Giải pt `[x+(2(3-x)]/(5)]/[12] = 1 + [1- ((9-2x))/12]/[5]`

2 bình luận về “Giải pt `[x+(2(3-x)]/(5)]/[12] = 1 + [1- ((9-2x))/12]/[5]`”

1. ${\displaystyle \frac{2+{\displaystyle \frac{2(3-x)}{5}}}{12}}=1+{\displaystyle \frac{1-{\displaystyle \frac{9-2x}{12}}}{5}}$

   <=> ${\displaystyle \frac{{\displaystyle \frac{5x+6-2x}{5}}}{12}}=1+{\displaystyle \frac{{\displaystyle \frac{12-9+2x}{12}}}{5}}$

   <=> ${\displaystyle \frac{{\displaystyle \frac{3x+6}{5}}}{12}}=1+{\displaystyle \frac{{\displaystyle \frac{3+2x}{12}}}{5}}$

   <=> ${\displaystyle \frac{3x+6}{60}}=1+{\displaystyle \frac{3+2x}{60}}$

   <=> ${\displaystyle \frac{3x+6}{60}}={\displaystyle \frac{60}{60}}+{\displaystyle \frac{3+2x}{60}}$

   <=>3x+6=60+3+2x

   <=>3x-2x=63-6

   <=>x=57

   Vậy x=57

   Trả lời
2. Giải đáp:

    \frac{x+\frac{2.(3-x)}{5}}{12}=1+\frac{1-\frac{9-2x}{12}}{5}

   <=>(x+\frac{6-2x}{5}):12=1+(1-\frac{9-2x}{12}):5

   <=>(\frac{5x}{5}+\frac{6-2x}{5}).\frac{1}{12}=1+(\frac{12}{12}-\frac{9-2x}{12}). 1/5

   <=>(\frac{5x+6-2x}{5}).\frac{1}{12}=1+(\frac{12-9+2x}{12}).\frac{1}{5}

   <=>\frac{3x+6}{5}.\frac{1}{12}=1+\frac{2x+3}{12}.\frac{1}{5}

   <=>\frac{3x+6}{60}=60/60+\frac{2x+3}{60}

   <=>3x+6=60+2x+3

   <=>3x-2x=63-6

   <=>x=57

   Vậy S={57}

   Trả lời