a) \(P=\frac{x-1}{2}:\left(\frac{x^2+2}{x^3-1}+\frac{x}{x^2+x+1}+\frac{1}{1-x}\right)\)
\(=\frac{x-1}{2}:\left(\frac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\right)\)
\(=\frac{x-1}{2}:\frac{x^2+2+\left(x^2-x\right)-\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)\(=\frac{x-1}{2}.\frac{\left(x-1\right)\left(x^2+x+1\right)}{x^2-2x+1}\)
\(=\frac{\left(x-1\right)^2\left(x^2+x+1\right)}{2\left(x-1\right)^2}=\frac{x^2+x+1}{2}\)
b) Ta thấy :
\(x^2+x+1=x^2+x+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
\(\frac{x^2+x+1}{2}>0\Rightarrow P=\left|P\right|\)
c) Lại có :
\(x^2+x+1\ge\frac{3}{4}\Rightarrow\frac{x^2+x+1}{2}\ge\frac{3}{8}\)
Dấu = xảy ra khi :
\(x^2+x+1=\frac{3}{4}\Leftrightarrow\left(x+\frac{1}{2}\right)^2=0\Leftrightarrow x=-\frac{1}{2}\)
Vậy MinP = 3/4 ⇔ x = -1/2
a, ĐKXĐ : \(\left\{{}\begin{matrix}x-1\ne0\\x^2+x+1\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne1\\\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ne0\end{matrix}\right.\)
=> \(x\ne1\)
- Ta có : \(P=\frac{x-1}{2}:\left(\frac{x^2+2}{x^3-1}+\frac{x}{x^2+x+1}+\frac{1}{1-x}\right)\)
=> \(P=\frac{x-1}{2}:\left(\frac{x^2+2}{x^3-1}+\frac{x}{x^2+x+1}-\frac{1}{x-1}\right)\)
=> \(P=\frac{x-1}{2}:\left(\frac{x^2+2}{x^3-1}+\frac{x\left(x-1\right)}{x^3-1}-\frac{x^2+x+1}{x^3-1}\right)\)
=> \(P=\frac{x-1}{2}:\left(\frac{x^2+2+x\left(x-1\right)-x^2-x-1}{x^3-1}\right)\)
=> \(P=\frac{x-1}{2}:\left(\frac{x^2+2+x^2-x-x^2-x-1}{x^3-1}\right)\)
=> \(P=\frac{x-1}{2}:\left(\frac{x^2-2x+1}{x^3-1}\right)\)
=> \(P=\frac{\left(x-1\right)\left(x^3-1\right)}{2\left(x-1\right)^2}=\frac{\left(x-1\right)\left(x-1\right)\left(x^2+x+1\right)}{2\left(x-1\right)^2}\)
=> \(P=\frac{x^2+x+1}{2}\)
b, Ta có : \(\left|P\right|=\left|\frac{x^2+x+1}{2}\right|=\frac{\left|x^2+x+1\right|}{2}\)
=> \(\left|P\right|=\frac{\left|x^2+x+\frac{1}{4}+\frac{3}{4}\right|}{2}=\frac{\left|\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\right|}{2}\)
- Ta thấy : \(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)
=> \(\frac{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}{2}>0\)
=> \(\left|\frac{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}{2}\right|=\frac{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}{2}\)
=> \(\left|P\right|=\frac{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}{2}\)
Vậy \(\left|P\right|=P=\frac{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}{2}\)
