\(\frac{2}{5\times7}+\frac{2}{7\times9}+\frac{2}{9\times11}+...+\frac{2}{13\times15}\)

\(=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{13}-\frac{1}{15}\)

\(=\frac{1}{5}-\frac{1}{15}\)

\(=\frac{2}{15}\)

= 2 x (1/ 5 x 7 + 1/ 7 x 9 + ... + 1/ 13 x 15)

= 2 x (1/5 - 1/7 + 1/7 - 1/9 + ... + 1/13 - 1/15)

= 2 x (1/5 - 1/15)

= 2 x (3/15 - 1/15)

= 2 x 2/15

= 4/15

Bạn Shino sai rồi nhé

Chọn D

\(\dfrac{7}{3}-\dfrac{13}{9}+3+\dfrac{4}{9}\\ =\left(\dfrac{7}{3}+3\right)+\left(\dfrac{4}{9}-\dfrac{13}{9}\right)\\ =\dfrac{16}{3}+\left(\dfrac{-9}{9}\right)\\ =\dfrac{16}{3}-1\\ =\dfrac{13}{3}\)

\(=\dfrac{7}{3}-\left(\dfrac{13}{9}-\dfrac{4}{9}\right)+3=\dfrac{7}{3}-1+3=\dfrac{7}{3}+2=\dfrac{13}{3}\)

14/6 + 1/9 + 19/13 + 17/9 + 7/13 + 4/6 = (14/6 + 4/6) + (1/9 + 17/9) + (19/13 + 7/13)

                                                            = 18/6 + 18/9 + 26/13

                                                             = 3 + 2 + 2

                                                               = 7 

\(\sqrt{13+\sqrt{30\sqrt{2+\sqrt{9+4\sqrt{2}}}}}=\sqrt{13+\sqrt{30\sqrt{2+\sqrt{8+2.2\sqrt{2}+1}}}}\) \(=\sqrt{13+\sqrt{30\sqrt{2+\sqrt{\left(2\sqrt{2}+1\right)^2}}}}=\sqrt{13+\sqrt{30\sqrt{2+2\sqrt{2}+1}}}\)\(=\sqrt{13+\sqrt{30\sqrt{\left(\sqrt{2}+1\right)^2}}}\)\(=\sqrt{13+\sqrt{30\left(\sqrt{2}+1\right)}}=\sqrt{13+\sqrt{30\sqrt{2}+30}}\)