\(3.\left|x-2\right|=x+1\)
\(\Rightarrow\left|x-2\right|=\dfrac{x+1}{3}\)
\(\Rightarrow\left[{}\begin{matrix}x-2=\dfrac{x+1}{3}\\x-2=-\left(\dfrac{x+1}{3}\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-2=\dfrac{x+1}{3}\\x-2=\dfrac{-x-1}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-2\right)3=x+1\\\left(x-2\right)3=-x-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x-6=x+1\\3x-6=-x-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x-x=1+6=7\\3x+x=-1+6=5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=7\\4x=5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=\dfrac{5}{4}\end{matrix}\right.\)
3x|x - 2| = x + 1
Xét các TH:
TH1: Với x \(\ge\) 2 ta có: 3x(x - 2) = x + 1
\(\Rightarrow\) 3x2 - 6x = x + 1
\(\Rightarrow\) 3x2 - 7x - 1 = 0
\(\Rightarrow\) 3(x2 - \(\dfrac{7}{3}\)x - \(\dfrac{1}{3}\)) = 0
\(\Rightarrow\) x2 - 2.\(\dfrac{7}{6}\)x + \(\dfrac{49}{36}\) - \(\dfrac{61}{36}\) = 0
\(\Rightarrow\) (x - \(\dfrac{7}{6}\))2 - \(\dfrac{61}{36}\) = 0
\(\Rightarrow\) (x - \(\dfrac{7}{6}-\dfrac{\sqrt{61}}{6}\))(x - \(\dfrac{7}{6}+\dfrac{\sqrt{61}}{6}\)) = 0
\(\Rightarrow\) x = \(\dfrac{7+\sqrt{61}}{6}\) (TM) và x = \(\dfrac{7-\sqrt{61}}{6}\) (KTM)
TH2: Với x < 2 ta có: 3x(2 - x) = x + 1
\(\Rightarrow\) 6x - 3x2 = x + 1
\(\Rightarrow\) 3x2 - 5x + 1 = 0
\(\Rightarrow\) 3(x2 - 2\(\dfrac{5}{6}\)x + \(\dfrac{25}{36}\) + \(\dfrac{11}{36}\)) = 0
\(\Rightarrow\) x2 - 2\(\dfrac{5}{6}\)x + \(\dfrac{25}{36}\) + \(\dfrac{11}{36}\) = 0
\(\Rightarrow\) (x - \(\dfrac{5}{6}\))2 + \(\dfrac{11}{36}\) = 0
Vì (x - \(\dfrac{5}{6}\))2 \(\ge\) 0 với mọi x
\(\Rightarrow\) (x - \(\dfrac{5}{6}\))2 + \(\dfrac{11}{36}\) > 0 với mọi x
\(\Rightarrow\) Không có giá trị của x nào thõa mãn
Vậy x = \(\dfrac{7+\sqrt{61}}{6}\)
Chúc bn học tốt! (Thử r nha ko có sai đâu :D)