Lời Giải
khai triển (−bsin(t)+bcos(t)+1−b)3
Lời Giải
−3b3sin(t)+3b3cos(t)−b3+3b2+b3cos3(t)+3b2cos2(t)−3b3cos2(t)−b3sin3(t)+3b2sin2(t)−b2sin(2t)+b3sin(2t)−3bsin(t)+3bcos(t)−b3cos2(t)sin(t)+b3sin2(t)cos(t)−3b3sin2(t)+6b2sin(t)−6b2cos(t)+b3sin(2t)sin(t)−b3cos(t)sin(2t)−4b2cos(t)sin(t)−3b+4b3cos(t)sin(t)+1
Các bước giải pháp
(−bsin(t)+bcos(t)+1−b)3
Rút gọn (−bsin(t)+bcos(t)+1−b)3:(−bsin(t)+bcos(t)+1−b)(−bsin(t)+bcos(t)+1−b)(−bsin(t)+bcos(t)+1−b)
=(−bsin(t)+bcos(t)+1−b)(−bsin(t)+bcos(t)+1−b)(−bsin(t)+bcos(t)+1−b)
Mở rộng (−bsin(t)+bcos(t)+1−b)(−bsin(t)+bcos(t)+1−b):b2+b2cos2(t)+b2sin2(t)+2b2sin(t)−2b2cos(t)−b2sin(2t)+2bcos(t)−2b−2bsin(t)+1
=(b2+b2cos2(t)+b2sin2(t)+2b2sin(t)−2b2cos(t)−b2sin(2t)+2bcos(t)−2b−2bsin(t)+1)(−bsin(t)+bcos(t)+1−b)
Phân phối dấu ngoặc đơn=b2(−bsin(t))+b2bcos(t)+b2⋅1+b2(−b)+b2cos2(t)(−bsin(t))+b2cos2(t)bcos(t)+b2cos2(t)⋅1+b2cos2(t)(−b)+b2sin2(t)(−bsin(t))+b2sin2(t)bcos(t)+b2sin2(t)⋅1+b2sin2(t)(−b)+2b2sin(t)(−bsin(t))+2b2sin(t)bcos(t)+2b2sin(t)⋅1+2b2sin(t)(−b)−2b2cos(t)(−bsin(t))−2b2cos(t)bcos(t)−2b2cos(t)⋅1−2b2cos(t)(−b)−b2sin(2t)(−bsin(t))−b2sin(2t)bcos(t)−b2sin(2t)⋅1−b2sin(2t)(−b)+2bcos(t)(−bsin(t))+2bcos(t)bcos(t)+2bcos(t)⋅1+2bcos(t)(−b)−2b(−bsin(t))−2bbcos(t)−2b⋅1−2b(−b)−2bsin(t)(−bsin(t))−2bsin(t)bcos(t)−2bsin(t)⋅1−2bsin(t)(−b)+1⋅(−bsin(t))+1⋅bcos(t)+1⋅1+1⋅(−b)
Rút gọn b2(−bsin(t))+b2bcos(t)+b2⋅1+b2(−b)+b2cos2(t)(−bsin(t))+b2cos2(t)bcos(t)+b2cos2(t)⋅1+b2cos2(t)(−b)+b2sin2(t)(−bsin(t))+b2sin2(t)bcos(t)+b2sin2(t)⋅1+b2sin2(t)(−b)+2b2sin(t)(−bsin(t))+2b2sin(t)bcos(t)+2b2sin(t)⋅1+2b2sin(t)(−b)−2b2cos(t)(−bsin(t))−2b2cos(t)bcos(t)−2b2cos(t)⋅1−2b2cos(t)(−b)−b2sin(2t)(−bsin(t))−b2sin(2t)bcos(t)−b2sin(2t)⋅1−b2sin(2t)(−b)+2bcos(t)(−bsin(t))+2bcos(t)bcos(t)+2bcos(t)⋅1+2bcos(t)(−b)−2b(−bsin(t))−2bbcos(t)−2b⋅1−2b(−b)−2bsin(t)(−bsin(t))−2bsin(t)bcos(t)−2bsin(t)⋅1−2bsin(t)(−b)+1⋅(−bsin(t))+1⋅bcos(t)+1⋅1+1⋅(−b):−3b3sin(t)+3b3cos(t)−b3+3b2+b3cos3(t)+3b2cos2(t)−3b3cos2(t)−b3sin3(t)+3b2sin2(t)−b2sin(2t)+b3sin(2t)−3bsin(t)+3bcos(t)−b3cos2(t)sin(t)+b3sin2(t)cos(t)−3b3sin2(t)+6b2sin(t)−6b2cos(t)+b3sin(2t)sin(t)−b3cos(t)sin(2t)−4b2cos(t)sin(t)−3b+4b3cos(t)sin(t)+1
=−3b3sin(t)+3b3cos(t)−b3+3b2+b3cos3(t)+3b2cos2(t)−3b3cos2(t)−b3sin3(t)+3b2sin2(t)−b2sin(2t)+b3sin(2t)−3bsin(t)+3bcos(t)−b3cos2(t)sin(t)+b3sin2(t)cos(t)−3b3sin2(t)+6b2sin(t)−6b2cos(t)+b3sin(2t)sin(t)−b3cos(t)sin(2t)−4b2cos(t)sin(t)−3b+4b3cos(t)sin(t)+1