\(2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge\frac{4}{a+b}+\frac{4}{b+c}+\frac{4}{c+a}\) ( Svac-xơ, Cauchy các kiểu -,- ) 

\(\Leftrightarrow\)\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{\frac{4}{a+b}+\frac{4}{b+c}+\frac{4}{c+a}}{2}=\frac{2}{a+b}+\frac{2}{b+c}+\frac{2}{c+a}\) ( đpcm ) 

... 

\(2VP=\frac{4}{a+b}+\frac{4}{b+c}+\frac{4}{c+a}\)

\(\le\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}+\frac{1}{a}=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=2VT\)

Từ đây,ta có: \(2VT\ge2VP\Rightarrow VT\ge VP^{\left(đpcm\right)}\)

Việt Nam vô địch

Việt Nam chiến thắng

Ai cổ vũ cho Đội tuyển Việt Nam thì k cho mik nha

Malaysia đúng không?

đề bài yêu cầu gì vậy bạn

Khó quá

dễ mà , đồ ngu , đúng là tép riu

P xác định khi \(\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)

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

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

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

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

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

\(=\frac{x-1}{\sqrt{x}}\)

P xác định khi \(\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)

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

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

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

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

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

\(=\frac{x-1}{\sqrt{x}}\)

ĐKXĐ: \(a,b\ge0;a\ge b\)

\(\sqrt{a-b}=\sqrt{a}-\sqrt{b}\)

\(\Rightarrow\left(\sqrt{a-b}\right)^2=\left(\sqrt{a}-\sqrt{b}\right)^2\)

\(\Leftrightarrow a-b=a-2.\sqrt{ab}+b\)

\(\Leftrightarrow2b=2\sqrt{ab}\)

\(\Leftrightarrow b=\sqrt{ab}\)

\(\Leftrightarrow b^2=\left(\sqrt{ab}\right)^2\)

\(\Leftrightarrow b^2=ab\)

\(\Leftrightarrow b^2-ab=0\)

\(\Leftrightarrow b\left(b-a\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}b=0\\b-a=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}b=0\\a=b\end{cases}}\)

b tự kết luận nhé~