\(a,8\left(x-1\right)-4=6\left(x+2\right)-2\\ \Leftrightarrow8x-8-4=6x+12-2\\ \Leftrightarrow8x-8-4-6x-12+2=0\\ \Leftrightarrow2x-22=0\\ \Leftrightarrow x=11\)
Vậy `S={11}`
\(b,\dfrac{x}{2}-\dfrac{2x+1}{3}=\dfrac{2x-1}{5}\\ \Leftrightarrow b,15x-10\left(2x+1\right)=6\left(2x-1\right)\\ \Leftrightarrow15x-20x-10=12x-6\\ \Leftrightarrow-5x-10-12x+6-0\\ \Leftrightarrow-17x-4=0\\ \Leftrightarrow x=-\dfrac{4}{17}\)
Vậy `S={-4/17}`
\(c,ĐKXĐ:x\ne\pm2\\ \dfrac{x+1}{x-2}-\dfrac{x}{x+2}+\dfrac{x^2+2}{x^2-4}=0\\ \Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\dfrac{x^2+2}{\left(x+2\right)\left(x-2\right)}=0\\ \Leftrightarrow\dfrac{x^2+3x+2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x^2-2x}{\left(x+2\right)\left(x-2\right)}+\dfrac{x^2+2}{\left(x+2\right)\left(x-2\right)}=0\\ \Leftrightarrow\dfrac{x^2+3x+2-x^2+2x+x^2+2}{\left(x-2\right)\left(x+2\right)}=0\\ \Rightarrow x^2+5x+4=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\left(tm\right)\\x=-4\left(tm\right)\end{matrix}\right.\)
Vậy `S={-1;4}`
a) 8x - 8 -4 = 6x +12 -2
8x - 6x = 10 + 4
2x = 14
x = 7
b) \(\dfrac{15x}{30}\) - \(\dfrac{10\left(2x+1\right)}{30}\) = \(\dfrac{6\left(2x-1\right)}{30}\)
15x - 20x - 10 = 12x - 6
-5x -12x = 10 - 6
-17x =4
x = -4/17
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