Question

Cho P= \(\left(1+\dfrac{\sqrt{a}}{a+1}\right)\) : \(\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{2\sqrt{a}}{a\sqrt{a}+\sqrt{a}-a-1}\right)\)

với a\(\ge\)0 ; a \(\ne\)1

a, Rút gọn P

b, Tìm a để P<1

c,Tìm P khi a= \(19-8\sqrt{3}\)

Answer 1

a: \(P=\dfrac{a+\sqrt{a}+1}{a+1}:\dfrac{a+1-2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+1\right)}\)

\(=\dfrac{a+\sqrt{a}+1}{a+1}\cdot\dfrac{\left(a+1\right)}{\sqrt{a}-1}=\dfrac{a+\sqrt{a}+1}{\sqrt{a}-1}\)

b: P<1

=>P-1<0

=>\(\dfrac{a+\sqrt{a}+1-\sqrt{a}+1}{\sqrt{a}-1}< 0\)

=>\(\dfrac{a+2}{\sqrt{a}-1}< 0\)

=>căn a-1<0

=>0<=a<1

c: Khi a=19-8căn 3=(4-căn 3)^2 thì \(P=\dfrac{19-8\sqrt{3}+4-\sqrt{3}+1}{4-\sqrt{3}-1}=\dfrac{24-9\sqrt{3}}{3-\sqrt{3}}=\dfrac{15-\sqrt{3}}{2}\)