Question

giup minh cau nay voi a

 

Answer 1

`a)` Với `x >= 0,x \ne 1,x \ne 9` có:

`P=(\sqrt{x}-[x+3]/[\sqrt{x}+1]):(\sqrt{x}/[\sqrt{x}+1]-[\sqrt{x}-9]/[1-x])`

`P=[\sqrt{x}(\sqrt{x}+1)-x-3]/[\sqrt{x}+1]:[\sqrt{x}(\sqrt{x}-1)+\sqrt{x}-9]/[(\sqrt{x}+1)(\sqrt{x}-1)]`

`P=[x+\sqrt{x}-x-3]/[\sqrt{x}+1].[(\sqrt{x}-1)(\sqrt{x}+1)]/[x-\sqrt{x}+\sqrt{x}-9]`

`P=[(\sqrt{x}-3)(\sqrt{x}-1)]/[(\sqrt{x}-3)(\sqrt{x}+3)]`

`P=[\sqrt{x}-1]/[\sqrt{x}+3]`

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`b)` Với `x >= 0,x \ne 1,x \ne 9` có:

`P <= 1/3<=>[\sqrt{x}-1]/[\sqrt{x}+3]-1/3 <= 0`

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

               Mà `3(\sqrt{x}+3) > 0 AA x >= 0`

   `=>2\sqrt{x}-6 <= 0`

 `<=>\sqrt{x} <= 3`

 `<=>x <= 9`

  Mà `x >= 0,x \ne 1,x \ne 9`

  `=>0 <= x < 9,x \ne 1`

`=>x in ZZ` lớn nhất là: `8`

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`c)` Với `x >= 0,x \ne 1,x \ne 9` có:

`P=[\sqrt{x}-1]/[\sqrt{x}+3]=[\sqrt{x}+3-4]/[\sqrt{x}+3]=1-4/[\sqrt{x}+3]`

 Vì `x >= 0<=>\sqrt{x} >= 0<=>\sqrt{x}+3 >= 3`

                          `<=>-4/[\sqrt{x}+3] >= -4/3`

                          `<=>P >= -1/3`

Dấu "`=`" xảy ra `<=>\sqrt{x}=0<=>x=0`