已知x2+xy=4,xy+y2=12,求代数式x2-y2与x2+2xy+y2的值各为多少格式正确

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/17 21:23:44
已知x2+xy=4,xy+y2=12,求代数式x2-y2与x2+2xy+y2的值各为多少格式正确
xn0_XQNQ9:Yjl!A;vBR#'NF^r2wr,Kn[.'qG$$N=(}D͡,^0:PO/,j1Uj&1Xך+գamҶIL#m}o] <'"&+v!L{O; yni6T`'Q:!ERQQF_4%k,P3\g(6wOlۄO&g4= #J

已知x2+xy=4,xy+y2=12,求代数式x2-y2与x2+2xy+y2的值各为多少格式正确
已知x2+xy=4,xy+y2=12,求代数式x2-y2与x2+2xy+y2的值各为多少
格式正确

已知x2+xy=4,xy+y2=12,求代数式x2-y2与x2+2xy+y2的值各为多少格式正确
X2+xy-(xy+y2)=4-12
x2+xy-xy-y2=-8
x2-y2=-8
x2+xy+xy+y2=4+12
x2+2xy+y2=16

x^2+xy=4----(1)
xy+y^2=12----(2)
(1)-(2)
(x^2+xy)-(xy+y^2)=4-12
x^2-y^2=-8
(1)+(2)
(x^2+xy)+(xy+y^2)=4+12
x^2+2xy+y^2=16

x2+xy=4 xy+y2=12
两式相加为 (x2+xy=4)+(xy+y2=12)=x2+2xy+y2=16
两式相减为 (x2+xy=4)-(xy+y2=12)=x2-y2=-8