Question

\(Q=\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}:\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)Cho biểu thức

a) rút gọn Q với a > 0 ; a khác 4 ; a khác 1

b) tính giá trị của a để Q dương

Answer 1

\(Q=\left(\frac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)

\(=\frac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\left(\frac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)

\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}=\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3\sqrt{a}\left(\sqrt{a}-1\right)}=\frac{\sqrt{a}-2}{3\sqrt{a}}\)

Để \(Q>0\Leftrightarrow\frac{\sqrt{a}-2}{3\sqrt{a}}>0\Rightarrow\sqrt{a}-2>0\Rightarrow a>4\)

Answer 2

\(Q=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)

\(=\left(\frac{\sqrt{a}-\sqrt{a}+1}{\left(\sqrt{a}-1\right).\sqrt{a}}\right):\left(\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\right)}{a-4}-\frac{\left(\sqrt{a}+2\right)\left(\sqrt{a}+1\right)}{a-1}\right)\)

\(=\left(\frac{\sqrt{a}-\sqrt{a}+1}{\left(\sqrt{a}-1\right).\sqrt{a}}\right):\left(\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\right)\left(a-1\right)-\left(\sqrt{a}+2\right)\left(\sqrt{a}+1\right)\left(a-4\right)}{\left(a-4\right)\left(a-1\right)}\right)\)

\(=\frac{1}{\left(\sqrt{a}-1\right).\sqrt{a}}:\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\right)\left(a-1-a+4\right)}{\left(a-4\right)\left(a-1\right)}\)

\(=\frac{1}{\left(\sqrt{a}-1\right).\sqrt{a}}:\frac{3\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\right)}{\left(a-4\right)\left(a-1\right)}\)

\(=\frac{1}{\left(\sqrt{a}-1\right).\sqrt{a}}.\frac{\left(a-4\right)\left(a-1\right)}{3\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\right)}\)

\(=\frac{a-4}{3\sqrt{a}\left(\sqrt{a}+2\right)}\) \(=\frac{a-4}{3a+6\sqrt{a}}\)

Để Q dương thì :

1) a - 4 \(\ge\) 0 và 3a+6\(\sqrt{a}\) \(\ge\) 0

<=> a \(\ge\) 4 và a \(\ge\) 0

<=> a > 4 (do a khác 4)

2) a - 4 \(\le\) 0 và 3a+6\(\sqrt{a}\) \(\le\) 0

<=> a \(\le\) 4 và a \(\le\) -2

<=> a \(\le\) -2 (ko thóa mãn)

Vậy: Q dương khi a > 4