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Bài 15:

1. Ta có:

`P=(\sqrt{x})/(\sqrt{x}-1)-3/(\sqrt{x}+1)-(6\sqrt{x}-4)/(x-1)`

`=(x+\sqrt{x}-3\sqrt{x}+3-6\sqrt{x}+4)/((\sqrt{x}+1)(\sqrt{x}-1))`

`=(x-8\sqrt{x}+7)/((\sqrt{x}+1)(\sqrt{x}-1))`

`=((x-\sqrt{x})-(7\sqrt{x}-7))/((\sqrt{x}+1)(\sqrt{x}-1))`

`=((\sqrt{x}-1)(\sqrt{x}-7))/((\sqrt{x}+1)(\sqrt{x}-1))`

`=(\sqrt{x}-7)/(\sqrt{x}+1)`

Vậy `P=(\sqrt{x}-7)/(\sqrt{x}+1)` với `x≥0, x\ne1`

2. Để `P<1/2⇔(\sqrt{x}-7)/(\sqrt{x}+1)<1/2` `ĐK: x≥0, x\ne1`

`⇔(2\sqrt{x}-14-\sqrt{x}-1)/(2(\sqrt{x}+1))<0`

`⇔(\sqrt{x}-15)/(2(\sqrt{x}+1))<0`

`⇔(\sqrt{x}-15)/(\sqrt{x}+1)<0`

Vì `\sqrt{x}+1≥1>0∀x,` `x` `tm` `ĐK`

`⇒\sqrt{x}-15<0`

`⇔\sqrt{x}<15`

`⇔x<225`

Kết hợp với `ĐK: x≥0, x\ne1` `⇒` $\begin{cases} 0≤x<225\\x\ne1 \end{cases}$

Vậy $\begin{cases} 0≤x<225\\x\ne1 \end{cases}$ là các giá trị cần tìm

Bài 16:

1. Ta có:

`P=[(a+3\sqrt{a}+2)/((\sqrt{a}+2)(\sqrt{a}-1))-(a+\sqrt{a})/(a-1)]:(1/(\sqrt{a}+1)+1/(\sqrt{a}-1))`

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

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

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

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

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

Vậy `P=(\sqrt{a}+1)/(2\sqrt{a})` với `a>0,a\ne1`

2. Ta có: `1/P-(\sqrt{a}+1)/8=(2\sqrt{a})/(\sqrt{a}+1)-(\sqrt{a}+1)/8` `(ĐK: a>0,a\ne1)`

`=(16\sqrt{a}-a-2\sqrt{a}-1)/(8(\sqrt{a}+1))`

`=(-a+14\sqrt{a}-1)/(8(\sqrt{a}+1))`

Để `1/P-(\sqrt{a}+1)/8≥1⇔(-a+14\sqrt{a}-1)/(8(\sqrt{a}+1))≥1`

`⇔(-a+14\sqrt{a}-1-8\sqrt{a}-8)/(8(\sqrt{a}+1))≥0`

`⇔(-a+6\sqrt{a}-9)/(8(\sqrt{a}+1))≥0`

`⇔(a-6\sqrt{a}+9)/(\sqrt{a}+1)≤0`

`⇔((\sqrt{a}-3)^2)/(\sqrt{a}+1)≤0` `(1)`

Vì `(\sqrt{a}-3)^2≥0∀a,` `a` `tm` `ĐK`

`\sqrt{a}+1≥1>0∀a,` `a` `tm` `ĐK`

`⇒((\sqrt{a}-3)^2)/(\sqrt{a}+1)≥0∀a,` `a` `tm` `ĐK`

Kết hợp với `(1)⇒((\sqrt{a}-3)^2)/(\sqrt{a}+1)=0`

`⇒\sqrt{a}-3=0`

`⇔a=9` `(tm)`

Vậy `a=9` là giá trị cần tìm.