2 в ст 1/6~1.12246204830...
60/(6^log(5))=17.1493185520
замена cos^2x=t x=0 t=1 x=П/4 t=1/2
-2cosxsinxdx=dt -sin2xdx=dt
∫-dt/√(t^2+4)=это табличный интеграл = -arsht/2
ответ -arsh(1/4)+arsh(1/2)
√28(√14-√7)-2√98=√4*√7(√7*√2-√7)-2*√2*√49=√2²*√7*√7(√2-1)-2*√2*√7²=2*√7²(√2-1)-2*7*√2=2*7*(√2-1)-14√2=14√2-14-14√2=-14
5ax-6bx-5ay+6by = 5ax-5ay+6by-6bx = 5а(х-у) + 6b(y-x) = 5а(х-у) - 6b(x-y) = (5a-6b)(x-y)