如图已知AD为△ABC的内角∠BAC的平分线,MN垂直平分AD于M点,交BC的延长线于点N,设BN=m,DN=n,CN=

2个回答

  • 先看看判别式.4nn-4mq=4(nn-mq)=4[(DC+CN)(DC+CN)-(BD+DC+CN)CN]

    =4[DC·DC+2DC·CN+CN·CN-BD·CN-DC·CN-CN·CN]=4[DC·DC+2DC·CN-BD·CN-DC·CN]

    =4[DC·DC+DC·CN-BD·CN]=4[DC·DC+(DC·CN-BD·CN)]=4[DC·DC-(BD-DC)·CN]

    ∵BD/DC=(BC+2CN)/(2CN)=BC/(2CN)+1,BC/(2CN)=BD/DC-1=(BD-DC)/DC,CN=BC·DC/[2(BD-DC)]=(BD+DC)DC/[2(BD-DC)],∴4nn-4mq=4[DC·DC-(BD-DC)·CN]=4{DC·DC-(BD-DC) (BD+DC)DC/[2(BD-DC)] }=4{DC·DC- (BD+DC)DC/2}=4DC(2DC-BD-DC)/2=2DC(DC-BD).又∵DC<BD,∴2DC(DC-BD)<0,即4nn-4mq<0,∴mxx-2nx+q=0无实数根.

    (请你仔细检查一下上面的过程或举一个具体例子检查一下这个结果)