来自傅清祥的问题
已知an-bm≠0,a≠0,ax2+bx+c=0,mx2+nx+p=0,求证:(cm-ap)2=(bp-cn)(an-bm).
已知an-bm≠0,a≠0,ax2+bx+c=0,mx2+nx+p=0,求证:(cm-ap)2=(bp-cn)(an-bm).
1回答
2020-08-04 11:54
已知an-bm≠0,a≠0,ax2+bx+c=0,mx2+nx+p=0,求证:(cm-ap)2=(bp-cn)(an-bm).
已知an-bm≠0,a≠0,ax2+bx+c=0,mx2+nx+p=0,求证:(cm-ap)2=(bp-cn)(an-bm).
证明:∵an-bm≠0
∴方程ax2+bx+c=0和方程mx2+nx+p=0有相等的根.
方程ax2+bx+c=0可化为x2+ba