Question

Bài `2`

cho : 

\(A_4=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2};\left(x\ge0;x\ne1\right)\)

`a)` rút gọn \(A_4\)

`b)CMR` : \(A_4>0\forall x\ge0;x\ne1\)

Answer 1

`a)` Với `x >= 0,x \ne 1` có:

`A_4=([x+2]/[x\sqrt{x}-1]+\sqrt{x}/[x+\sqrt{x}+1]+1/[1-\sqrt{x}]):[\sqrt{x}-1]/2`

`A_4=[x+2+\sqrt{x}(\sqrt{x}-1)-x-\sqrt{x}-1]/[(\sqrt{x}-1)(x+\sqrt{x}+1)]. 2/[\sqrt{x}-1]`

`A_4=[x+2+x-\sqrt{x}-x-\sqrt{x}-1]/[(\sqrt{x}-1)(x+\sqrt{x}+1)]. 2/[\sqrt{x}-1]`

`A_4=[x-2\sqrt{x}+1]/[(\sqrt{x}-1)(x+\sqrt{x}+1)]. 2/[\sqrt{x}-1]`

`A_4=[2(\sqrt{x}-1)^2]/[(\sqrt{x}-1)^2(x+\sqrt{x}+1)]`

`A_4=2/[x+\sqrt{x}+1]`

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`b)` Với `x >= 0,x \ne 1` có: `A_4=2/[x+\sqrt{x}+1]`

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

 Vì `x >= 0<=>(\sqrt{x}+1/2)^2 >= 0`

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

      Hay `x+\sqrt{x}+1 > 0`

     Mà `2 > 0`

  `=>2/[x+\sqrt{x}+1] > 0`

  Hay `A_4 > 0 AA x >= 0,x \ne 1`

  `=>Đpcm`

Answer 2

a: \(A=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)

\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\cdot\dfrac{2}{x+\sqrt{x}+1}=\dfrac{2}{x+\sqrt{x}+1}\)

b: \(x+\sqrt{x}+1>=1>0\)

nên A>0 với x>=0; x<>1