Sửa đề: \(H=\left(\frac{2a+\sqrt{a}-1}{1-a}-\frac{2a\sqrt{a}-\sqrt{a}+a}{1-a\cdot\sqrt{a}}\right)\cdot\frac{a-\sqrt{a}}{1-2\sqrt{a}}\)
ĐKXĐ: a>=0; a<>1; a<>1/4
a: Ta có: \(\frac{2a+\sqrt{a}-1}{1-a}-\frac{2a\sqrt{a}-\sqrt{a}+a}{1-a\cdot\sqrt{a}}\)
\(=\frac{\left(2\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}+\frac{\sqrt{a}\left(2a+\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\)
\(=\frac{2\sqrt{a}-1}{1-\sqrt{a}}+\frac{\sqrt{a}\left(2\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\)
\(=-\left(2\sqrt{a}-1\right)\left(\frac{1}{\sqrt{a}-1}-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\left(a+\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)
\(=-\left(2\sqrt{a}-1\right)\cdot\frac{a+\sqrt{a}+1-a-\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}=\frac{-\left(2\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}_{}+1\right)}\)
Ta có: \(H=\left(\frac{2a+\sqrt{a}-1}{1-a}-\frac{2a\sqrt{a}-\sqrt{a}+a}{1-a\cdot\sqrt{a}}\right)\cdot\frac{a-\sqrt{a}}{1-2\sqrt{a}}\)
\(=\frac{-\left(2\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\cdot\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{1-2\sqrt{a}}=\frac{\sqrt{a}}{a+\sqrt{a}+1}\)
b: \(H=\frac27\)
=>\(\frac{\sqrt{a}}{a+\sqrt{a}+1}=\frac27\)
=>\(2a+2\sqrt{a}+2=7\sqrt{a}\)
=>\(2a-5\sqrt{a}+2=0\)
=>\(2a-4\sqrt{a}-\sqrt{a}+2=0\)
=>\(\left(\sqrt{a}-2\right)\left(2\sqrt{a}-1\right)=0\)
=>\(\left[\begin{array}{l}\sqrt{a}-2=0\\ 2\sqrt{a}-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}a=4\left(nhận\right)\\ a=\frac14\left(loại\right)\end{array}\right.\)
c: \(H-\frac13=\frac{\sqrt{a}}{a+\sqrt{a}+1}-\frac13=\frac{3\sqrt{a}-a-\sqrt{a}-1}{3\left(a+\sqrt{a}+1\right)}\)
\(=\frac{-a+2\sqrt{a}-1}{3\left(a+\sqrt{a}+1\right)}=\frac{-\left(\sqrt{a}-1\right)^2}{3\left(a+\sqrt{a}+1\right)}<0\forall a\) thỏa mãn ĐKXĐ
=>\(H<\frac13\)
