\(a,1+\dfrac{2x-5}{6}=\dfrac{3x}{4}\\ \Leftrightarrow\dfrac{12}{12}+\dfrac{2\left(2x-5\right)}{12}-\dfrac{9x}{12}=0\\ \Leftrightarrow\dfrac{12+4x-10-9x}{12}=0\\ \Leftrightarrow-5x+2=0\\ \Leftrightarrow x=\dfrac{2}{5}\)
\(b,\dfrac{5x-4}{2}=\dfrac{16x+1}{7}\\ \Leftrightarrow7\left(5x-4\right)=2\left(16x+1\right)\\ \Leftrightarrow35x-28=32x+2\\ \Rightarrow3x=30\\ \Leftrightarrow x=10\)
\(c,\dfrac{2\left(3x+1\right)+1}{4}-5=\dfrac{2\left(3x-1\right)}{5}-\dfrac{3x+2}{10}\\ \Leftrightarrow c,\dfrac{5\left(6x+3\right)}{20}-\dfrac{100}{20}-\dfrac{8\left(3x-1\right)}{20}+\dfrac{2\left(3x+2\right)}{20}=0\\ \Leftrightarrow30x+15-100-24x+8+6x+4=0\\ \Leftrightarrow12x-73=0\\ \Leftrightarrow x=\dfrac{73}{12}\)
\(d,\dfrac{x+1}{2005}+\dfrac{x+1}{2003}=\dfrac{x+1}{2001}+\dfrac{x+1}{1999}\\ \Leftrightarrow\dfrac{x+1}{2005}+\dfrac{x+1}{2003}-\dfrac{x+1}{2001}-\dfrac{x+1}{1999}=0\\ \Leftrightarrow\left(x+1\right)\left(\dfrac{1}{2005}+\dfrac{1}{2003}-\dfrac{1}{2001}-\dfrac{1}{1999}\right)=0\)
\(\Leftrightarrow x+1=0\) (vì \(\dfrac{1}{2005}+\dfrac{1}{2003}-\dfrac{1}{2001}-\dfrac{1}{1999}\ne0\))
\(\Leftrightarrow x=-1\)
a) 1 + \(\dfrac{2x-5}{6}\) = \(\dfrac{3x}{4}\)
⇔ \(\dfrac{12+4x-10}{12}\) = \(\dfrac{9x}{12}\)
⇔ 12 + 4x - 10 = 9x
⇔ 4x - 9x = 10 - 12
⇔ -5x = -2
⇔ x = \(\dfrac{2}{5}\)
b) \(\dfrac{5x-4}{2}\) = \(\dfrac{16x+1}{7}\)
⇔ \(\dfrac{35x-28}{14}\) = \(\dfrac{32x+2}{14}\)
⇔ 35x - 28 = 32x + 2
⇔ 35x - 32x = 28 + 2
⇔ 3x = 30
⇔ x = 30 : 3 = 10
