là 000

\(\frac{ }{ }\frac{ }{ }\)

\(\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\cdot\sqrt{3-\sqrt{5}}\)

\(=\left(3+\sqrt{5}\right)\cdot\left(\sqrt{5}-1\right)\cdot\sqrt{2}\cdot\sqrt{3-\sqrt{5}}\)

\(=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\sqrt{6-2\sqrt{5}}\)

\(=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\cdot\sqrt{\left(\sqrt{5}-1\right)^2}\)

\(=\left(3+\sqrt{5}\right)\cdot\left(\sqrt{5}-1\right)^2\)

\(=\left(3+\sqrt{5}\right)\left(6-2\sqrt{5}\right)\)

\(=2\cdot\left(3+\sqrt{5}\right)\cdot\left(3-\sqrt{5}\right)\)

\(=2\cdot\left(9-5\right)\)

\(=2-4=8\)

thế x đâu

ai trả lời mình k đúng cho

\(a+b+c\le\sqrt{3\left(a^2+b^2+c^2\right)}\)

\(\Leftrightarrow\left(a+b+c\right)^2\le3\left(a^2+b^2+c^2\right)\)

\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca\le3a^2+3b^2+3c^2\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca\ge0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\) ( luôn đúng )

Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)

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