a)\(x=2\sqrt{1,2+\dfrac{2.5+4}{5}}=2\sqrt{\dfrac{6}{5}+\dfrac{14}{5}}\)
\(x=2\sqrt{\dfrac{20}{5}}=2\sqrt{4}\)
\(x=2\sqrt{2^2}=2.\left|2\right|=2.2=4\)
b)\(\left(3-\sqrt{2}\right).x=\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=>x=\dfrac{\sqrt{\left(3-\sqrt{2}\right)^2}}{3-\sqrt{2}}\)
\(x=\dfrac{\left|3-\sqrt{2}\right|}{3-\sqrt{2}}\)
\(x=\dfrac{3-\sqrt{2}}{3-\sqrt{2}}=1\)
Vậy \(x=\dfrac{3-\sqrt{2}}{3-\sqrt{2}}=1\)
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