设二次方程anx²–an+1x+1=0有两个实数跟a和b,并且满足6a–2ab+6b=3.试用an表示an+1
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![设二次方程anx²–an+1x+1=0有两个实数跟a和b,并且满足6a–2ab+6b=3.试用an表示an+1](/uploads/image/z/5229811-19-1.jpg?t=%E8%AE%BE%E4%BA%8C%E6%AC%A1%E6%96%B9%E7%A8%8Banx%26%23178%3B%E2%80%93an%2B1x%2B1%3D0%E6%9C%89%E4%B8%A4%E4%B8%AA%E5%AE%9E%E6%95%B0%E8%B7%9Fa%E5%92%8Cb%2C%E5%B9%B6%E4%B8%94%E6%BB%A1%E8%B6%B36a%E2%80%932ab%2B6b%3D3.%E8%AF%95%E7%94%A8an%E8%A1%A8%E7%A4%BAan%2B1)
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设二次方程anx²–an+1x+1=0有两个实数跟a和b,并且满足6a–2ab+6b=3.试用an表示an+1
设二次方程anx²–an+1x+1=0有两个实数跟a和b,并且满足6a–2ab+6b=3.试用an表示an+1
设二次方程anx²–an+1x+1=0有两个实数跟a和b,并且满足6a–2ab+6b=3.试用an表示an+1
6a–2ab+6b=3
6(a+b) -2ab=3
6*(an+1)/(an) -2*1/an =3
6*(an+1) =( 3+2*1/an)*(an)
an+1 = (3an +2)/6