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Ta có: \(\left(a+b+c\right)^3-27abc=\frac{7a+b+c}{2}\left(b-c\right)^2+\frac{7b+c+a}{2}\left(c-a\right)^2+\frac{7c+a+b}{2}\left(a-b\right)^2\)
Va` \(a+b+c-\sqrt{3\left(a^2+b^2+c^2\right)}=\frac{-\left(a-b\right)^2-\left(b-c\right)^2-\left(c-a\right)^2}{a+b+c+\sqrt{3\left(a^2+b^2+c^2\right)}}\)
\(BDT\Leftrightarrow\frac{\left(a+b+c\right)^3}{abc}-27+54\left(\frac{a+b+c}{\sqrt{3\left(a^2+b^2+c^2\right)}}-1\right)\ge0\)
\(\Leftrightarrow\frac{\left(a+b+c\right)^3-27abc}{abc}+54\left(\frac{a+b+c-\sqrt{3\left(a^2+b^2+c^2\right)}}{\sqrt{3\left(a^2+b^2+c^2\right)}}\right)\ge0\)
\(\Leftrightarrow\frac{Σ\left(\frac{7c+a+b}{2}\left(a-b\right)^2\right)}{abc}-\frac{\frac{Σ54\left(a-b\right)^2}{a+b+c+\sqrt{3\left(a^2+b^2+c^2\right)}}}{\sqrt{3\left(a^2+b^2+c^2\right)}}\ge0\)
\(\LeftrightarrowΣ\left(a-b\right)^2\left(\frac{\frac{7c+a+b}{2}}{abc}-\frac{\frac{54}{a+b+c+\sqrt{3\left(a^2+b^2+c^2\right)}}}{\sqrt{3\left(a^2+b^2+c^2\right)}}\right)\ge0\) *Đúng*
"=" <=> a=b=c :v
Ta có: \(\frac{1}{3}\left(\sqrt{6}+\sqrt{5}\right)^2-\frac{1}{4}\sqrt{120}-2\sqrt{\frac{15}{2}}\)
\(=\frac{1}{3}\left(11+2\sqrt{30}\right)-\frac{\sqrt{30}}{2}-\sqrt{30}\)
\(=\frac{11}{3}+\frac{2}{3}\sqrt{30}-\frac{\sqrt{30}}{2}-\sqrt{30}\)
\(=\frac{11}{3}-\frac{5}{6}\sqrt{30}\)
\(=\frac{22-5\sqrt{30}}{6}\)
Ta có: \(\left(\frac{1}{2}\sqrt{\frac{2}{3}}-\frac{3}{4}\sqrt{54}+\frac{1}{3}\sqrt{\frac{8}{3}}\right)\div\sqrt{\frac{81}{6}}\)
\(=\left(\frac{\sqrt{6}}{6}-\frac{9\sqrt{6}}{4}+\frac{2\sqrt{6}}{9}\right)\div\frac{3\sqrt{6}}{2}\)
\(=-\frac{67\sqrt{6}}{36}\cdot\frac{2}{3\sqrt{6}}\)
\(=-\frac{67}{54}\)
a) nhân ra thôi b
\(=\frac{\left(2\sqrt{10}-5\right)\left(9+\sqrt{10}\right)}{71}=\frac{18\sqrt{10}-45+20-5\sqrt{10}}{71}=\frac{-25+13\sqrt{10}}{71}.\)
b)cách khác nhé !\(\frac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}=\frac{\sqrt{3}\left(3\sqrt{3}-2\right)}{\sqrt{2}\left(3\sqrt{3}-2\right)}=\frac{\sqrt{3}}{\sqrt{2}}=\frac{\sqrt{6}}{2}.\)
\(P=\dfrac{3a-12\sqrt{a}+a+4\sqrt{a}-4a-8}{\left(\sqrt{a}+4\right)\left(\sqrt{a}-4\right)}:\dfrac{\sqrt{a}+4-2\sqrt{a}-5}{\left(\sqrt{a}+4\right)}\)
\(=\dfrac{-8\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+4\right)\left(\sqrt{a}-4\right)}\cdot\dfrac{\sqrt{a}+4}{-\left(\sqrt{a}+1\right)}=\dfrac{8}{\sqrt{a}-4}\)