已知方程x^2+px+q=0的两根是a,b.求证:一元二次方程qx^2+p(1+q)x+(1+q)^2=0的根为a+1/b和b+1/a如题.
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![已知方程x^2+px+q=0的两根是a,b.求证:一元二次方程qx^2+p(1+q)x+(1+q)^2=0的根为a+1/b和b+1/a如题.](/uploads/image/z/800209-1-9.jpg?t=%E5%B7%B2%E7%9F%A5%E6%96%B9%E7%A8%8Bx%5E2%2Bpx%2Bq%3D0%E7%9A%84%E4%B8%A4%E6%A0%B9%E6%98%AFa%2Cb.%E6%B1%82%E8%AF%81%EF%BC%9A%E4%B8%80%E5%85%83%E4%BA%8C%E6%AC%A1%E6%96%B9%E7%A8%8Bqx%5E2%2Bp%281%2Bq%29x%2B%281%2Bq%29%5E2%3D0%E7%9A%84%E6%A0%B9%E4%B8%BAa%2B1%2Fb%E5%92%8Cb%2B1%2Fa%E5%A6%82%E9%A2%98.)
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已知方程x^2+px+q=0的两根是a,b.求证:一元二次方程qx^2+p(1+q)x+(1+q)^2=0的根为a+1/b和b+1/a如题.
已知方程x^2+px+q=0的两根是a,b.求证:一元二次方程qx^2+p(1+q)x+(1+q)^2=0的根为a+1/b和b+1/a
如题.
已知方程x^2+px+q=0的两根是a,b.求证:一元二次方程qx^2+p(1+q)x+(1+q)^2=0的根为a+1/b和b+1/a如题.
方程x^2+px+q=0的两根是a,b
所以a+b=-p,ab=q
则a+1/b+b+1/a
=a+b+(a+b)/ab
=-p-p/q
=-p(1+1/q)
=-p(q+1)/q
(a+1/b)(b+1/a)
=ab+1+1+1/ab
=q+1/q+2
=(q^2+2q+1)/q
=(q+1)^2/q
所以a+1/b和b+1/a是方程x^2+p(q+1)x/q+(q+1)^2/q的跟
即qx^2+p(1+q)x+(1+q)^2=0
最直接的就是a+1/b和b+1/a代入一元二次方程qx^2+p(1+q)x+(1+q)^2,再利用方程x^2+px+q=0的两根是a,b,得qx^2+p(1+q)x+(1+q)^2=0,即证