Mình cần gấp mình cảm ơn

`a)` Rút Gọn `B5`

`B5=(15\sqrt{x}-11)/(x+2\sqrt{x}-3)+(3\sqrt{x}-2)/(1-\sqrt{x})-(2\sqrt{x}+3)/(3+\sqrt{x})` với `x>=0;x\ne1`

`=(15\sqrt{x}-11)/(x+2\sqrt{x}-3)-(3\sqrt{x}-2)/(\sqrt{x}-1)-(2\sqrt{x}+3)/(\sqrt{x}+3)`

`=(15\sqrt{x}-11)/(x+2\sqrt{x}-3)-((3\sqrt{x}-2)(\sqrt{x}+3))/((\sqrt{x}-1)(\sqrt{x}+3))-((2\sqrt{x}+3)(\sqrt{x}-1))/((\sqrt{x}+3)(\sqrt{x}-1))`

`=(15\sqrt{x}-11)/(x+2\sqrt{x}-3)-(3x+9\sqrt{x}-2\sqrt{x}-6)/((\sqrt{x}-1)(\sqrt{x}+3))-((2\sqrt{x}+3)(\sqrt{x}-1))/((\sqrt{x}+3)(\sqrt{x}-1))`

`=(15\sqrt{x}-11)/(x+2\sqrt{x}-3)-(3x+7\sqrt{x}-6)/((\sqrt{x}-1)(\sqrt{x}+3))-((2\sqrt{x}+3)(\sqrt{x}-1))/((\sqrt{x}+3)(\sqrt{x}-1))`

`=(15\sqrt{x}-11)/(x+2\sqrt{x}-3)-(3x+7\sqrt{x}-6)/((\sqrt{x}-1)(\sqrt{x}+3))-(2x-2\sqrt{x}+3\sqrt{x}-3)/((\sqrt{x}+3)(\sqrt{x}-1))`

`=(15\sqrt{x}-11)/(x+2\sqrt{x}-3)-(3x+7\sqrt{x}-6)/((\sqrt{x}-1)(\sqrt{x}+3))-(2x+\sqrt{x}-3)/((\sqrt{x}+3)(\sqrt{x}-1))`

`=(15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3)/((\sqrt{x}-1)(\sqrt{x}+3))`

`=(7\sqrt{x}-5x-2)/((\sqrt{x}-1)(\sqrt{x}+3))`

`=(-(\sqrt{x}-1)(5\sqrt{x}-2))/((\sqrt{x}-1)(\sqrt{x}+3))`

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

Vậy `B5=(2-5\sqrt{x})/(\sqrt{x}+3)` với `x>=0;x\ne1`

`b)` `B5=1/2`

`=>(2-5\sqrt{x})/(\sqrt{x}+3)=1/2`

`<=>2.(2-5\sqrt{x})=\sqrt{x}+3`

`<=>4-10\sqrt{x}=\sqrt{x}+3`

`<=>-10\sqrt{x}-\sqrt{x}=-4+3`

`<=>-11\sqrt{x}=-1`

`<=>\sqrt{x}=1/11`

`=>x=1/121`

Kết Hợp Với ĐKXĐ ta thấy `x=1/121` thoả mãn

Vậy `x=1/121` khi `B5=1/2`