Question

1/ 3 x (x  -  \(\dfrac{1}{2}\)) - 3 x (x - \(\dfrac{1}{3}\)) - (x - \(\dfrac{1}{2}\)) = 0

2/ \(\dfrac{-2x+1}{3}\) + \(\dfrac{1}{2}\) = \(\dfrac{-1}{3}\)

Answer 1

1) \(3x\left(x-\dfrac{1}{2}\right)-3x\left(x-\dfrac{1}{3}\right)-\left(x-\dfrac{1}{2}\right)=0\)

\(\Rightarrow3x^2-\dfrac{3}{2}x-3x^2+x-x+\dfrac{1}{2}=0\)

\(\Rightarrow\dfrac{3}{2}x=\dfrac{1}{2}\Rightarrow x=\dfrac{1}{3}\)

2) \(\dfrac{-2x+1}{3}+\dfrac{1}{2}=-\dfrac{1}{3}\)

\(\Rightarrow\dfrac{-4x+2}{6}=-\dfrac{5}{6}\)

\(\Rightarrow-4x+2=-5\)

\(\Rightarrow-4x=-7\Rightarrow x=\dfrac{7}{4}\)

Answer 2

1. 3x\(\left(x-\dfrac{1}{2}\right)\) - 3x\(\left(x-\dfrac{1}{3}\right)\) - \(\left(x-\dfrac{1}{2}\right)\) = 0

<=> 3x² - \(\dfrac{3x}{2}\) - 3x² + x - x + \(\dfrac{1}{2}\) = 0

<=> \(\dfrac{6x^2}{2}-\dfrac{3x}{2}-\dfrac{6x^2}{2}+\dfrac{2x}{2}-\dfrac{2x}{2}+\dfrac{1}{2}=0\) 

<=> 6x² - 3x - 6x² + 2x - 2x + 1 = 0

<=> -3x + 1 = 0

<=> -3x = -1

<=> x = \(\dfrac{1}{3}\)

Answer 3

2. \(\dfrac{-2x+1}{3}+\dfrac{1}{2}=\dfrac{-1}{3}\)

<=> \(\dfrac{2\left(-2x+1\right)}{6}+\dfrac{3}{6}=\dfrac{-2}{6}\)

<=> 2(-2x + 1) + 3 = -2

<=> -4x + 2 + 3 = -2

<=> 2 + 3 + 2 = 4x

<=> 7 = 4x

<=> x = \(\dfrac{7}{4}\)

Answer 4

1: Ta có: \(3x\left(x-\dfrac{1}{2}\right)-3x\left(x-\dfrac{1}{3}\right)-\left(x-\dfrac{1}{2}\right)=0\)

\(\Leftrightarrow3x^2-\dfrac{3}{2}x-3x^2+x-x+\dfrac{1}{2}=0\)

\(\Leftrightarrow x\cdot\dfrac{-3}{2}=-\dfrac{1}{2}\)

hay \(x=\dfrac{1}{2}:\dfrac{3}{2}=\dfrac{1}{3}\)

2: Ta có: \(\dfrac{-2x+1}{3}+\dfrac{1}{2}=-\dfrac{1}{3}\)

\(\Leftrightarrow-4x+2+3=-2\)

\(\Leftrightarrow-4x=-7\)

hay \(x=\dfrac{7}{4}\)