\(\dfrac{1+a}{1+\sqrt{a}+1}=\dfrac{\sqrt{6}}{1+\sqrt{6}}\\ \Leftrightarrow\left(1+\sqrt{6}\right)\left(1+a\right)=\sqrt{6}\left(1+\sqrt{a}+a\right)\\ \Leftrightarrow1+a+\sqrt{6}+\sqrt{6}a=\sqrt{6}+\sqrt{6}a+\sqrt{6}a\\ \Leftrightarrow1+a=\sqrt{6}a\\ \Leftrightarrow\sqrt{6}a=1+a\\ \Leftrightarrow6a=1+2a+a^2\\ \Leftrightarrow6a-1-2a-a^2=0\\ \Leftrightarrow4a-1-a^2=0\\ \Leftrightarrow-a^2+4a-1=0\\ \Leftrightarrow a^2-4a+1=0\)
\(a=\dfrac{-\left(-4\right)\pm\sqrt{\left(-4\right)^2-4.1.1}}{2.1}\\ a=\dfrac{4\pm\sqrt{16-4}}{2}\\ a=\dfrac{4\pm\sqrt{12}}{2}\\ a=\dfrac{4\pm2\sqrt{3}}{2}\)
\(\left[{}\begin{matrix}a=\dfrac{4+2\sqrt{3}}{2}=2+\sqrt{3}\\a=\dfrac{4-2\sqrt{3}}{2}=2-\sqrt{3}\end{matrix}\right.\)
Vậy......
Sr bạn cách làm mình hơi khó hiểu chút :v
a)ĐK: \(a\ge0\)
\(\Leftrightarrow\left(1+a\right)\left(1+\sqrt{6}\right)=\sqrt{6}\left(a+\sqrt{a}+1\right)\)
\(\Leftrightarrow1+\sqrt{6}+a+a\sqrt{6}=a\sqrt{6}+\sqrt{6a}+\sqrt{6}\)
\(\Leftrightarrow1+a=\sqrt{6a}\)
\(\Leftrightarrow a^2+2a+1-6a=0\)
\(\Leftrightarrow a^2-4a+1=0\)
\(\Leftrightarrow\left(a-2\right)^2-3=0\)
\(\Leftrightarrow\left(a-2\right)^2=3\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-2=\sqrt{3}\\a-2=-\sqrt{3}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=\sqrt{3}+2\\a=2-\sqrt{3}\end{matrix}\right.\)
b)
MK sẽ chứng minh tương đương:
\(\Leftrightarrow\dfrac{1+a}{1+\sqrt{a}+a}-\dfrac{2}{3}>0\)
\(\Leftrightarrow\dfrac{a-2\sqrt{a}+1}{1+\sqrt{a}+a}>0\)
\(\Leftrightarrow\dfrac{\left(\sqrt{a}-1\right)^2}{1+\sqrt{a}+a}>0\)
Ta có:
\(\left(\sqrt{a}-1\right)^2\ge0\left(1\right)\)
\(1+\sqrt{a}+a=\left(\sqrt{a}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Mà \(\left(\sqrt{a}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\Leftrightarrow1+\sqrt{a}+a>0\left(2\right)\)
Từ (1), (2)
=>\(\dfrac{\left(\sqrt{a}-1\right)^2}{1+\sqrt{a}+a}>0\)
=>\(\dfrac{1+a}{1+\sqrt{a}+a}>\dfrac{2}{3}\)
