\(a,|x+1|+|x-1|=4\)\((*)\)

• \(TH_1:\hept{\begin{cases}x+1\ge0\\x-1\ge0\end{cases}}\Leftrightarrow x\ge1\)

Khi đó pt \((*)\) \(\Leftrightarrow x+1+x-1=4\Leftrightarrow x=2\left(tm\right)\)

• \(TH_2:\hept{\begin{cases}x+1\ge0\\x-1< 0\end{cases}}\Leftrightarrow-1\le x< 1\)

Khi đó pt \((*)\) \(\Leftrightarrow x+1+1-x=4\Leftrightarrow0=2\left(vl\right)\)

• \(TH_3:\hept{\begin{cases}x+1< 0\\x-1\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x< -1\\x\ge1\end{cases}}\left(l\right)\)
• \(TH_4:\hept{\begin{cases}x+1< 0\\x-1>0\end{cases}}\Leftrightarrow x< -1\)

Khi đó pt \((*)\) \(\Leftrightarrow-x-1+1-x=4\)

\(\Leftrightarrow x=-2\left(tm\right)\)

Vậy pt có 2 \(n_0\) \(x=\pm2\)

\(b,\frac{|2x-1|}{x+1}=\frac{1}{2}\left(Đkxđ:x\ne-1\right)\)

\(\Rightarrow2|2x-1|=x+1\)

• \(TH_1:x\ge\frac{1}{2}\Leftrightarrow2\left(2x-1\right)=x+1\)

\(\Leftrightarrow4x-2=x+1\)

\(\Leftrightarrow x=1\left(tm\right)\)

• \(TH_2:x< \frac{1}{2}\Leftrightarrow2\left(1-2x\right)=x+1\)

\(\Leftrightarrow2-4x=x+1\)

\(\Leftrightarrow x=\frac{1}{5}\left(l\right)\)

Vậy pt có 1 \(n_0\) \(x=1\)

\(P=a-\frac{ab^2}{1+b^2}+b-\frac{bc^2}{1+c^2}+c-\frac{ca^2}{1+a^2}\)

    \(\ge a-\frac{ab^2}{2b}+b-\frac{bc^2}{2c}+c-\frac{ca^2}{2c}\) (AM-GM)

      \(\ge a-\frac{ab}{2}+b-\frac{bc}{2}+c-\frac{ac}{2}\ge\left(a+b+c\right)-\frac{\left(a+b+c\right)^2}{6}\ge3-\frac{3}{2}=\frac{3}{2}\)

Vay MinP=3/2 dau = xay ra khi a=b=c=1