Question

Tim x : x^3-x^2-x+1=0

Chung minh 2x-2x^2-1<0 voi moi x

Giup vs

Answer 1

Bài 1:

\(x^3-x^2-x+1=0\)

\(\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x^2-1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)^2\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

Vậy x = 1 hoặc x = -1

Bài 2:
\(2x-2x^2-1=-2\left(x^2-x+\dfrac{1}{2}\right)\)

\(=-2\left(x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{4}\right)\)

\(=-2\left(x^2-\dfrac{1}{2}\right)^2-\dfrac{1}{2}< 0\)

\(\Rightarrowđpcm\)