Question

4) A = 3.\(\sqrt{2}\)+ 4.\(\sqrt{8}\)-\(\sqrt{18}\)

5) ( \(\sqrt{12}-\sqrt{48}-\sqrt{108}-\sqrt{192}\) ) :2\(\sqrt{3}\)

Answer 1

4. \(A=3\sqrt{2}+4\sqrt{8}-\sqrt{18}=3\sqrt{2}+8\sqrt{2}=3\sqrt{2}=8\sqrt{2}\)

5. \(\dfrac{\sqrt{12}-\sqrt{48}-\sqrt{108}-\sqrt{192}}{2\sqrt{3}}=\dfrac{2\sqrt{3}-4\sqrt{3}-6\sqrt{3}-8\sqrt{3}}{2\sqrt{3}}=\dfrac{-16\sqrt{3}}{2\sqrt{3}}=-8\sqrt{3}\)

Answer 2

4: Ta có: \(A=3\sqrt{2}+4\sqrt{8}-\sqrt{18}\)

\(=3\sqrt{2}+8\sqrt{2}-3\sqrt{2}\)

\(=8\sqrt{2}\)

5: Ta có: \(\left(\sqrt{12}-\sqrt{48}-\sqrt{108}-\sqrt{192}\right):2\sqrt{3}\)

\(=\left(2\sqrt{3}-4\sqrt{3}-6\sqrt{3}-8\sqrt{3}\right):2\sqrt{3}\)

\(=-16\sqrt{3}:2\sqrt{3}=-8\)