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12 tháng 9 2023

a) \(\left(2\sqrt{2}-3\right)^2\)

\(=\left(2\sqrt{2}\right)^2-2\cdot2\sqrt{2}\cdot3+3^2\)

\(=4\cdot2-12\sqrt{2}+9\)

\(=17-12\sqrt{2}\)

b) \(\sqrt{\left(\dfrac{1}{\sqrt{2}}-\dfrac{1}{2}\right)^2}\)

\(=\left|\dfrac{1}{\sqrt{2}}-\dfrac{1}{2}\right|\)

\(=\dfrac{1}{\sqrt{2}}-\dfrac{1}{2}\)

\(=\dfrac{\sqrt{2}}{2}-\dfrac{1}{2}\)

\(=\dfrac{\sqrt{2}-1}{2}\)

c) \(\sqrt{\left(0,1-\sqrt{0,1}\right)^2}\)

\(=\left|0,1-\sqrt{0,1}\right|\)

\(=0,1-\sqrt{0,1}\)

12 tháng 9 2023

a) \(\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}\)

\(=\left|\sqrt{5}-\sqrt{2}\right|+\left|\sqrt{5}+\sqrt{2}\right|\)

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

\(=\sqrt{5}+\sqrt{5}\)

\(=2\sqrt{5}\)

b) \(\sqrt{\left(\sqrt{2}-1\right)^2}-\sqrt{\left(\sqrt{2}-5\right)^2}\)

\(=\left|\sqrt{2}-1\right|-\left|\sqrt{2}-5\right|\)

\(=\sqrt{2}-1-\left(5-\sqrt{2}\right)\)

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

\(=2\sqrt{2}-6\)

25 tháng 7 2023

\(\dfrac{4}{\sqrt{5}-\sqrt{2}}+\dfrac{3}{\sqrt{5}-2}-\dfrac{2}{\sqrt{3}-2}-\dfrac{\sqrt{3}-1}{6}\)

\(=\dfrac{4\left(\sqrt{2}+\sqrt{5}\right)}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{2}+\sqrt{5}\right)}+\dfrac{3\left(\sqrt{5}+2\right)}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}-\dfrac{2\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}-\dfrac{\sqrt{3}-1}{6}\)

\(=\dfrac{4\left(\sqrt{2}+\sqrt{5}\right)}{\left(\sqrt{5}\right)^2-\left(\sqrt{2}\right)^2}+\dfrac{3\left(\sqrt{5}+2\right)}{\left(\sqrt{5}\right)^2-2^2}-\dfrac{2\left(\sqrt{3}+2\right)}{\left(\sqrt{3}\right)^2-2^2}-\dfrac{\sqrt{3}-1}{6}\)

\(=\dfrac{4\left(\sqrt{2}+\sqrt{5}\right)}{3}+\dfrac{3\left(\sqrt{5}+2\right)}{1}-\dfrac{2\left(\sqrt{3}+2\right)}{-1}-\dfrac{\sqrt{3}-1}{6}\)

\(=\dfrac{8\left(\sqrt{2}+\sqrt{5}\right)}{6}+\dfrac{18\left(\sqrt{5}+2\right)}{6}+\dfrac{12\left(\sqrt{3}+2\right)}{6}-\dfrac{\sqrt{3}-1}{6}\)

\(=\dfrac{8\sqrt{2}+8\sqrt{5}+18\sqrt{5}+36+12\sqrt{3}+24-\sqrt{3}+1}{6}\)

\(=\dfrac{8\sqrt{2}+26\sqrt{5}+11\sqrt{3}+61}{6}\)

\(=\dfrac{4\left(\sqrt{5}+\sqrt{2}\right)}{3}+\dfrac{3\left(\sqrt{5}+2\right)}{1}+\dfrac{2\left(2+\sqrt{3}\right)}{1}-\dfrac{\sqrt{3}-1}{6}\)

\(=\dfrac{4\sqrt{5}+4\sqrt{2}+9\sqrt{5}+18}{3}+\dfrac{4+2\sqrt{3}}{1}-\dfrac{\sqrt{3}-1}{6}\)

\(=\dfrac{2\left(13\sqrt{5}+4\sqrt{2}+18\right)+24+12\sqrt{3}-\sqrt{3}+1}{6}\)

\(=\dfrac{26\sqrt{5}+4\sqrt{2}+36+25+11\sqrt{3}}{6}\)

\(=\dfrac{61+11\sqrt{3}+26\sqrt{5}+4\sqrt{2}}{6}\)

a: Ta có: \(3\sqrt{2}\cdot5\sqrt{6}\cdot4\sqrt{12}\)

\(=\sqrt{18\cdot25\cdot6\cdot16\cdot12}\)

\(=\sqrt{518400}\)

=720

b: Ta có: \(\left(\sqrt{7}-\sqrt{2}\right)^2+2\sqrt{14}\)

\(=9-2\sqrt{14}+2\sqrt{14}\)

=9

c: Ta có: \(\left(1+\sqrt{5}+\sqrt{6}\right)\left(1+\sqrt{5}-\sqrt{6}\right)\)

\(=6+2\sqrt{5}-6\)

\(=2\sqrt{5}\)

16 tháng 8 2021

câu a làm tắt thế bạn

 

2 tháng 5 2018

\(\sqrt{36}+\sqrt{9}-\sqrt{49}\)

\(=6+3-7\)

\(=2\)

\(\sqrt{2}\cdot\left(\sqrt{50}-3\sqrt{2}\right)\)

\(=\sqrt{2}\cdot\left(5\sqrt{2}-3\sqrt{2}\right)\)

\(=\sqrt{2}\cdot2\sqrt{2}\)

\(=4\)

2 tháng 5 2018

a) \(T=\sqrt{36}+\sqrt{9}-\sqrt{49}\)

    \(=6+3-7\)

      \(=2\)

b)  \(B=\sqrt{2\left(\sqrt{50}-3\sqrt{2}\right)}\)

         \(=\sqrt{10\sqrt{2}-6\sqrt{2}}\)

           \(=\sqrt{\left(10-6\right)\sqrt{2}}\)

           \(=\sqrt{4\sqrt{2}}\)

            \(\approx2,39\)

17 tháng 12 2018

a, (\(\sqrt{128}\)-\(\sqrt{50}\)+\(\sqrt{98}\)):\(\sqrt{2}\)

=(8-5+3)

=10

b, (\(\sqrt{48}\)+\(\sqrt{27}\)-\(\sqrt{192}\)):2\(\sqrt{3}\)

=(2+1,5-4)

=-0,5

c, \(\dfrac{1}{8}\)-3\(\sqrt{2}\) +\(\dfrac{1}{8}\)+3\(\sqrt{2}\)

=\(\dfrac{1}{4}\)

d, \(\sqrt{\left(1-\sqrt{5}\right)^2}-\sqrt{5}\)

=-1

18 tháng 12 2018

bn có thể giải chi tiết ra hộ mk được ko

\(A=2\sqrt{2}+\sqrt{3}\)

\(B=\dfrac{2\sqrt{2}}{1+\sqrt{2-\sqrt{3}}}=\dfrac{4}{2+\sqrt{3}-1}=\dfrac{4}{\sqrt{3}+1}=2\sqrt{3}-2\)

=>A>B