Đáp án:

`a)` Với `x=16` t/m đk `x >= 0`

Thay `x=16` vào `A` có: `A=[\sqrt{16}-1]/[\sqrt{16}+1]=3/5`

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`b)` Với `x >= 0` có:

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

`B=[x+2]/[(\sqrt{x}+1)(x-\sqrt{x}+1)]+\sqrt{x}/[x-\sqrt{x}+1]-1/[\sqrt{x}+1]`

`B=[x+2+\sqrt{x}(\sqrt{x}+1)-(x-\sqrt{x}+1)]/[(\sqrt{x}+1)(x-\sqrt{x}+1)]`

`B=[x+2+x+\sqrt{x}-x+\sqrt{x}-1]/[(\sqrt{x}+1)(x-\sqrt{x}+1)]`

`B=[x+2\sqrt{x}+1]/[(\sqrt{x}+1)(x-\sqrt{x}+1)]`

`B=[(\sqrt{x}+1)^2]/[(\sqrt{x}+1)(x-\sqrt{x}+1)]`

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

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`c)` Với `x >= 0` có:

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

               `<=>[\sqrt{x}-1]/[x-\sqrt{x}+1]-1/2 < 0`

               `<=>[2\sqrt{x}-2-x+\sqrt{x}-1]/[2(x-\sqrt{x}+2)] < 0`

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

                   Mà `2(x-\sqrt{x}+2) > 0`

   `=>-x+3\sqrt{x}-3 < 0`

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

`<=>(\sqrt{x}-3/2)^2+3/4 > 0`

`<=>(\sqrt{x}-3/2)^2 > -3/4` (LĐ `AA x >= 0`)

Vậy `x >= 0` thì `A.B < 1/2`

Giải thích các bước giải:

Đáp án:

Giải thích các bước giải:

 `a)`Thay `x=16` vào `A:`

`A=(\sqrt{16}-1)/(\sqrt{16}+1)`

`=(4-1)/(4+1)`

`=3/5`

Vậy `A=3/5<=>x=16`

\(b)B=\dfrac{x+2}{x\sqrt{x}+1}+\dfrac{\sqrt{x}}{x-\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}.ĐKXĐ:x\geq 0\\=\dfrac{x+2}{(\sqrt{x}+1)(x-\sqrt{x}+1)}+\dfrac{\sqrt{x}}{x-\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\\=\dfrac{x+2+\sqrt{x}(\sqrt{x}+1)-x+\sqrt{x}-1}{(\sqrt{x}+1)(x-\sqrt{x}+1)}\\=\dfrac{x+2+x+\sqrt{x}-x+\sqrt{x}-1}{(\sqrt{x}+1)(x-\sqrt{x}+1)}\\=\dfrac{x+2\sqrt{x}+1}{(\sqrt{x}+1)(x-\sqrt{x}+1)}\\=\dfrac{(\sqrt{x}+1)^2}{(\sqrt{x}+1)(x-\sqrt{x}+1)}\\=\dfrac{\sqrt{x}+1}{x-\sqrt{x}+1}(đpcm)\\c)A.B\\=\dfrac{\sqrt{x}-1}{\sqrt{x}-1}.\dfrac{\sqrt{x}+1}{x-\sqrt{x}+1}\\=\dfrac{\sqrt{x}-1}{x-\sqrt{x}+1}\)

Xét hiệu: : `(\sqrt{x}-1)/(x-\sqrt{x}+1)-1/2`

`=(2\sqrt{x}-2-x+\sqrt{x}-1)/(x-\sqrt{x}+1)`

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

Ta có: 

`-x+3\sqrt{x}-3`

`=-(x-3\sqrt{x}+3)`

`=-[(x-3\sqrt{x}+9/4)+3/4]`

`=-(\sqrt{x}-3/2)^2 -3/4`

Ta có: `x-\sqrt{x}+1`

`=(x-\sqrt{x}+1/4)+3/4`

`=(\sqrt{x}-1/2)^2 +3/4`

Do :`{((\sqrt{x}-3/2)^2>=0AAx>=0=>-(\sqrt{x}-3/2)^2 <=0=>-(\sqrt{x}-3/2)^2 -3/4<= -3/4<0),((\sqrt{x}-1/2)^2 >=0AAx>=0=>(\sqrt{x}-1/2)^2 +3/4>=3/4>0):}`

`=>(-(\sqrt{x}-3/2)^2 -3/4)/((\sqrt{x}-1/2)^2 +3/4)<0`

`<=>(\sqrt{x}-1)/(x-\sqrt{x}+1)-1/2<0`

`<=>A.B<1/2(đpcm)`

Vậy `A.B<1/2` với `x>=0`