Question

[Wanna.B Linah]

Cho biểu thức P=\(\frac{x-1}{2}:\left(\frac{x^2+2}{x^3-1}+\frac{x}{x^2+x+1}+\frac{1}{1-x}\right)\)

a, Rút gọn P

b, So sánh P với /P/

c, Tìm giá trị nhỏ nhất của P

Answer 1

[Natsu Dragneel]

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

Answer 2

[Nguyễn Ngọc Lộc]

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 \(\left|P\right|=P=\frac{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}{2}\)