Wykaż, że... (pierwiastki)

[Karlajn88]

\(\displaystyle{ \sqrt{13-2\sqrt{30}}=\sqrt{10}-\sqrt{3}}\)

[yorgin]

\(\displaystyle{ \sqrt{13-2\sqrt{30}}=\sqrt{10-2\cdot\sqrt{10}\cdot\sqrt{3}+3}=
\sqrt{(\sqrt{10})^2-2\cdot\sqrt{10}\cdot\sqrt{3}+(\sqrt{3})^2}=\sqrt{(\sqrt{10}-\sqrt{3})^2}=
|\sqrt{10}-\sqrt{3}|=\sqrt{10}-\sqrt{3}}\)

[Karlajn88]

\(\displaystyle{ \sqrt{13-2\sqrt{30}}=\sqrt{10-2\cdot\sqrt{10}\cdot\sqrt{3}+3}=
\sqrt{(\sqrt{10})^2-2\cdot\sqrt{10}\cdot\sqrt{3}+(\sqrt{3})^2}=\sqrt{(\sqrt{10}-\sqrt{3})^2}=
|\sqrt{10}-\sqrt{3}|=\sqrt{10}-\sqrt{3}}\)

Nie rozumiem jak doszłeś z tego:
\(\displaystyle{ \sqrt{(\sqrt{10})^2-2\cdot\sqrt{10}\cdot\sqrt{3}+(\sqrt{3})^2}}\)
tu:
\(\displaystyle{ \sqrt{(\sqrt{10}-\sqrt{3})^2}}\)

[yorgin]

Hmm
\(\displaystyle{ a^2-2ab+b^2=(a-b)^2\\
a=\sqrt{10}\\
b=\sqrt{3}}\)