Question

Answer 1

`a)A=[x+\sqrt{x}+64]/\sqrt{x}`         `(x > 0)`

`<=>A=\sqrt{x}+1+64/\sqrt{x}`

Với `x > 0=>` Áp dụng BĐT cosi có: `\sqrt{x}+64/\sqrt{x} >= 2\sqrt{\sqrt{x}. 64/\sqrt{x}}=16`

     `<=>\sqrt{x}+64/\sqrt{x}+1 >= 17`

  Hay `A >= 17`

Dấu "`=`" xảy ra `<=>\sqrt{x}=64/\sqrt{x}<=>x=64` (t/m)

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`b)B=[x+8]/[\sqrt{x}-1]`     `(x > 1)`

`<=>B=[x-1+9]/[\sqrt{x}-1]`

`<=>B=[(\sqrt{x}-1)(\sqrt{x}+1)+9]/[\sqrt{x}-1]`

`<=>B=\sqrt{x}+1+9/[\sqrt{x}-1]`

`<=>B=\sqrt{x}-1+9/[\sqrt{x}-1]+2`

Với `x > 1=>` Áp dụng BĐT cosi có: `\sqrt{x}-1+9/[\sqrt{x}-1] >= 2\sqrt{9}=6`

       `<=>\sqrt{x}-1+9/[\sqrt{x}-1]+2 >= 8`

   Hay `B >= 8`

Dấu "`=`" xảy ra `<=>\sqrt{x}-1=9/[\sqrt{x}-1]<=>(\sqrt{x}-1)^2=9`

                               `<=>|\sqrt{x}-1|=3<=>[(\sqrt{x}-1=3),(\sqrt{x}-1=-3):}`

                                        `<=>[(\sqrt{x}=4<=>x=16),(\sqrt{x}=-2(VN)):}`

         `=>x=16` (t/m)