là đấy

kết quả là2

a, |3x - 2| = x  => 3x - 2 = x hoặc 3x - 2 = -x    và x \(\ge\) 0 (vì |3x - 2| \(\ge\)0)

+) TH₁: 3x - 2 = x  => 3x - x = 2  => 2x = 2  => x = 1 (thỏa mãn)

+) TH₂: 3x - 2 = -x  => 3x + x = 2  => 4x = 2  => x = 0,5 (thỏa mãn)

Vậy x = 1 hoặc x = 0,5

b, |x - 2| = 2x + 1  => x - 2 = 2x + 1 và 2x + 1 \(\ge\)0 (vì |x - 2| \(\ge\)0)  =>  x \(\ge\) -0,5 hoặc x - 2 = -(2x + 1) và -(2x + 1) \(\ge\)0 (vì |x - 2| \(\ge\)0)  =>  x \(\ge\) -0,5

+)TH₁: x - 2 = 2x + 1  => x - 2x = 1 + 2   => -x = 3  => x = -3 (loại)

+)TH₂: x - 2 = -(2x + 1)  => x - 2 = -2x - 1  => x + 2x = -1 + 2  => 3x = 1  => x = \(\frac{1}{3}\)(chọn)

Vậy: x = \(\frac{1}{3}\)

\(\left(\dfrac{3x}{4}+5\right)-\left(\dfrac{2x}{3}-4\right)-\left(\dfrac{x}{6}+1\right)=\left(\dfrac{1}{3}+4\right)-\left(\dfrac{1}{3}x-3\right)\)

\(\Leftrightarrow\dfrac{3x}{4}-\dfrac{2x}{3}-\dfrac{x}{6}+5+4-1=\dfrac{13}{3}-\dfrac{1}{3}x+9\)

\(\Leftrightarrow\dfrac{9x-8x-2x}{12}+8=\dfrac{13-x}{3}+\dfrac{27}{3}\)

\(\Leftrightarrow\dfrac{-x}{12}+\dfrac{96}{12}=\dfrac{40-x}{3}\Leftrightarrow\dfrac{96-x}{12}=\dfrac{160-4x}{12}\)

\(\Rightarrow96-160=-4x+x\Leftrightarrow-64=-3x\Leftrightarrow x=\dfrac{64}{3}\)

a) 3x-1 = x-2

x = -1/2

b) 3x-1 = -x+2 

x = 3/4

3/4x + 5-(2/3x-4)-(1/6+1)=(1/3x+4)-(1/3x-3)

=3/4x+5-2/3x-4-1/6x+1=1/3x+4-1/3x-3

=-1/12=7

x=84

Đ/S...