Question

giải các phương trình sau :

\(\frac{x+1}{1}+\frac{2x+3}{3}+\frac{3x+5}{5}+...+\frac{20x+39}{39}=22+\frac{4}{3}+\frac{6}{5}+...+\frac{40}{39}\)

 

Answer 1

 <=>\(\left(\frac{x}{1}+\frac{2x}{3}+\frac{3x}{5}+...+\frac{20x}{39}\right)+\left(\frac{1}{1}+\frac{3}{3}+\frac{5}{5}+...+\frac{39}{39}\right)=20+2.\left(\frac{1}{1}+\frac{2}{3}+\frac{3}{5}+...+\frac{20}{39}\right)\)<=> 

\(\left(\frac{1}{1}+\frac{2}{3}+\frac{3}{5}+...+\frac{20}{39}\right).x+20=20+2.\left(\frac{1}{1}+\frac{2}{3}+\frac{3}{5}+...+\frac{20}{39}\right)\)

<=> \(\left(\frac{1}{1}+\frac{2}{3}+\frac{3}{5}+...+\frac{20}{39}\right).x=2.\left(\frac{1}{1}+\frac{2}{3}+\frac{3}{5}+...+\frac{20}{39}\right)\)<=> x = 2 

Answer 2

(x+1) / 1 + (2x+3) / 3 + (3x+5) / 5+ ... + (20x + 39) / 39

= 22 + 4 /3 + 6 / 5 +... + 40 /39 
<=> x+ 1+ 2x / 3 +1 + 3x / 5+1+...+20x / 39+1 = 22+4 / 3+6 / 5+8 / 7+...+38 / 37+40 / 39 
<=> (1+2 / 3+3 / 5+4 / 7+...+19 / 37+20 / 39)x + 20 = 22+4/3+6/5+8/7+...+38/37+40/39 
<=> (1+2/3+3/5+4/7+...+19/37+20/39)x = 2(1 + 2/3 + 3/5 + 4/7 +...+ 19/37 + 20/39) 
<=> x = 2

Answer 3

 [2x-3]=3x-5