Tìm `x` để `B=2A`. - Điều kiện : `x>=0;x\ne4;x\ne9/6`. - `B = sqrt{x}/(sqrt{x}-2)+(sqrt{x}-10)/(x-4)` `->` Sau khi rút gọn : `(sqrt{x}+5)/(sqrt{x}+2)` - `A : sqrt{x}/(4sqrt{x}-3)`

Để `B=2A`               `Đk:x>=0;x\ne4;x\ne9/6`

`<=>(\sqrt{x}+5)/(\sqrt{x}+2)=2.(\sqrt{x})/(4\sqrt{x}-3)`

`<=>(\sqrt{x}+5)/(\sqrt{x}+2)=(2\sqrt{x})/(4\sqrt{x}-3)`

`<=>(\sqrt{x}+5).(4\sqrt{x}-3)=(\sqrt{x}+2).(2\sqrt{x})`

`<=>4x-3\sqrt{x}+20\sqrt{x}-15=2x+4\sqrt{x}`

`<=>2x+13\sqrt{x}-15=0`

`<=>x+(13)/(2)\sqrt{x}-(15)/2=0`

`<=>x+(15)/(2)\sqrt{x}-\sqrt{x}-(15)/2=0`

`<=>\sqrt{x}(\sqrt{x}+(15)/(2))-(\sqrt{x}+(15)/2)=0`

`<=>(\sqrt{x}+(15)/(2))(\sqrt{x}-1)=0`

`<=>` \(\left[ \begin{array}{l}\sqrt{x}+\dfrac{15}{2}=0\\\sqrt{x}-1=0\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}\sqrt{x}=\dfrac{-15}{2}(l)*\\\sqrt{x}=1\end{array} \right.\) 

`=>x=1(tm)`

Vậy `x=1` thì `B=2A`

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$*$ Vì `\sqrt{x}>=0`