1. giải pt
a. 5(x-3)-4=2(x-1)+7
b. \(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)
c.\(\frac{2\left(x+5\right)}{3}+\frac{x+12}{2}-\frac{5\left(x-2\right)}{6}=\frac{x}{3}+11\)
d. \(\frac{x-10}{1994}+\frac{x-8}{1996}+\frac{x-6}{1998}+\frac{x-4}{2000}+\frac{x-2}{2002}\)\(=\frac{x-2002}{2}+\frac{x-2000}{4}+\frac{x-1998}{6}+\frac{x-1996}{8}+\frac{x-1994}{10}\)
e. \(\frac{x-85}{15}+\frac{x-74}{13}+\frac{x-67}{11}+\frac{x-64}{9}=10\)
\( a)5\left( {x - 3} \right) - 4 = 2\left( {x - 1} \right) + 7\\ \Leftrightarrow 5x - 15 - 4 = 2x - 2 + 7\\ \Leftrightarrow 5x - 19 = 2x + 5\\ \Leftrightarrow 5x - 2x = 5 + 19\\ \Leftrightarrow 3x = 24\\ \Leftrightarrow x = 8\\ b)\dfrac{{8x - 3}}{4} - \dfrac{{3x - 2}}{2} = \dfrac{{2x - 1}}{2} + \dfrac{{x + 3}}{4}\\ \Leftrightarrow 8x - 3 - \left( {3x - 2} \right).2 = \left( {2x - 1} \right).2 + x + 3\\ \Leftrightarrow 8x - 3 - 6x + 4 = 4x - 2 + x + 3\\ \Leftrightarrow 2x + 1 = 5x + 1\\ \Leftrightarrow 2x - 5x = 0\\ \Leftrightarrow - 3x = 0\\ \Leftrightarrow x = 0 \)
\( c)\dfrac{{2\left( {x + 5} \right)}}{3} + \dfrac{{x + 12}}{2} - \dfrac{{5\left( {x - 2} \right)}}{6} = \dfrac{x}{3} + 11\\ \Leftrightarrow 4\left( {x + 5} \right) + 3\left( {x + 12} \right) - \left[ {5\left( {x - 2} \right)} \right] = 2x + 66\\ \Leftrightarrow 4x + 20 + 3x + 36 - 5x + 10 = 2x + 66\\ \Leftrightarrow 2x + 66 = 2x + 66\\ \Leftrightarrow 0x = 0\left( {VSN} \right)\\ \Leftrightarrow x = 0 \)
\(d)\dfrac{x-10}{1994}+\dfrac{x-8}{1996}+\dfrac{x-6}{1998}+\dfrac{x-4}{2000}+\dfrac{x-2}{2002}=\dfrac{x-2002}{2}+\dfrac{x-2000}{4}+\dfrac{x-1998}{6}+\dfrac{x-1996}{8}+\dfrac{x-1994}{10}\\ \Leftrightarrow \dfrac{x-10}{1994}-1+\dfrac{x-8}{1996}-1+\dfrac{x-6}{1998}-1+\dfrac{x-4}{2000}-1+\dfrac{x-2}{2002}-1=\dfrac{x-2002}{2}-1+\dfrac{x-2000}{4}-1+\dfrac{x-1998}{6}-1+\dfrac{x-1996}{8}-1+\dfrac{x-1994}{10}-1\\ \Leftrightarrow \dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}+\dfrac{x-2004}{2000}\dfrac{x-2004}{2002}=\dfrac{x-2004}{2}+\dfrac{x-2004}{4}+\dfrac{x-2004}{6}+\dfrac{x-2004}{8}+\dfrac{x-2004}{10}\\ \Leftrightarrow \dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}+\dfrac{x-2004}{2000}\dfrac{x-2004}{2002}-\dfrac{x-2004}{2}-\dfrac{x-2004}{4}-\dfrac{x-2004}{6}-\dfrac{x-2004}{8}-\dfrac{x-2004}{10}=0\\ \Leftrightarrow \left(x-2004\right)\left(\dfrac{1}{1994}+\dfrac{1}{1996}+\dfrac{1}{1998}+\dfrac{1}{2000}+\dfrac{1}{2002}-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{6}-\dfrac{1}{8}-\dfrac{1}{10}=0\right)\\ \Leftrightarrow x-2004=0\\ \Leftrightarrow x=2004\)
a, 5(x-3)-4=2(x-1)+7
<=>\(5x-15-4=2x-2+7\)
\(\Leftrightarrow5x-2x=15+4-2+7\)
\(\Leftrightarrow3x=24\)
\(\Leftrightarrow x=8\)
b, \(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)
\(\Leftrightarrow\frac{8x-3}{4}-\frac{2\left(3x-2\right)}{4}=\frac{2\left(2x-1\right)}{4}+\frac{x+3}{4}\)
\(\Rightarrow8x-3-6x+4=4x-2+x+3\)
\(\Leftrightarrow8x-6x-4x-x=3+4-2+3\)
\(\Leftrightarrow-3x=8\)
\(\Leftrightarrow x=\frac{-8}{3}\)
c,\(\frac{2\left(x+5\right)}{3}+\frac{x+12}{2}-\frac{5\left(x-2\right)}{6}=\frac{x}{3}+11\)
<=>\(\frac{4\left(x+5\right)}{6}+\frac{3\left(x+12\right)}{6}-\frac{5\left(x-2\right)}{6}=\frac{2x}{6}+\frac{66}{6}\)
\(\Leftrightarrow\frac{4x+20}{6}+\frac{3x+36}{6}-\frac{5x-10}{6}=\frac{2x}{6}+\frac{66}{6}\)
\(\Rightarrow4x+20+3x+36-5x+10=2x+66\)
\(\Leftrightarrow4x+3x-5x-2x=66-20-36-10\)
\(\Leftrightarrow0=0\)
