Question

пожалуйста!!!!!((( оч надо

Answer 1

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