Answer:
[tex]\sf -11+7\sqrt{2}[/tex]
Step-by-step explanation:
Given expression:
[tex]\sf \dfrac{3-\sqrt{32}}{1+\sqrt{2} }[/tex]
Rewrite 32 as 16 · 2:
[tex]\sf \implies \dfrac{3-\sqrt{16 \cdot 2}}{1+\sqrt{2} }[/tex]
Apply radical rule [tex]\sf \sqrt{a \cdot b}=\sqrt{a}\sqrt{b}[/tex]
[tex]\sf \implies \dfrac{3-\sqrt{16}\sqrt{2}}{1+\sqrt{2} }[/tex]
As [tex]\sf \sqrt{16}=4[/tex]:
[tex]\sf \implies \dfrac{3-4\sqrt{2}}{1+\sqrt{2} }[/tex]
Multiply by the conjugate:
[tex]\sf \implies \dfrac{3-4\sqrt{2}}{1+\sqrt{2} } \times \dfrac{1-\sqrt{2} }{1-\sqrt{2} }[/tex]
[tex]\sf \implies \dfrac{(3-4\sqrt{2})(1-\sqrt{2})}{(1+\sqrt{2})(1-\sqrt{2})}[/tex]
[tex]\sf \implies \dfrac{3-3\sqrt{2}-4\sqrt{2}+4\sqrt{2}\sqrt{2}}{1-\sqrt{2}+\sqrt{2}-\sqrt{2}\sqrt{2}}[/tex]
As [tex]\sf \sqrt{2}\sqrt{2}=\sqrt{4}=2[/tex]:
[tex]\sf \implies \dfrac{3-3\sqrt{2}-4\sqrt{2}+4 \cdot 2}{1-\sqrt{2}+\sqrt{2}-2}[/tex]
[tex]\sf \implies \dfrac{3-7\sqrt{2}+8}{1-2}[/tex]
[tex]\sf \implies \dfrac{11-7\sqrt{2}}{-1}[/tex]
[tex]\sf \implies -11+7\sqrt{2}[/tex]
