Vì tam giác DEF vuông tại D
=>\(DE^2+DF^2=EF^2\)
=> EF=30(cm)
Vì Diện tích DEF=\(\dfrac{DI.EF}{2}=\dfrac{DE.DF}{2}\)

                    => DI= 14.4

vì tam giác DIE vuông tại I

=>\(DI^2+EI^2=ED^2\)

=>EI=10.8

=>DI=19.2

b)\(\dfrac{D}{\sqrt{3}}=\dfrac{\sqrt{\sqrt{3}+1}+1-\sqrt{\sqrt{3}+1}+1}{\sqrt{3}+1-1}\)

    \(\dfrac{D}{\sqrt{3}}=\dfrac{2}{\sqrt{3}}\)

     D=2

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

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

   =\(\dfrac{4\sqrt{6}}{6}-\dfrac{3\sqrt{6}}{6}=\dfrac{\sqrt[]{6}}{6}\)

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

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

       =\(\sqrt{2-1}\)

       =1

3)     =(5+4\(\sqrt{2}\))(\(9-4\left(1+\sqrt{2}\right)\))

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

        =25-16.2
        =-7

2)    =\(\sqrt{14.6+14\sqrt{35}}\)-\(\sqrt{10.6+10.\sqrt{35}}\)

       =\(\sqrt{64+14\sqrt{35}}\)-\(\sqrt{60+10\sqrt{35}}\)

       =\(\sqrt{49+14\sqrt{35}+35}-\sqrt{25+10\sqrt{35}+35}\)

       =\(\sqrt{\left(7+\sqrt{35}\right)^2}-\sqrt{\left(5+\sqrt{35}\right)^2}\)

       =\(7+\sqrt{35}-5-\sqrt{35}\)

       =2