Câu hỏi của Hang Daov

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 \(x=\dfrac{3-\sqrt{2}}{3-\sqrt{2}}=1\)Câu trả lời: 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 \(x=\dfrac{3-\sqrt{2}}{3-\sqrt{2}}=1\)