LEVER

\(\frac{3}{a+b\sqrt{3}}-\frac{2}{a-b\sqrt{3}}=7-20\sqrt{3}\)\(\Leftrightarrow\frac{3\left(a-b\sqrt{3}\right)-2\left(a+b\sqrt{3}\right)}{a^2-3b^2}=7-20\sqrt{3}\)

\(\Leftrightarrow\frac{a-5\sqrt{3}b}{a^2-3b^2}=7-20\sqrt{3}\)\(\Leftrightarrow\frac{a-5\sqrt{3}b}{a^2-3b^2}=\frac{7-20\sqrt{3}}{49-48}\Leftrightarrow\frac{a-5\sqrt{3}b}{a^2-3b^2}=\frac{7-20\sqrt{3}}{7^2-3.4^2}\Leftrightarrow\hept{\begin{cases}a=7\\b=4\end{cases}}\)

Ta có: \(\frac{1+A}{1-A}=\frac{1+\frac{x-y}{x+y}}{1-\frac{x-y}{x+y}}=\frac{\frac{2x}{x+y}}{\frac{2y}{x+y}}=\frac{x}{y}\)

Tương tự: \(\frac{1+B}{1-B}=\frac{y}{z};\frac{1+C}{1-C}=\frac{z}{x}\)

\(\Rightarrow\frac{\left(1+A\right)\left(1+B\right)\left(1+C\right)}{\left(1-A\right)\left(1-B\right)\left(1-C\right)}=1\Rightarrowđpcm\)

Ta có: \(\left(x-y\right)^2\ge0\Leftrightarrow x^2+y^2\ge2xy\Leftrightarrow\left(x+y\right)^2\ge4xy\)

\(\Rightarrow4.2011a\left(2011a-2\right)\le\left(2011a+2011a-2\right)^2=4\left(2011a-1\right)^2\)

\(\Leftrightarrow2011a\left(2011a-2\right)\le\left(2011a-1\right)^2\)

\(\Leftrightarrow\frac{2011a\left(2011a-2\right)}{\left(2011a-1\right)^2}\le1\)

\(\Leftrightarrow\frac{1}{a}-\frac{2011a\left(2011a-2\right)}{\left(2011a-1\right)^2}\ge\frac{1}{a}-1\)\(\Leftrightarrow\frac{1}{a\left(2011a-1\right)^2}\ge\frac{1}{a}-1\)

Tương tự: \(\frac{1}{b\left(2011b-1\right)^2}\ge\frac{1}{b}-1;\frac{1}{c\left(2011c-1\right)^2}\ge\frac{1}{c}-1\)

\(\Leftrightarrow\frac{1}{a\left(2011a-1\right)^2}+\frac{1}{b\left(2011b-1\right)^2}+\frac{1}{c\left(2011c-1\right)^2}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-3=2011-3=2008\)

Sai thì thôi nhá bẹn!

Dễ thôi, ta có:

Kẻ đường cao BH ta được: \(BC^2=BH^2+HC^2\)

\(\Leftrightarrow a^2=\left(AB^2-AH^2\right)+\left(AC-AH\right)^2\)

\(=c^2-AH^2+b^2-2\cdot b\cdot AH+AH^2\)

\(=b^2+c^2-2\cdot AH\cdot b\)

\(=b^2+c^2-2ab\cdot\cos A\)

Trước tiên ta chứng minh : \(2\left(a^2+b^2\right)\ge\left(a+b\right)^2\); ở đó, a,b tùy ý. Thật vậy:

\(2\left(a^2+b^2\right)S\ge\left(a+b\right)^2\Leftrightarrow a^2+b^2\ge2ab\Leftrightarrow\left(a-b\right)^2\ge0\)

Ta có: \(x\left(x-1\right)+\frac{1}{4}+y\left(y-1\right)+\frac{1}{4}=\left(x-\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2\ge\frac{1}{2}\text{[}\left(x-\frac{1}{2}\right)+\left(y-\frac{1}{2}\right)\text{]}^2\)\(=\frac{1}{2}\left(x+y-1\right)^2\ge\frac{1}{2}\left(6-1\right)^2=\frac{25}{2}\Rightarrow x\left(x-1\right)+y\left(y-1\right)\ge\frac{25}{2}-\frac{1}{2}=12\)

Khi x=y=3 thì => đpcm

Hơi khác cách của mình nhưng đúng r.

\(x=\sqrt[3]{13-7\sqrt{6}}+\sqrt[3]{13+7\sqrt{6}}\Rightarrow x^3=26-15x\)

\(x^3+15x-25=1\Rightarrow\left(x^3+15x-25\right)^{2013}=1\)

Vậy P(x)=1 với .....