Đáp án:

 `T _{min}=2\sqrt{3}` khi `x=4`

Giải thích các bước giải:

`A={25\sqrt{x}+6}/{x-36}-{\sqrt{x}-1}/{6-\sqrt{x}}+{2\sqrt{x}}/{\sqrt{x}+6}` $(x\ge 0; x\ne 36)$

`A={25\sqrt{x}+6+(\sqrt{x}-1)(\sqrt{x}+6)+2\sqrt{x}.(\sqrt{x}-6)}/{(\sqrt{x}-6)(\sqrt{x}+6)}`

`A={25\sqrt{x}+6+x+5\sqrt{x}-6+2x-12\sqrt{x}}/{(\sqrt{x}-6)(\sqrt{x}+6)}`

`A={3x+18\sqrt{x}}/{(\sqrt{x}-6)(\sqrt{x}+6)}`

`A={3\sqrt{x}.(\sqrt{x}+6)}/{(\sqrt{x}-6)(\sqrt{x}+6)}`

`A={3\sqrt{x}}/{\sqrt{x}-6}`

$\\$

`B={x-6\sqrt{x}}/{\sqrt{x}-1}` $(x\ne 1; x\ge 0)$

`B={\sqrt{x}.(\sqrt{x}-6)}/{\sqrt{x}-1}`

$\\$

`T=\sqrt{A.B}`

`T=\sqrt{{3\sqrt{x}}/{\sqrt{x}-6}.{\sqrt{x}.(\sqrt{x}-6)}/{\sqrt{x}-1}}`

`T=\sqrt{{3x}/{\sqrt{x}-1}}`

`T` xác định khi `{3x}/{\sqrt{x}-1}\ge 0`

Mà `x\ge 0=>\sqrt{x}-1> 0`

`<=>\sqrt{x}>1<=>x>1`

Ta có:

`\qquad {3x}/{\sqrt{x}-1}`

`={3(x-2\sqrt{x}+1)+3+6(\sqrt{x}-1)}/{\sqrt{x}-1}`

`={3(\sqrt{x}-1)^2}/{\sqrt{x}-1}+3/{\sqrt{x}-1}+{6(\sqrt{x}-1)}/{\sqrt{x}-1}`

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

`\ge 2\sqrt{3.(\sqrt{x}-1). 3/{\sqrt{x}-1}}+6` (BĐT Cosi)

`\ge 2\sqrt{9}+6=12`

`=>T=\sqrt{{3x}/{\sqrt{x}-1}} \ge \sqrt{12}=2\sqrt{3}`

Dấu `=` xảy ra khi:

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

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

`<=>(\sqrt{x}-1)^2=1`

`<=>`$\left[\begin{array}{l}\sqrt{x}-1=1\\\sqrt{x}-1=-1\end{array}\right.$

`<=>`$\left[\begin{array}{l}\sqrt{x}=2\\\sqrt{x}=0\end{array}\right.$

`<=>`$\left[\begin{array}{l}x=4( T M)\\x=0(loại)\end{array}\right.$

Vậy $GTNN$ của $T$ bằng `2\sqrt{3}` khi `x=4`

Đáp án-Giải thích các bước giải:

a) Thay` x=16 in` ĐKXĐ vào `B` ta có:

`B=(16-6.\sqrt16)/(\sqrt16 -1)=(16-6.4)/(4-1)=-8/3`

Vậy `B=-8/3` tại `x=16.`

b) ĐKXĐ: `x>=0;x\ne 1;x\ne 36`

`A=(25\sqrtx +6)/(x-36) -(\sqrtx -1)/(6-\sqrtx) +(2\sqrtx)/(\sqrtx +6)`

`A=(25\sqrtx +6)/((\sqrtx +6)(\sqrtx-6))+(\sqrtx-1)/(\sqrtx -6)+(2\sqrtx)/(\sqrtx +6)`

`A=(25\sqrtx +6+(\sqrtx-1)(\sqrtx+6)+2\sqrtx(\sqrtx-6))/((\sqrtx+6)(\sqrtx-6))`

`A=(25\sqrtx+6+x+5\sqrtx-6+2x-12\sqrtx)/((\sqrtx+6)(\sqrtx-6))`

`A=(3x+18\sqrtx)/((\sqrtx+6)(\sqrtx-6))`

`A=(3\sqrtx(\sqrtx+6))/((\sqrtx+6)(\sqrtx-6))`

`A=(3\sqrtx)/(\sqrtx-6)`

Vậy` A=(3\sqrtx)/(\sqrtx-6)` với `x>=0;x\ne1;x\ne 36.`

c) `T=\sqrt(A.B)=\sqrt((3\sqrtx)/(\sqrtx-6). (x-6\sqrtx)/(\sqrtx -1))`

`T=\sqrt((3\sqrtx)/(\sqrtx-6). (\sqrtx(\sqrtx-6))/(\sqrtx -1))`

`T=\sqrt((3x)/(\sqrtx-1))`

Điều kiện: `(3x)/(\sqrtx-1)>=0`

Mà `x>=0=> \sqrtx-1>0`

`<=>x>1`

Ta có: `(3x)/(\sqrtx -1)={3(x-2\sqrt{x}+1)+3+6(\sqrt{x}-1)}/{\sqrt{x}-1}`

`={3(\sqrt{x}-1)^2}/{\sqrt{x}-1}+3/{\sqrt{x}-1}+{6(\sqrt{x}-1)}/{\sqrt{x}-1}`

`=3(\sqrt{x}-1)+3/{\sqrt{x}-1}+6>= 2\sqrt( 3.(\sqrtx -1) .3/(\sqrtx -1)) +6`( BĐT Cosi)

`>=2\sqrt9 +6=12`

`=> T>=\sqrt 12 =2\sqrt3`

Dấu "=" xảy ra `<=> 3(\sqrtx-1)=3/(\sqrtx-1)`

`=>3(\sqrtx-1)^2=3`

`<=>(\sqrtx -1)^2=1`

`<=>|\sqrtx-1|=1`

`<=>`\(\left[ \begin{array}{l}\sqrt{x}-1=1\\\sqrt{x}-1=-1\end{array} \right.\)

`<=>`\(\left[ \begin{array}{l}\sqrt{x}=2\\\sqrt{x}=0\end{array} \right.\)

`<=>`\(\left[ \begin{array}{l}x=4(tm)\\x=0(loại)\end{array} \right.\)

Vậy `T_{min}=2\sqrt3` tại `x=4.`