\(\frac{4}{1+\sqrt{2}+\sqrt{3}}=\frac{4\left(1-\sqrt{2}+\sqrt{3}\right)}{\left(1+\sqrt{2}+\sqrt{3}\right)\left(1-\sqrt{2}+\sqrt{3}\right)}=\frac{4-4\sqrt{2}+4\sqrt{3}}{1^2-\left(\sqrt{2}+\sqrt{3}\right)^2}=\frac{4-4\sqrt{2}+4\sqrt{3}}{4+2\sqrt{6}}=\frac{4\sqrt{6}-8\sqrt{3}+12\sqrt{2}}{16}=\frac{\sqrt{6}-8\sqrt{3}+12\sqrt{2}}{4}\)
4(1+√2-√3)/(1+√2+√3)(1+√2-√3)
=4(1+√2-√3)/(1+√2)2-3
=4(1+√2-√3)/1+2√2+2-3
=4(1+√2-√3)/2√2
=1+√2-√3/√2