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a) Đk: a>=0, \(a\ne1\)
\(A=\frac{a}{\sqrt{a}-1}-\frac{\sqrt{a}\left(2\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}\)
\(=\frac{a}{\sqrt{a}-1}-\frac{2\sqrt{a}-1}{\sqrt{a}-1}\)
\(=\frac{a-2\sqrt{a}+1}{\sqrt{a}-1}=\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}-1}=\sqrt{a}-1\)
b) Ta có : \(a=3+\sqrt{8}=3+2\sqrt{2}=\left(1+\sqrt{2}\right)^2\)
\(A=\sqrt{a}-1=\sqrt{\left(1+\sqrt{2}\right)^2}-1=1+\sqrt{2}-1=\sqrt{2}\)
c) \(A=\sqrt{a}-1>0\Leftrightarrow\sqrt{a}>1\Leftrightarrow a>1\)
\(A=\sqrt{a}-1=0\Leftrightarrow\sqrt{a}=1\Leftrightarrow a=1\)(loại vì a khác 1 theo điều kiện )
a: \(B=\left(\dfrac{2a+\sqrt{a}-1}{1-a}-\dfrac{2a\sqrt{a}+a-\sqrt{a}}{1-a\sqrt{a}}\right)\)
\(=\dfrac{\left(\sqrt{a}+1\right)\left(2\sqrt{a}-1\right)}{\left(\sqrt{a}+1\right)\left(1-\sqrt{a}\right)}-\dfrac{\sqrt{a}\left(2a+\sqrt{a}-1\right)}{\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}\)
\(=\dfrac{2a\sqrt{a}+2a+2\sqrt{a}-a-\sqrt{a}-1-2a\sqrt{a}-a+\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}\)
\(=\dfrac{2\sqrt{a}-1}{\left(1-\sqrt{a}\right)\left(1+a+\sqrt{a}\right)}\)
\(A=1-\dfrac{2\sqrt{a}-1}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\cdot\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\)
\(=1-\dfrac{\sqrt{a}}{a+\sqrt{a}+1}=\dfrac{a+1}{a+\sqrt{a}+1}\)
c: \(A-\dfrac{2}{3}=\dfrac{a+1}{a+\sqrt{a}+1}-\dfrac{2}{3}\)
\(=\dfrac{3a+3-2a-2\sqrt{a}-2}{3\left(a+\sqrt{a}+1\right)}=\dfrac{a-2\sqrt{a}+1}{3\left(a+\sqrt{a}+1\right)}>0\)
=>A>2/3
\(A=1+\left(\dfrac{2a+\sqrt{a}-1}{1-a}-\dfrac{2a\sqrt{a}-\sqrt{a}+a}{1-a\sqrt{a}}\right).\dfrac{a-\sqrt{a}}{2\sqrt{a}-1}\\ =1+\left(\dfrac{2a+2\sqrt{a}-\sqrt{a}-1}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}-\dfrac{2a\sqrt{a}-\sqrt{a}+a}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}\right).\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\)
\(=1+\dfrac{2\sqrt{a}-1+2a+2a\sqrt{a}-a-2a\sqrt{a}+\sqrt{a}-a}{-\left(\sqrt{a}-1\right)\left(1+\sqrt{a}+a\right)}\)
\(=1+\dfrac{2\sqrt{a}-1+0}{1+\sqrt{a}+a}.\dfrac{\sqrt{a}\left(-1\right)}{2\sqrt{a}-1}\\ =1+\dfrac{1}{1+\sqrt{a}+a}.\sqrt{a}.\left(-1\right)\)
\(=1-\dfrac{\sqrt{a}}{1+\sqrt{a}+a}\\ =\dfrac{1+\sqrt{a}+a-\sqrt{a}}{1+\sqrt{a}+a}\\ =\dfrac{1+a}{1+\sqrt{a}+a}\)
a)
\(A=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\) (ĐKXĐ \(a>0\))
\(\Leftrightarrow A=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-2\sqrt{a}\)
\(\Leftrightarrow a+\sqrt{a}-2\sqrt{a}=a-\sqrt{a}\) (Với \(a>0\))
b)
Để A = 2 \(\Rightarrow a-\sqrt{a}=2\)
\(\Leftrightarrow\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)=0\)
\(\Leftrightarrow a=4\left(tm\right)\)
Vậy a = 4 thì A = 2 .
c)
\(A=a-\sqrt{a}=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\) Với \(\forall a>0\)
Vậy GTNN của A là \(-\dfrac{1}{4}\) khi a = \(\dfrac{1}{4}\) .
a: \(P=a+\sqrt{a}-2\sqrt{a}-1+1=a-\sqrt{a}\)
b: Để A=2 thì a-căn a-2=0
=>a=4
c: \(P=a-\sqrt{a}+\dfrac{1}{4}-\dfrac{1}{4}=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}>=-\dfrac{1}{4}\)
Dấu = xảy ra khi a=1/4