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[tex]\it \sqrt{12}= \sqrt{4\cdot3} =2\sqrt3
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\sqrt{48}=\sqrt{16\cdot3}=4\sqrt3[/tex]
Acum, ecuația devine:
[tex]\it |2x-2\sqrt3| =4\sqrt3 \Leftrightarrow |2(x-\sqrt3)|=4\sqrt3 \Leftrightarrow|2||x-\sqrt3|=4\sqrt3 \Leftrightarrow
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\Leftrightarrow 2|x-\sqrt3| =4\sqrt3|_{:2} \Leftrightarrow |x-\sqrt3|=2\sqrt3 \Leftrightarrow x-\sqrt3=\pm2\sqrt3[/tex]
[tex]\it I) \ x-\sqrt3=-2\sqrt3|_{+\sqrt3}\ \Leftrightarrow\ x_1=-\sqrt3
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II) \ x-\sqrt3= 2\sqrt3|_{+\sqrt3}\ \Leftrightarrow\ x_2=3\sqrt3[/tex]
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