`a) A= \frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{2x+\sqrt{x}}{\sqrt{x}}+\frac{2x-2}{\sqrt{x}-1}`
`ĐKXĐ: x>0, x\ne1.`
`A= \frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{2x+\sqrt{x}}{\sqrt{x}}+\frac{2x-2}{\sqrt{x}-1}`
`A= \frac{\sqrt{x}(x\sqrt{x}-1)}{x+\sqrt{x}+1}-\frac{\sqrt{x}(2\sqrt{x}+1)}{\sqrt{x}}+\frac{2(x-1)}{\sqrt{x}-1}`
`A= \frac{\sqrt{x}(\sqrt{x}-1)(x+\sqrt{x}+1)}{x+\sqrt{x}+1}-(2\sqrt{x}+1)+\frac{2(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}-1}`
`A= \sqrt{x}(\sqrt{x}-1)-(2\sqrt{x}+1)+2\sqrt{x}+2`
`A= x- \sqrt{x} + 1`
`A=( x- \sqrt{x} + 1/4 ) + 3/4`
`A=( \sqrt{x}-1/2)^2+3/4\ge3/4`
Dấu ''='' xảy ra khi `\sqrt{x}-1/2=0`
`⇔ \sqrt{x} = 0 + 1/2 = 1/2`
`⇔ x = 1/4.`
Vậy $Min_A$ `=3/4 ⇔ x = 1/4.`
`b) B= \frac{2\sqrt{x}}{A} = \frac{2\sqrt{x}}{x- \sqrt{x} + 1}`
Ta có: `x>0; x- \sqrt{x} + 1= ( \sqrt{x}-1/2)^2+3/4\ge3/4 > 0 ⇒ B>0.`
Có: `2\sqrt{x}=2x-2\sqrt{x}+2-2x+4\sqrt{x}-2=2x-2\sqrt{x}+2-(2x-4\sqrt{x}+2)`
Thay `2\sqrt{x}=2x-2\sqrt{x}+2-(2x-4\sqrt{x}+2)` vào `B` ta được:
`B= \frac{2x-2\sqrt{x}+2-(2x-4\sqrt{x}+2)}{x- \sqrt{x} + 1}=2- \frac{2(\sqrt{x}-1)^2}{x- \sqrt{x} + 1}<2` (vì `x>0` )
`⇒0<B<2`
Mà `B` nguyên nên `B` chỉ có thể là `1`
`⇔\frac{2\sqrt{x}}{x- \sqrt{x} + 1}=1`
`⇒2\sqrt{x}=x-\sqrt{x}+1`
`⇔ x - 3\sqrt{x} +1=0`
`⇒`\(\left[ \begin{array}{l}\sqrt{x}=\frac{3+\sqrt{5}}{2}\\\sqrt{x}=\frac{3-\sqrt{5}}{2}\end{array} \right.\) `⇒`\(\left[ \begin{array}{l}x=\frac{7+3\sqrt{5}}{2}\\x=\frac{7-3\sqrt{5}}{2}\end{array} \right.\)
Vậy `B` nguyên `(B=1)` khi `x= \frac{7±3\sqrt{5}}{2}.`
