√(4-2 √3)/1- √3 √ (2x-5)^2=1

Đáp án:

 `=1`

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`x=3` hoặc `x=2`

Giải thích các bước giải:

Ta có` :`

`(\sqrt{4-2\sqrt{3}})/(1-\sqrt{3})`

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

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

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

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

`=-1`

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`\sqrt{(2x-5)^2}=1`

`⇔|2x-5|=1`

`⇔`\(\left[ \begin{array}{l}2x-5=1\\2x-5=-1\end{array} \right.\) 

`⇔`\(\left[ \begin{array}{l}2x=6\\2x=4\end{array} \right.\) 

`⇔`\(\left[ \begin{array}{l}x=3\\x=2\end{array} \right.\) 

Vậy `x=3` hoặc `x=2`

a) `(\sqrt{4-2\sqrt{3}})/(1-\sqrt{3})`

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

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

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

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

`=-1`

b) `\sqrt{(2x-5)^2}=1` `(x\inR)`

`<=>|2x-5|=1`

`<=>`\(\left[ \begin{array}{l}2x-5=1\\2x-5=-1\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=3(tm)\\x=2(tm)\end{array} \right.\) 

Vậy `S={3,2}`