\(-\dfrac{6\sqrt{2}-\sqrt{\left(9-8\sqrt{2}\right)\cdot2}}{2}\)

\(\sqrt{2.36}+\sqrt{2.\dfrac{9}{4}}-\sqrt{2.16}-\sqrt{2.81}=6\sqrt{2}+\dfrac{3}{2}\sqrt{2}-4\sqrt{2}-9\sqrt{2}=\dfrac{-11}{2}\sqrt{2}\)

\(\sqrt{4\dfrac{1}{2}}+\sqrt{32}-\sqrt{72}+\sqrt{162}\\ =\sqrt{\dfrac{4\cdot2+1}{2}}+\sqrt{4^2\cdot2}-\sqrt{6^2\cdot2}+\sqrt{9^2\cdot2}\\ =\sqrt{\dfrac{9}{2}}+4\sqrt{2}-6\sqrt{2}+9\sqrt{2}\\ =\dfrac{3}{\sqrt{2}}+7\sqrt{2}\\ =\dfrac{3}{\sqrt{2}}+\dfrac{7\sqrt{2}\cdot\sqrt{2}}{\sqrt{2}}\\ =\dfrac{17}{\sqrt{2}}\)

\(=\sqrt{\dfrac{9}{2}}+4\sqrt{2}-6\sqrt{2}+9\sqrt{2}\)

\(=\dfrac{3}{2}\sqrt{2}+7\sqrt{2}=\dfrac{17}{2}\sqrt{2}\)

\(\sqrt{4\dfrac{1}{2}}+\sqrt{32}-\sqrt{72}+\sqrt{162}\)

\(=\sqrt{\dfrac{9}{2}}+\sqrt{4^2.2}-\sqrt{6^2.2}+\sqrt{9^2.2}\)

\(=\dfrac{3}{\sqrt{2}}+4\sqrt{2}-6\sqrt{2}+9\sqrt{2}\)

\(=\dfrac{3\sqrt{2}}{2}+7\sqrt{2}=\dfrac{3\sqrt{2}+14\sqrt{2}}{2}=\dfrac{17\sqrt{2}}{2}\)

6: Ta có: \(\left(3\sqrt{2}-\sqrt{3}\right)\left(3\sqrt{2}+\sqrt{3}\right)\)

=18-3

=15

7: Ta có: \(\sqrt{72}+\sqrt{4\dfrac{1}{2}}-\sqrt{32}-\sqrt{162}\)

\(=6\sqrt{2}+\dfrac{3}{2}\sqrt{2}-4\sqrt{2}-9\sqrt{2}\)

\(=-\dfrac{11}{2}\sqrt{2}\)

a: \(A=\left(1-\sqrt{7}\right)\cdot\left(1+\sqrt{7}\right)=1-7=-6\)

b: \(B=3\sqrt{3}+8\sqrt{3}-15\sqrt{3}=-4\sqrt{3}\)

c: \(C=4\sqrt{2}-5\sqrt{2}+3\sqrt{2}=2\sqrt{2}\)

\(\sqrt{2\cdot36}+\sqrt{2\cdot\dfrac{9}{4}}-\sqrt{2\cdot16}-\sqrt{2\cdot81}=6\sqrt{2}+\dfrac{3}{2}\sqrt{2}-4\sqrt{2}-9\sqrt{2}=\dfrac{-11}{2}\sqrt{2}\)

a, A = \(3\sqrt{3}+4\sqrt{12}-5\sqrt{27}\)

= \(3\sqrt{3}+8\sqrt{3}-15\sqrt{3}\)

= \(-4\sqrt{3}\)

b, B = \(\sqrt{32}-\sqrt{50}+\sqrt{18}\)

= \(4\sqrt{2}-5\sqrt{2}+3\sqrt{2}\)

= \(2\sqrt{2}\)

a) \(=\sqrt{\frac{9}{2}}-\sqrt{16.2}+\sqrt{36.2}-\sqrt{81.2}\)

\(=\frac{3}{2}\sqrt{2}-4\sqrt{2}+6\sqrt{2}-9\sqrt{2}\)

\(=\left(\frac{3}{2}-4+6-9\right)\sqrt{2}=\frac{-11}{2}\sqrt{2}\)

b) \(=\frac{\sqrt{5}+3-\sqrt{5}+3}{\left(\sqrt{5}-3\right)\left(\sqrt{5}+3\right)}.\frac{\sqrt{3}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}\)

\(=\frac{6}{5-9}.\left(-\sqrt{3}\right)=\frac{3}{2}\sqrt{3}\)

c) \(=\left(\frac{a-1-4\sqrt{a}+\sqrt{a}+1}{a-1}\right):\frac{\sqrt{a}\left(\sqrt{a}-2\right)}{a-1}\)

\(=\frac{a-3\sqrt{a}}{a-1}.\frac{a-1}{\sqrt{a}\left(\sqrt{a}-2\right)}\)

\(=\frac{\sqrt{a}\left(\sqrt{a}-3\right)}{\sqrt{a}\left(\sqrt{a}-2\right)}=\frac{\sqrt{a}-3}{\sqrt{a}-2}\)

Câu a : \(\sqrt{200}-\sqrt{32}+\sqrt{72}-\sqrt{162}\)

\(=10\sqrt{2}-4\sqrt{2}+6\sqrt{2}-9\sqrt{2}\)

\(=3\sqrt{2}\)

Câu b : \(\sqrt{63}+\sqrt{175}+\sqrt{112}\)

\(=3\sqrt{7}+5\sqrt{7}+4\sqrt{7}\)

\(=12\sqrt{7}\)

Học tốt !!!!!!!!!!!!!

a: \(=10\sqrt{2}-4\sqrt{2}+6\sqrt{2}=12\sqrt{2}\)

b: \(=5\sqrt{7}-4\sqrt{7}+3\sqrt{7}=4\sqrt{7}\)

c: \(=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{6}=\dfrac{1}{6}\sqrt{6}\)

d: \(=8\sqrt{5}-15\sqrt{5}+15\sqrt{5}-3\sqrt{5}=5\sqrt{5}\)

e: \(=\sqrt{5}+\dfrac{2}{5}\sqrt{5}+\sqrt{5}=2.4\sqrt{5}\)

f: \(=\dfrac{1}{5}\sqrt{5}+\dfrac{3}{2}\sqrt{2}+\dfrac{5}{2}\sqrt{2}=\dfrac{1}{5}\sqrt{5}+4\sqrt{2}\)