T=4,06731601

Em tham khảo đề bài và bài làm tại link: Câu hỏi của Trân Vũ Mai Ngọc - Toán lớp 9 - Học toán với OnlineMath

\(S=\left(a^2+\frac{1}{4}\right)+\left(b^2+\frac{1}{4}\right)+\left(c^2+\frac{1}{4}\right)+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\frac{3}{4}\)

\(\ge a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{3}{4}=\left(a+\frac{1}{4a}\right)+\left(b+\frac{1}{4b}\right)+\left(c+\frac{1}{4c}\right)-\frac{3}{4}\)

\(\ge1+1+1-\frac{3}{4}=\frac{9}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=\frac{1}{2}\)

à quên tách ra mà quên đoạn sau :v thêm vào tí nhé 

\(S\ge\left(a+\frac{1}{4a}\right)+\left(b+\frac{1}{4b}\right)+\left(c+\frac{1}{4c}\right)+\frac{3}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\frac{3}{4}\)

\(\ge2\sqrt{\frac{a}{4a}}+2\sqrt{\frac{b}{4b}}+2\sqrt{\frac{c}{4c}}+\frac{3}{4}.\frac{9}{a+b+c}-\frac{3}{4}\ge1+1+1+\frac{3}{4}.\frac{9}{\frac{3}{2}}-\frac{3}{4}=\frac{27}{4}\)

b, Đặt \(\sqrt[3]{x}=t\)

Ta có: \(\sqrt[3]{x^2}-8\sqrt[3]{x}=20\)

\(\Leftrightarrow t^2-8t=20\Leftrightarrow t^2-8t-20=0\)

\(\Leftrightarrow\left(t+2\right)\left(t-10\right)=0\)

\(\orbr{\begin{cases}t=-2\\t=10\end{cases}\Leftrightarrow\orbr{\begin{cases}\sqrt[3]{x}=-2\\\sqrt[3]{x}=10\end{cases}\Leftrightarrow}}\orbr{\begin{cases}x=-8\\x=1000\end{cases}}\)

a)\(2\sqrt{5}-\sqrt{125}-\sqrt{80}+\sqrt{605}\)

= \(2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}\)

= \(4\sqrt{5}\)

b) \(\sqrt{15-\sqrt{216}}+\sqrt{33-12\sqrt{6}}\)

= \(\sqrt{3\left(5-2\sqrt{6}\right)}-\sqrt{33-12\sqrt{6}}\)

= \(\sqrt{3\left(5-2\sqrt{6}\right)}-\sqrt{3\left(11-4\sqrt{6}\right)}\)

\(a,2\sqrt{5}-\sqrt{125}-\sqrt{80}+\sqrt{605}\)

\(=2\sqrt{5}-\sqrt{5^2.5}-\sqrt{4^2.5}+\sqrt{11^2.5}\)

\(=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}\)

\(=4\sqrt{5}\)

\(b,\sqrt{15-\sqrt{216}}+\sqrt{33-12\sqrt{6}}\)

\(=\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)

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

\(=|3-\sqrt{6}|+|3-2\sqrt{6}|\)

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

\(=\sqrt{6}\)

\(x+2y=\sqrt{\left(\frac{1}{\sqrt{2}}.\sqrt{2}x+\frac{2}{\sqrt{3}}.\sqrt{3}y\right)^2}\le\sqrt{\left(\frac{1}{2}+\frac{4}{3}\right)\left(2x^2+3y^2\right)}=\sqrt{\frac{22}{3}}\)

\(3=x^2+y^2+z^2\ge\frac{\left(x+y+z\right)^2}{3}\)\(\Leftrightarrow\)\(x+y+z\le3\)

\(x^3+y^3+z^3=\frac{x^4}{x}+\frac{y^4}{y}+\frac{z^4}{z}\ge\frac{\left(x^2+y^2+z^2\right)^2}{x+y+z}\ge\frac{3^2}{3}=3\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=z=1\)

Ta có:\(x;y;z>0\Leftrightarrow x^3;y^3;z^3\ge0\Leftrightarrow x^3\ge x^2;y^3\ge y^2;z^3\ge z^2\)

\(\Leftrightarrow x^3+y^3+z^3\ge x^2+y^2+z^2hay:x^3+y^3+z^3\ge3\)