Решите уравнение (с модулем) |x-2014|=2015

Х-2014=2015; или х-2014=-2015; х=2015+2014; или х=-2015+2014; х=4029 или х=-1;

Verse 1 a a5 it seems the more we talk asus4 a the less i have to say f#m d5 let's put our differences aside a a5 i wanted to make you proud asus4 a but i just got in your way f#m d5 i found a place that i can hide pre-chorus e now everything is changing d5 a but i still feel the same e are we running out of time chorus a what do i have to do d5 to try to make you see e that this is who i am d and it's all that i can be verse 2 a a5 i tried to find myself asus4 a looking inside your eyes f#m d5 you were all that i wanted to be a a5 there must be something else asus4 a behind all the lies f#m d5 that you wanted me to believe pre-chorus e now everything is changing d5 a but i still feel the same e are we running out of time chorus a what do i have to do d5 to try to make you see e that this is who i am d and it's all that i can be a what do i have to do d5 to try to make you see e trying to be like you d a e isn't good enough for me bridge d a e and i won't let you go d a e and i won't let you down d a e i won примерно вот так; ) по -крайней мере я так бы играла) меня, если не сложно)

Ac=16  см db=10 см so=14 см as-?   bs-?   cs-?   ds-? рассмотрим  δasc  высота so делит основание ac пополам  ⇒  δasc -равнобедренный  ⇒  as=sc, аналогично с δbsd, bs=sd. рассмотрим  δaso - прямоугольный  (so⊥ao) по теореме пифагора: as²=ao²+so² ao=ac/2 (о - центр ас)  ao=8 см as²=8²+14²=64+196=260 см² as=√260 см as=sc=√260 см рассмотрим  δbso - прямоугольный (so⊥bo) по теореме пифагора: bs²=bo²+so²bo=bd/2  (о - центр bd)bo=10/2  см=5  см  bs²=5²+14²=25+196=219  см² bs=√219 см bs=sd=√219 см ответ:   as=sc=√260 см;   bs=sd=√219 см