5. (x - 3)(1 - x) + (x - 2)2 = (1 - x)2 - 2(1 + x)
<=> (x - 3) - x(x - 3) + (x - 2)x + 2(x - 2) = (1 - x) - x(1 - x) - 2 - 2x
<=> x - 3 - x2 + 3x + x2 - 2x + 2x - 4 = 1 - x - x + x2 - 2 - 2x
<=> x - x2 + 3x + x2 - 2x + 2x + x + x - x2 + 2x = 1 - 2 + 3 + 4
<=> 7x = 6
<=> x = \(\dfrac{6}{7}\) (thỏa mãn)
@Nguyễn Hoàng Vũ
4. 1\(\dfrac{1}{2}\).(x - 2) - \(\dfrac{x-5}{3}\) = 3\(\dfrac{1}{3}\).(1 - 2x) - \(\dfrac{5.\left(x+1\right)}{6}\)
<=> \(\dfrac{3}{2}\).(x - 2) - \(\dfrac{x-5}{3}\) = \(\dfrac{10}{3}\).(1 - 2x) - \(\dfrac{5x+5}{6}\)
<=> \(\dfrac{3}{2}x-3-\dfrac{x}{3}+\dfrac{5}{3}=\dfrac{10}{3}-\dfrac{20}{3}x-\dfrac{5x}{6}-\dfrac{5}{6}\)
<=> \(\dfrac{3}{2}x-\dfrac{x}{3}+\dfrac{20}{3}x-\dfrac{5x}{6}=\dfrac{10}{3}-\dfrac{5}{6}-3+\dfrac{5}{3}\)
<=> 7x = \(\dfrac{7}{6}\)
<=> x = \(\dfrac{1}{6}\)
@Nguyễn Hoàng Vũ
1. a, 3x + |x - 2| = 8
<=> |x - 2| = 8 - 3x
Xét 2 TH :
TH1: x - 2 = 8 - 3x
<=> x + 3x = 8 + 2
<=> 4x = 10
<=> x = \(\dfrac{5}{2}\) (thỏa mãn)
TH2: x - 2 = -(8 - 3x)
<=> x - 2 = -8 + 3x
<=> -2 + 8 = 3x - x
<=> 6 = 2x
<=> x = 3 (thỏa mãn)
b, 5 - |x - 1| = 4
<=> |x - 1| = 1
<=> \(\left[{}\begin{matrix}x-1=1\\x-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\) (thỏa mãn)
@Nguyễn Hoàng Vũ