`P=(x+7)/(3\sqrt{x})`

Thế `x=1/9` vào `P`, ta có:

`(1/9+7):(3\sqrt{1/9})`

`=(64)/9:(3.1/3)`

`=(64)/9:1`

`=64/9`

Vậy khi `x=1/9` thì `P=64/9`

 $Q=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{11\sqrt{x}-3}{x-9}$

$=\dfrac{2\sqrt{x}(\sqrt{x}-3)+(\sqrt{x}+1)(\sqrt{x}+3)+11\sqrt{x}-3}{(\sqrt{x}-3)(\sqrt{x}+3)}$
$=\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3+11\sqrt{x}-3}{(\sqrt{x}-3)(\sqrt{x}+3)}$

$=\dfrac{3x+9\sqrt{x}}{(\sqrt{x}-3)(\sqrt{x}+3)}$

$=\dfrac{3\sqrt{x}(\sqrt{x}+3)}{(\sqrt{x}-3)(\sqrt{x}+3)}$
$=\dfrac{3\sqrt{x}}{\sqrt{x}-3}$

$c)x>9$

$M=P.Q$

$<=>M=\dfrac{3\sqrt{x}}{\sqrt{x}-3}.\dfrac{x+7}{3\sqrt{x}}$

$=\dfrac{x+7}{\sqrt{x}-3}$

$=\dfrac{x-9+16}{\sqrt{x}-3}$

$=\dfrac{(\sqrt{x}-3)(\sqrt{x}+3)+16}{\sqrt{x}-3}$

$=\sqrt{x}+3+\dfrac{16}{\sqrt{x}-3}$

$=\sqrt{x}-3+\dfrac{16}{\sqrt{x}-3}+6$

Áp dụng BĐT cô-si

`->(\sqrt{x}-3)+\frac{16}{\sqrt{x}-3}>=2\sqrt{(\sqrt{x}-3).\frac{16}{\sqrt{x}-3}}=2\sqrt{16}=2.4=8`

$=>\sqrt{x}-3+\frac{16}{\sqrt{x}-3}+6>=8+6=14$

Dấu "=" xảy ra khi `x=49`

`<=>GTNNNN_M=14` khi và chỉ khi `x=49`

Đáp án + Giải thích các bước giải:

`a.`

Khi `x=1/9=>P=\frac{1/9+7}{3sqrt(1/9)}=\frac{64}{9}`

`b.`

`Q=\frac{2sqrtx}{sqrtx+3}+\frac{sqrtx+1}{sqrtx-3}+\frac{11sqrtx-3}{x-9}`

`=\frac{2sqrtx(sqrtx-3)+(sqrtx+1)(sqrtx+3)+11sqrtx-3}{(sqrtx-3)(sqrtx+3)}`

`=\frac{2x-6sqrtx+x+4sqrtx+3+11sqrtx-3}{(sqrtx-3)(sqrtx+3)}`

`=\frac{3x+9sqrtx}{(sqrtx-3)(sqrtx+3)}`

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

`=\frac{3sqrt(x)}{sqrtx-3}`

`c.`

`M=P.Q=\frac{x+7}{3sqrtx}.\frac{3sqrtx}{sqrtx-3}=(x+7)/(sqrtx-3)`

`=>M=(x+7)/(sqrtx-3)=(x-9+15)/(sqrtx-3)`

`=sqrtx+3+16/(sqrtx-3)=sqrtx-3+16/(sqrtx-3)+6`

Áp dụng BĐT Côsi ta có:

`M=sqrtx-3+16/(sqrtx-3)+6>=2sqrt((sqrtx-3). 16/(sqrtx-3))+6=14 `

Dấu "=" xảy ra khi `sqrtx-3=16/(sqrtx-3)<=>x=49`

Vậy giá trị nhỏ nhất của `M=14` khi `x=49`