Question

[tex]\sqrt{3+2\sqrt{2} } -\sqrt{2}=...\\\\(a) \:4\sqrt{2} \\(b) \:3+\sqrt{2} \\(c)\:\sqrt{2} \\(d)\:1\\(e)\:0[/tex]

Answer 1

Verified answer

Gunakan sifat:

[tex]\sqrt{(a+b)+2\sqrt{a\cdot b} } =\sqrt{a} +\sqrt{b}[/tex]

Maka:

[tex]\sqrt{3+2\sqrt{2} } -\sqrt{2}\\\\=\sqrt{(1+2)+2\sqrt{1\cdot2} } -\sqrt{2}\\\\=\sqrt{1} +\sqrt{2}-\sqrt{2} \\ \\=\sqrt{1} \\\\=1[/tex]

Hasilnya 1 (d).