có: \(M=\dfrac{a\sqrt{a}-b\sqrt{b}}{a-b}-\dfrac{a}{\sqrt{a}+\sqrt{b}}+\dfrac{b}{\sqrt{a}-\sqrt{b}}\)
\(=\dfrac{a\sqrt{a}-b\sqrt{b}-a\sqrt{a}+a\sqrt{b}+b\sqrt{a}+b\sqrt{b}}{\left(\sqrt{a}+b\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\dfrac{\sqrt{ab}}{\sqrt{a}-\sqrt{b}}\)
a ĐKXĐ \(a\ge0,a\ne\dfrac{1}{4},a\ne1\)
\(\Rightarrow P=1+\left(\dfrac{\left(2\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}-\dfrac{\sqrt{a}\left(2\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}\right)\cdot\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\)
= \(1+\left(\dfrac{\left(-1\right)\left(2\sqrt{a}-1\right)}{\sqrt{a}-1}+\dfrac{\sqrt{a}\left(2\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\right)\cdot\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{2\sqrt{a}-1}\)
= \(1+\left(-1+\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{a+\sqrt{a}+1}\right)\sqrt{a}\)
= \(1-\sqrt{a}+\dfrac{a\sqrt{a}+a}{a+\sqrt{a}+1}\) = \(\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)+a\sqrt{a}+a}{a+\sqrt{a}+1}=\dfrac{1-a\sqrt{a}+a\sqrt{a}+a}{a+\sqrt{a}+1}=\dfrac{a+1}{a+\sqrt{a}+1}\)
b Xét hiệu \(P-\dfrac{2}{3}=\dfrac{a+1}{a+\sqrt{a}+1}-\dfrac{2}{3}=\dfrac{3a+3-2a-2\sqrt{a}-2}{a+\sqrt{a}+1}=\dfrac{a-2\sqrt{a}+1}{a+\sqrt{a}+1}=\dfrac{\left(\sqrt{a}-1\right)^2}{a+\sqrt{a}+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}>0\) \(\Rightarrow P>\dfrac{2}{3}\)
c Ta có \(P=\dfrac{\sqrt{6}}{\sqrt{6}+1}\Rightarrow\dfrac{a+1}{a+\sqrt{a}+1}=\dfrac{\sqrt{6}}{\sqrt{6}+1}\) \(\Rightarrow\left(a+1\right)\left(\sqrt{6}+1\right)=\sqrt{6}\left(a+\sqrt{a}+1\right)\Leftrightarrow a\sqrt{6}+a+\sqrt{6}+1=a\sqrt{6}+\sqrt{6a}+\sqrt{6}\)
\(\Leftrightarrow a-\sqrt{6a}+1=0\Leftrightarrow a-\sqrt{6a}+\dfrac{6}{4}-\dfrac{2}{4}=0\Leftrightarrow\left(\sqrt{a}-\dfrac{\sqrt{6}}{2}\right)^2=\dfrac{1}{2}\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{a}=\dfrac{\sqrt{6}+1}{2}\\\sqrt{a}=\dfrac{1-\sqrt{6}}{2}\left(L\right)\end{matrix}\right.\) (Do \(\sqrt{a}\ge0\)) \(\Rightarrow a=\dfrac{\left(\sqrt{6}+1\right)^2}{4}=\dfrac{7+2\sqrt{6}}{4}\left(TM\right)\)
Vậy...
a: \(\left(\dfrac{1}{a-\sqrt{a}}+\dfrac{1}{\sqrt{a}-1}\right):\dfrac{\sqrt{a}+1}{a-2\sqrt{a}+1}\)
\(=\dfrac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\)
\(=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
a) \(M=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
b) \(M=\dfrac{\sqrt{a}-1}{\sqrt{a}}=1-\dfrac{1}{\sqrt{a}}< 1\)
c) \(M=\dfrac{\sqrt{a}-1}{\sqrt{a}}=\dfrac{\sqrt{3-2\sqrt{2}}-1}{\sqrt{3-2\sqrt{2}}}=\dfrac{\sqrt{\left(\sqrt{2}-1\right)^2}-1}{\sqrt{\left(\sqrt{2}-1\right)^2}}=\dfrac{\sqrt{2}-1-1}{\sqrt{2}-1}=\dfrac{\sqrt{2}-2}{\sqrt{2}-1}\)
\(a,M=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ b,M=1-\dfrac{1}{\sqrt{a}}< 1\\ c,a=3-2\sqrt{2}\Leftrightarrow\sqrt{a}=\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{2}-1\\ \Leftrightarrow M=\dfrac{\sqrt{2}-1-1}{\sqrt{2}-1}=\dfrac{\sqrt{2}-2}{\sqrt{2}-1}=\dfrac{-\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=-\sqrt{2}\)
\(1.ĐKXĐ\left\{{}\begin{matrix}a>0\\a\ne1\end{matrix}\right.\)
\(M=\frac{a+1}{\sqrt{a}}+\frac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
\(\Leftrightarrow M=\frac{a+1}{\sqrt{a}}+\frac{a+\sqrt{a}+1}{\sqrt{a}}-\frac{a-\sqrt{a}+1}{\sqrt{a}}\)
\(\Leftrightarrow M=\frac{a+2\sqrt{a}+1}{\sqrt{a}}\)
2.
có \(M=\frac{a+2\sqrt{a}+1}{\sqrt{a}}\)
\(\Leftrightarrow M=4+\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}}\)
vì \(a>0\) và \(a\ne0\)
\(\Rightarrow\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}}>0\Leftrightarrow4+\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}}>4\)
Vậy M > 4
3.
M > 4\(\Leftrightarrow0\le N< \frac{3}{2}=1,5\)
Để N nhận giá trị nguyên
\(\Rightarrow N=1\)
\(\Rightarrow\frac{6}{M}=1\Leftrightarrow M=6\)
\(\Rightarrow\frac{a+2\sqrt{a}+1}{\sqrt{a}}=6\Leftrightarrow a+2\sqrt{a}+1=6\sqrt{a}\)
\(\Leftrightarrow a-4\sqrt{a}+1=0\)
Có \(\Delta`=2+1=3>0\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{a}=2+\sqrt{3}\left(tm\right)\\\sqrt{a}=2-\sqrt{3}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow a=7+4\sqrt{3}\)
Vậy \(a=7+4\sqrt{3}\) thì N nhận giá trị nguyên
