Question

tính giá trị biểu thức

A=\(\sqrt{20}-3\sqrt{45}+\sqrt{500}\)

B=\(\left(\dfrac{3}{2}\sqrt{6}+2\sqrt{\dfrac{2}{3}}-4\sqrt{\dfrac{3}{2}}\right).\left(3\sqrt{\dfrac{2}{3}}-\sqrt{12}-\sqrt{6}\right)\)

giúp mình với nha mọi người mình đang cần gấp

Answer 1

A=\(\sqrt{20}-3\sqrt{45}+\sqrt{500}\)

=\(2\sqrt{5}-9\sqrt{5}+10\sqrt{5}\)

=\(3\sqrt{5}\)

B=\(\left(\dfrac{3}{2}\sqrt{6}+2\sqrt{\dfrac{2}{3}}-4\sqrt{\dfrac{3}{2}}\right)\times\left(3\sqrt{\dfrac{2}{3}}-\sqrt{12}-\sqrt{6}\right)\)

=\(\left(\dfrac{3\sqrt{6}}{2}+\dfrac{2\sqrt{6}}{3}-\dfrac{4\sqrt{6}}{2}\right)\times\left(\sqrt{6}-\sqrt{12}-\sqrt{6}\right)\)

=\(\left(\dfrac{9\sqrt{6}}{6}+\dfrac{4\sqrt{6}}{6}-\dfrac{12\sqrt{6}}{6}\right)\times\left(-\sqrt{12}\right)\)

=\(\dfrac{9\sqrt{6}+4\sqrt{6}-12\sqrt{6}}{6}\times\left(-\sqrt{12}\right)\)

=\(\dfrac{\sqrt{6}}{6}\left(-\sqrt{12}\right)\)

=\(\dfrac{-\left(\sqrt{6\times12}\right)}{6}\)

=\(-\sqrt{2}\)