Question

C=(1-a^2):\([\)(\(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}\)+\(\sqrt{a}\))(\(\dfrac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\))\(]\)+1

a)Rút gọn C

b) tính C khi a =9

c)Tính a để /C/=C

Answer 1

a) điều kiện xác định : \(x\ge0;x\ne1\)

ta có : \(C=\left(1-a^2\right):\left[\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\right]+1\)

\(\Leftrightarrow C=\left(1-a^2\right):\left[\left(\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{\left(1+\sqrt{a}\right)\left(1-\sqrt{a}+a\right)}{1+\sqrt{a}}-\sqrt{a}\right)\right]+1\) \(\Leftrightarrow C=\left(1-a^2\right):\left[\left(1+\sqrt{a}+a+\sqrt{a}\right)\left(1-\sqrt{a}+a-\sqrt{a}\right)\right]+1\) \(\Leftrightarrow C=\left(1-a^2\right):\left[\left(1+\sqrt{a}\right)^2\left(1-\sqrt{a}\right)^2\right]+1\) \(\Leftrightarrow C=\dfrac{\left(1-a\right)\left(1+a\right)}{\left(1-a\right)^2}+1=\dfrac{1+a}{1-a}+1=\dfrac{1+a+1-a}{1-a}\) \(\Leftrightarrow C=\dfrac{2}{1-a}\) b) thế \(a=9\) vào \(C\) ta có : \(C=\dfrac{2}{1-9}=\dfrac{2}{-8}=\dfrac{-1}{4}\) c) để \(\left|C\right|=C\) thì \(C\ge0\Leftrightarrow\dfrac{2}{1-a}\ge0\Leftrightarrow1-a>0\Leftrightarrow a< 1\) vậy \(a< 1\)