Cho biểu thức: `A = 2/(\sqrtx - 3) + (2\sqrtx)/(x-4\sqrtx + 3) + \sqrtx/(\sqrtx - 1)` a) Rút gọn A b) Tìm x để `A = \sqrt3`

Đáp án:

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

b) `x=(98+55\sqrt{3})/2`

Giải thích các bước giải:

a)

`x>=0;x\ne1;9`

Xét `x-4\sqrt{x}+3`

`=x-\sqrt{x}-3\sqrt{x}+3`

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

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

Ta có:

`A=2/(\sqrt{x}-3)+(2\sqrt{x})/(x-4\sqrt{x}+3)+\sqrt{x}/(\sqrt{x}-1)`

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

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

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

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

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

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

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

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

b) 

`A=\sqrt{3}`

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

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

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

`->3(x-6\sqrt{x}+9)=x+4\sqrt{x}+4`

`->3x-18\sqrt{x}+27=x+4\sqrt{x}+4`

`->2x-22\sqrt{x}+23=0`

`->2(x-11\sqrt{x}+121/4)-75/2=0`

`->2(\sqrt{x}-11/2)^2=75/2`

`->(\sqrt{x}-11/2)^2=75/4`

`->`\(\left[ \begin{array}{l}\sqrt{x}-\dfrac{11}{2}=\dfrac{5\sqrt{3}}{2}\\\sqrt{x}-\dfrac{11}{2}=\dfrac{-5\sqrt{3}}{2}\end{array} \right.\) 

`->` \(\left[ \begin{array}{l}\sqrt{x}=\dfrac{5\sqrt{3}+11}{2}\\\sqrt{x}=\dfrac{-5\sqrt{3}+11}{2}\end{array} \right.\) 

`->` \(\left[ \begin{array}{l}x=\dfrac{98+55\sqrt{3}}{2}\\x=\dfrac{98-55\sqrt{3}}{2}\end{array} \right.\) 

Do phép bình phương chưa tương đương nên thử lại, nghiệm duy nhất là `(98+55\sqrt{3})/2`