E=\(|x-2005|+|2006-x|\ge|x-2005+2006-x|=1\)

Dấu = xảy ra khi \(\orbr{\begin{cases}x\ge2006\\x\le2005\end{cases}}\)

\(\left(\sqrt{\frac{3}{4}}-\sqrt{3}+5\cdot\sqrt{\frac{4}{3}}\right)\sqrt{12}.\)

\(=\left(\frac{\sqrt{3}}{\sqrt{4}}-\sqrt{3}+5\cdot\frac{\sqrt{4}}{\sqrt{3}}\right)\sqrt{12}.\)

\(=\left(\frac{\sqrt{3}}{2}-\sqrt{3}+5\cdot\frac{2}{\sqrt{3}}\right)\sqrt{12}.\)

\(=\left(\frac{1}{2}\cdot\sqrt{3}-\sqrt{3}+5\cdot\frac{2}{\sqrt{3}}\right)\sqrt{12}.\)

\(=\left(-\frac{1}{2}\sqrt{3}+\frac{10}{\sqrt{3}}\right)\sqrt{12}\)

\(=\left(-\frac{1}{2}\sqrt{3}+\frac{10}{3}\sqrt{3}\right)\sqrt{12}\)

\(=\frac{17}{6}\sqrt{3}\sqrt{12}=\frac{17}{6}\sqrt{36}=\frac{17}{6}\cdot6=17\)

\(\Rightarrow\left(\sqrt{\frac{3}{4}}-\sqrt{3}+5\cdot\sqrt{\frac{4}{3}}\right)\sqrt{12}=17\)

Ta co:

\(P\ge21\left(a^2+b^2+c^2\right)+12\left(a+b+c\right)^2+\frac{2017.9}{2}\)

\(=21\left(a^2+b^2+c^2\right)+12\left(a+b+c\right)^2+\frac{18153}{2}\)

\(\Leftrightarrow\frac{P}{\left(a+b+c\right)^2}\ge21\left[\left(\frac{a}{a+b+c}\right)^2+\left(\frac{b}{a+b+c}\right)^2+\left(\frac{c}{a+b+c}\right)^2\right]+12+\frac{\frac{18153}{2}}{\left(a+b+c\right)^2}\)

Dat \(\left(\frac{a}{a+b+c};\frac{b}{a+b+c};\frac{c}{a+b+c}\right)\rightarrow\left(x;y;z\right)\)

\(\Rightarrow x+y+z=1\)

\(\Rightarrow\left(a+b+c\right)^2=\frac{a^2}{x^2}\)

BDT tro thanh:

\(\frac{P}{\left(a+b+c\right)^2}\ge21\left(x^2+y^2+z^2\right)+12+\frac{18153}{2\left(a+b+c\right)^2}\)

\(\Leftrightarrow\frac{P}{\frac{a^2}{x^2}}\ge21\left(x^2+y^2+z^2\right)+12+\frac{18153}{2\left(a+b+c\right)^2}\ge21.\frac{\left(x+y+z\right)^2}{3}+12+\frac{18153}{8}\)

\(\Leftrightarrow\frac{x^2P}{a^2}\ge7+12+\frac{18153}{8}\)

Ta lai co:\(x=\frac{a}{a+b+c}\ge\frac{a}{2}\Rightarrow a^2\le4x^2\)

Suy ra:\(\frac{x^2P}{a^2}\ge\frac{x^2P}{4x^2}=\frac{P}{4}\)

\(\Rightarrow\frac{P}{4}\ge\frac{18503}{8}\)

\(\Leftrightarrow P\ge\frac{18503}{2}\)

Dau '=' xay ra khi \(a=b=c=\frac{2}{3}\)

Vay \(P_{min}=\frac{18503}{2}\)khi \(a=b=c=\frac{2}{3}\)

x-1=x^2-2x+1 (x>=1)

x^2-3x+2=0

(x-1)(x-2)=0

 => x=1

    hoặc x=2

ĐK \(x\supseteq1\)

\(x-1=\left(x-1\right)^2\)

\(\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+2}}}}}\)

\(=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+2}}}}\)

\(=\sqrt{2+\sqrt{2+\sqrt{2+2}}}\)

\(=\sqrt{2+\sqrt{2+2}}\)

\(=\sqrt{2+2}=2\)

sai đề òi bn ơi