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!

A B C H

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 .....

giải hộ mik với

đáp án là \(\frac{7}{13}\)

a) \(M=\frac{x+1+\sqrt{x}}{x+1}:\left(\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{\sqrt{x}\left(x+1\right)-\left(x+1\right)}\right)\)

\(=\frac{x+\sqrt{x}+1}{x+1}:\left(\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right)\)

\(=\frac{x+\sqrt{x}+1}{x+1}:\frac{x+1-2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\)\(=\frac{x+\sqrt{x}+1}{x+1}.\frac{\left(x+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)^2}=\frac{x+\sqrt{x}+1}{\sqrt{x}-1}\)

b) \(M>3\Rightarrow\frac{x+\sqrt{x}+1}{\sqrt{x}-1}>3\Leftrightarrow\frac{x+\sqrt{x}+1}{\sqrt{x}-1}-3>0\)

\(\Leftrightarrow\frac{x+\sqrt{x}+1-3\left(\sqrt{x}-1\right)}{\sqrt{x}-1}>0\Leftrightarrow\frac{x+\sqrt{x}+1-3\sqrt{x}+3}{\sqrt{x}-1}>0\)\(\Leftrightarrow\frac{x-2\sqrt{x}+4}{\sqrt{x}-1}>0\)

Ta có: \(x-2\sqrt{x}+4=x-2\sqrt{x}+1+3=\left(\sqrt{x}-1\right)+3>0\)\(\Rightarrow\sqrt{x}-1>0\Leftrightarrow\sqrt{x}>1\Leftrightarrow x>1\)

Vậy x>1

c) \(M=7\Rightarrow\frac{x+\sqrt{x}+1}{\sqrt{x}-1}=7\Rightarrow x+\sqrt{x}+1=7\left(\sqrt{x}-1\right)\)

\(\Leftrightarrow x+\sqrt{x}+1=7\sqrt{x}-7\Leftrightarrow x-6\sqrt{x}+8=0\)\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}-4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-2=0\\\sqrt{x}-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=2\\\sqrt{x}=4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=4\\x=16\end{cases}\left(tm\right)}}\)

Vậy \(x\in\text{{}4;16\)