Question

2c-1/4c+2 + 2c+1/6c-3 + 4c/4c²-1​

Answer 1

$\dfrac{2c-1}{4c+2}+\dfrac{2c+1}{6c-3}+\dfrac{4c}{4c^2-1}=\dfrac{2c-1}{2(2c+1)}+\dfrac{2c+1}{3(2c-1)}+\dfrac{4c}{(2c-1)(2c+1)}=$

$=\dfrac{3(2c-1)(2c-1)+2(2c+1)(2c+1)+4c\cdot6}{2\cdot3\cdot(2c-1)(2c+1)}=$

$=\dfrac{3(4c^2-4c+1)+2(4c^2+4c+1)+24c}{6(2c-1)(2c+1)}=$

$=\dfrac{12c^2-12c+3+8c^2+8c+2+24c}{6(2c-1)(2c+1)}=\dfrac{20c^2+20c+5}{6(2c-1)(2c+1)}=\dfrac{5(4c^2+4c+1)}{6(2c-1)(2c+1)}=$

$=\dfrac{5(2c+1)^2}{6(2c-1)(2c+1)}=\dfrac{5(2c+1)}{6(2c-1)}=\dfrac{10c+5}{12c-6}$