Answer :
First a few reminders:
• For a ≠ 0 and b ≥ 0, [tex]\sqrt[a]{b}=b^{\frac1a}[/tex]
• 4 = 2² and 8 = 2³
• For real numbers a, b, c, we have [tex](a^b)^c=a^{bc}[/tex]
• For any real numbers a and n (so long as both are not zero), we have [tex]\frac1{a^n}=a^{-n}[/tex]
• For any numbers a, b, c, [tex]a^b\times a^c=a^{b+c}[/tex]
So we have
[tex]\sqrt2=2^{\frac12}[/tex]
[tex]\dfrac1{\sqrt2}=2^{-\frac12}[/tex]
[tex]\dfrac1{\sqrt8}=\dfrac1{(2^3)^{\frac12}}=2^{-\frac32}[/tex]
Then the given expression can be rewritten as
[tex]2\sqrt2-\dfrac3{\sqrt2}+\dfrac4{\sqrt8}=2\times2^{\frac12}-3\times2^{-\frac12}+2^2\times2^{-\frac32}[/tex]
[tex]2\sqrt2-\dfrac3{\sqrt2}+\dfrac4{\sqrt8}=2^{1+\frac12}-3\times2^{-\frac12}+2^{2-\frac32}[/tex]
[tex]2\sqrt2-\dfrac3{\sqrt2}+\dfrac4{\sqrt8}=2^{\frac32}-3\times2^{-\frac12}+2^{\frac12}[/tex]
Pull out a factor of [tex]2^{-\frac12}[/tex]:
[tex]2\sqrt2-\dfrac3{\sqrt2}+\dfrac4{\sqrt8}=2^{-\frac12}\left(2^{\frac32+\frac12}-3+2^{\frac12+\frac12}\right)[/tex]
Simplify this:
[tex]2\sqrt2-\dfrac3{\sqrt2}+\dfrac4{\sqrt8}=2^{-\frac12}\left(2^2-3+2^1\right)=2^{-\frac12}(4-3+2)=3\times2^{-\frac12}=\dfrac3{\sqrt2}[/tex]
Rationalize this last result to get the square root out of the denominator:
[tex]2\sqrt2-\dfrac3{\sqrt2}+\dfrac4{\sqrt8}=\dfrac3{\sqrt2}\times\dfrac{\sqrt2}{\sqrt2}=\dfrac{3\sqrt2}{(\sqrt2)^2}=\boxed{\dfrac{3\sqrt2}2}[/tex]