Question

Answer 1

c:Ta có: \(C=\left(\dfrac{1}{4}\cdot\sqrt{32}-2\sqrt{18}+9\cdot\sqrt{\dfrac{2}{9}}\right)\cdot\left(-2\sqrt{2}\right)\)

\(=\left(\dfrac{1}{4}\cdot4\sqrt{2}-2\cdot3\sqrt{2}+9\cdot\dfrac{\sqrt{2}}{3}\right)\cdot\left(-2\sqrt{2}\right)\)

\(=\left(-2\sqrt{2}\right)^2=8\)

d: ta có: \(D=\sqrt{4-2\sqrt{3}}-\sqrt{\left(2\sqrt{3}-5\right)^2}+\dfrac{2\sqrt{3}}{1-\sqrt{3}}\)

\(=\sqrt{3}-1-2\sqrt{3}+5-3-\sqrt{3}\)

=1