在RT三角形ABC中,斜边AB=5,两直角边BC,AC之长是一元二次方程x²-(2m-1)+4(m-1)=0的两根求实数m的值。2.设关于x²-2mx+m²-2m+1=0的两实数跟为x1,x2已知x1\x2+x2\x1=7,求m的值3.已知关于x²-(2k+3)x-2k&
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/28 12:52:43
![在RT三角形ABC中,斜边AB=5,两直角边BC,AC之长是一元二次方程x²-(2m-1)+4(m-1)=0的两根求实数m的值。2.设关于x²-2mx+m²-2m+1=0的两实数跟为x1,x2已知x1\x2+x2\x1=7,求m的值3.已知关于x²-(2k+3)x-2k&](/uploads/image/z/1105335-63-5.jpg?t=%E5%9C%A8RT%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%AD%2C%E6%96%9C%E8%BE%B9AB%3D5%2C%E4%B8%A4%E7%9B%B4%E8%A7%92%E8%BE%B9BC%2CAC%E4%B9%8B%E9%95%BF%E6%98%AF%E4%B8%80%E5%85%83%E4%BA%8C%E6%AC%A1%E6%96%B9%E7%A8%8Bx%26%23178%3B-%282m-1%29%2B4%28m-1%29%3D0%E7%9A%84%E4%B8%A4%E6%A0%B9%E6%B1%82%E5%AE%9E%E6%95%B0m%E7%9A%84%E5%80%BC%E3%80%822.%E8%AE%BE%E5%85%B3%E4%BA%8Ex%26%23178%3B-2mx%2Bm%26%23178%3B-2m%2B1%3D0%E7%9A%84%E4%B8%A4%E5%AE%9E%E6%95%B0%E8%B7%9F%E4%B8%BAx1%2Cx2%E5%B7%B2%E7%9F%A5x1%5Cx2%2Bx2%5Cx1%3D7%2C%E6%B1%82m%E7%9A%84%E5%80%BC3.%E5%B7%B2%E7%9F%A5%E5%85%B3%E4%BA%8Ex%26%23178%3B-%282k%2B3%29x-2k%26)
xT[OQ+)=nەMhb "7#W"RDPZ}_pRXA_|ڙo|͙i[T}vGF_.O]{yQ Ba&VӍ3}u_-崉'jeFZ_.܃ݼPT[UY?ӊ' -W#zT8T+sDV-hxS-U(D;\QpB
W01P&