[tex]|AB|=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}[/tex]
a)
[tex]|AB|=\sqrt{(1-(-11))^2+(-4-1)^2}=\sqrt{12^2+(-5)^2}=\sqrt{144+25}=\\=\sqrt{169}=13[/tex]
b)
[tex]\dfrac{1}{\sqrt{2}-1}=\dfrac{1\cdot(\sqrt{2}+1)}{(\sqrt{2}-1)\cdot(\sqrt{2}+1)}=\dfrac{\sqrt{2}+1}{2-1}=\dfrac{\sqrt{2}+1}{1}=\sqrt{2}+1[/tex]
[tex]|AB|=\sqrt{(3+\sqrt{2}-(\sqrt{2}+1))^2+(-1-(-3))^2}=\sqrt{(3+\sqrt{2}-\sqrt{2}-1)^2+(-1+3)^2}=\\=\sqrt{2^2+2^2}=\sqrt{4+4}=\sqrt{8}=\sqrt{4\cdot2}=2\sqrt{2}[/tex]
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[tex]|AB|=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}[/tex]
a)
[tex]|AB|=\sqrt{(1-(-11))^2+(-4-1)^2}=\sqrt{12^2+(-5)^2}=\sqrt{144+25}=\\=\sqrt{169}=13[/tex]
b)
[tex]\dfrac{1}{\sqrt{2}-1}=\dfrac{1\cdot(\sqrt{2}+1)}{(\sqrt{2}-1)\cdot(\sqrt{2}+1)}=\dfrac{\sqrt{2}+1}{2-1}=\dfrac{\sqrt{2}+1}{1}=\sqrt{2}+1[/tex]
[tex]|AB|=\sqrt{(3+\sqrt{2}-(\sqrt{2}+1))^2+(-1-(-3))^2}=\sqrt{(3+\sqrt{2}-\sqrt{2}-1)^2+(-1+3)^2}=\\=\sqrt{2^2+2^2}=\sqrt{4+4}=\sqrt{8}=\sqrt{4\cdot2}=2\sqrt{2}[/tex]