cho Q= ($\frac{1}{√a-1}$ – $\frac{1}{\sqrt{a}}$) : ( $\frac{\sqrt{a}+1}{\sqrt{a}-2}$ – $\frac{\sqrt{a}+2}{\sqrt{a}-1}$)
a>0, a∦4, a∦1
a) Rút gọn.
b) Tìm a để Q>0.
cho Q= ($\frac{1}{√a-1}$ – $\frac{1}{\sqrt{a}}$) : ( $\frac{\sqrt{a}+1}{\sqrt{a}-2}$ – $\frac{\sqrt{a}+2}{\sqrt{a}-1}$)
a>0, a∦4, a∦1
a) Rút gọn.
b) Tìm a để Q>0.
Đáp án:
Giải thích các bước giải:
$\eqalign{
& Q = ({{\sqrt a } \over {(\sqrt a – 1)\sqrt a }} – {{\sqrt a – 1} \over {(\sqrt a – 1)\sqrt a }}) – ({{a – 1} \over {(\sqrt a – 2)(\sqrt a – 1)}} – {{a – 4} \over {(\sqrt a – 2)(\sqrt a – 1)}}) \cr
& = {1 \over {(\sqrt a – 1)\sqrt a }} – {3 \over {(\sqrt a – 2)(\sqrt a – 1)}} \cr
& = {{(\sqrt a – 2) – 3\sqrt a } \over {(\sqrt a – 1)(\sqrt a – 2)\sqrt a }} = {{ – 2(\sqrt a + 1)} \over {(\sqrt a – 1)(\sqrt a – 2)\sqrt a }} \cr} $
b,
Để Q>0 thì ${(\sqrt a – 1)(\sqrt a – 2)\sqrt a }$<0 vì ${ - 2(\sqrt a + 1)}$ luoon bé hơn 0
=> $\eqalign{
& (\sqrt a – 1)(\sqrt a – 2) < 0 \cr & 1 < \sqrt a < 2 \cr} $ =>1