Question

A = \(\left(\dfrac{1}{\sqrt{a}-3}+\dfrac{1}{\sqrt{a}+3}\right).\left(1+\dfrac{3}{\sqrt{a}}\right)\)

a/Rút gọn biểu thức A

b/Tìm a để A = \(\dfrac{1}{4}\)

c/Tìm a để A > \(\dfrac{1}{2}\)

Answer 1

`a)` Với `a > 0,a \ne 9` có:

`A=(1/[\sqrt{a}-3]+1/[\sqrt{a}+3])(1+3/\sqrt{a})`

`A=[\sqrt{a}+3+\sqrt{a}-3]/[(\sqrt{a}-3)(\sqrt{a}+3)].[\sqrt{a}+3]/\sqrt{a}`

`A=[2\sqrt{a}]/[\sqrt{a}(\sqrt{a}-3)]=2/[\sqrt{a}-3]`

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`b)` Với `a > 0,a \ne 9` có:

`A=1/4<=>2/[\sqrt{a}-3]=1/4`

          `<=>\sqrt{a}-3=8<=>\sqrt{a}=11<=>a=121` (t/m)

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`c)` Với `a > 0,a \ne 9` có:

`A > 1/2<=>2/[\sqrt{a}-3]-1/2 > 0`

            `<=>[4-\sqrt{a}+3]/[2(\sqrt{a}-3)] > 0`

            `<=>[7-\sqrt{a}]/[\sqrt{a}-3] > 0`

`@TH1:{(7-\sqrt{a} > 0),(\sqrt{a}-3 > 0):}<=>{(a < 49),(a > 9):}<=>9 < a < 49`

`@TH2:{(7-\sqrt{a} < 0),(\sqrt{a}-3 < 0):}<=>{(a > 49),(a < 9):}=>` Không có `a` t/m