已知A=-y²+ay-1,B=2y²+3ay-2y-1,且多项式2A+B的值与字母y的取值无关,求a
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/14 18:38:18
![已知A=-y²+ay-1,B=2y²+3ay-2y-1,且多项式2A+B的值与字母y的取值无关,求a](/uploads/image/z/5320978-34-8.jpg?t=%E5%B7%B2%E7%9F%A5A%3D%EF%BC%8Dy%26%23178%3B%EF%BC%8Bay%EF%BC%8D1%2CB%3D2y%26%23178%3B%EF%BC%8B3ay%EF%BC%8D2y%EF%BC%8D1%2C%E4%B8%94%E5%A4%9A%E9%A1%B9%E5%BC%8F2A%EF%BC%8BB%E7%9A%84%E5%80%BC%E4%B8%8E%E5%AD%97%E6%AF%8Dy%E7%9A%84%E5%8F%96%E5%80%BC%E6%97%A0%E5%85%B3%2C%E6%B1%82a)
xQJ@AR&C A$I ]iBDQ0Lt_f&!n{Ϲs
T\]tyI_Y]+BM!@"<=bB.ey:dwOx4b{?nA#q'(|R>!XIJ<'HE/uL1}L:̳ߙ&5bՕ b2y:tO~/X]\A'M4ҶéDҔ
-OR&F?tsm@J4~/%ЖćS
已知A=-y²+ay-1,B=2y²+3ay-2y-1,且多项式2A+B的值与字母y的取值无关,求a
已知A=-y²+ay-1,B=2y²+3ay-2y-1,且多项式2A+B的值与字母y的取值无关,求a
已知A=-y²+ay-1,B=2y²+3ay-2y-1,且多项式2A+B的值与字母y的取值无关,求a
2A+B
=2(-y²+ay-1)+(2y²+3ay-2y-1)
=-2y²+2ay-2+2y²+3ay-2y-1
=(5a-2)y-3
因它的取值与y无关,可得含y的项的系数为0,于是有:
5a-2=0
解得:a=2/5
2A+B
=2(-y²+ay-1)+(2y²+3ay-2y-1)
=-2y²+2ay-2+2y²+3ay-2y-1
=5ay-2y-3=(5a-2)y-3
因为多项式2A+B的值与字母y的取值无关
所以5a-2=0
所以a=2/5