`bb(b o t)`

`bb(a.)`

`B= (sqrtx)/(sqrtx+3) - (sqrtx+3)/(sqrtx-3) - ( x + 11sqrtx +6)/(9-x)`

`ĐK  : x >=0 ; x ne  9`

`B = (sqrtx(sqrtx-3) - (sqrtx+3)^2  + (x + 11sqrtx +6))/((sqrtx-3)(sqrtx+3))`

`B = (x -3sqrtx- x - 6sqrtx - 9 + x + 11sqrtx +6)/((sqrtx+3)(sqrtx-3))`

`B = ( x + 2sqrtx -3)/((sqrtx+3)(sqrtx-3))`

`B = (x + 3sqrtx - sqrtx-3)/((sqrtx+3)(sqrtx-3))`

`B = (sqrtx(sqrtx +3) - (sqrtx+3))/((sqrtx+3)(sqrtx-3))`

`B =((sqrtx-1)(sqrtx+3))/((sqrtx+3)(sqrtx - 3))`

`B = (sqrtx-1)/(sqrtx-3)`

Vậy `B  = (sqrtx-1)/(sqrtx -3)` với `x >=0 ; x ne 9`

`bb(b.)`

`M = A:B =(sqrtx-1)/(sqrtx+3) : (sqrtx-1)/(sqrtx-3)`

`M = (sqrtx-1)/(sqrtx+3) . (sqrtx-3)/(sqrtx-1)`

`M = (sqrtx-3)/(sqrtx+3)`

Ta có `:`

`M(sqrtx+3) =sqrt{x-5} - 2` `đk :x >=5`

`(sqrtx-3)/(sqrtx+3) . (sqrtx+3) =sqrt{x-5} -2`

`sqrtx -3 = sqrt{x-5} -2`

`sqrtx -3 +2 = sqrt{x-5}`

`sqrtx -1 = sqrt{x-5}` 

`x - 2sqrtx + 1 = x-5`

`x -2sqrtx - x = -5 - 1`

`-2sqrtx = -6`

`sqrtx = 3`

`x= 9 (ktm)`

Vậy không có `bbx` thoả mãn `M(sqrtx+3) =sqrt{x-5} - 2`

Bài `14.`

`bb(1.)`

`B = (sqrtx-1)/(sqrtx-2)  - (5sqrtx - 8)/(x -2sqrtx)`

`ĐK :x >0 ; x ne 4 ; x ne 16`

`B = (sqrtx-1)/(sqrtx-2) - (5sqrtx -8)/(sqrtx(sqrtx-2))`

`B = (sqrtx(sqrtx-1) - (5sqrtx - 8))/(sqrtx(sqrtx-2))`

`B = (x -sqrtx -5sqrtx +8)/(sqrtx(sqrtx-2))`

`B=(x - 2sqrtx - 4sqrtx +8)/(sqrtx(sqrtx-2))`

`B=(sqrtx(sqrtx-2) - 4(sqrtx-2))/(sqrtx(sqrtx-2))`

`B=((sqrtx -4)(sqrtx-2))/(sqrtx(sqrtx-2))`

`B=(sqrtx-4)/sqrtx`

`bb(2.)`

`M =A.B`

`M = (x+sqrtx+1)/(sqrtx-4) . (sqrtx-4)/sqrtx`

`M = (x+sqrtx+1)/sqrtx`

Ta có `:`

`M . sqrtx = 4sqrtx +2sqrt{x-5} - 3`

`(x+sqrtx+1)/sqrtx .sqrtx = 4sqrtx +2sqrt{x-5} - 3`

`x +sqrtx +1 = 4sqrtx + 2sqrt{x-5} - 3`

`x +sqrtx - 4sqrtx = 2sqrt{x-5}  -3 - 1`

`x - 3sqrtx = 2sqrt{x-5} - 4`

`(x^2 - 9x)/(x +3sqrtx) = 2(sqrt{x-5} - 2)`

`(x(x-9))/(x +3sqrtx) - 2(x -5 - 4)/(sqrt{x-5} +2) = 0`

`(x-9)(x/(x+3sqrtx) - 2/(sqrt{x-5} +2))=0`

`TH1 : x - 9 = 0 to x = 9 (tmđk)`

`TH2 : x/(x+3sqrtx) - 2/(sqrt{x-5} +2) =0`

`to x/(sqrtx(sqrtx+3)) = 2/(sqrt{x-5} +2 )`

`sqrtx/(sqrtx+3) =2/(sqrt{x-5} +2)`

`sqrtx(sqrt{x-5} +2) = 2 (sqrtx+3)`

`sqrt{x(x-5)} +2sqrtx =2sqrtx + 6`

`sqrt{x(x-5)} =6`

`x(x-5) =36`

`x^2 - 5x - 36 =0`

`(x-9)(x+4)=0`

`x-9=0` hoặc `x+4 =0`

`x =9 (tm)` hoặc `x=-4 ( ktm)`

Vậy `x =9 .`