证明 根号(11...1[2n个1] - 22...2[n个2]) = 33...3[n个3]清晰点,最好有[ ]内的内容
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/05 11:52:55
![证明 根号(11...1[2n个1] - 22...2[n个2]) = 33...3[n个3]清晰点,最好有[ ]内的内容](/uploads/image/z/1115910-54-0.jpg?t=%E8%AF%81%E6%98%8E+%E6%A0%B9%E5%8F%B7%2811...1%5B2n%E4%B8%AA1%5D+-+22...2%5Bn%E4%B8%AA2%5D%29+%3D+33...3%5Bn%E4%B8%AA3%5D%E6%B8%85%E6%99%B0%E7%82%B9%2C%E6%9C%80%E5%A5%BD%E6%9C%89%5B+%5D%E5%86%85%E7%9A%84%E5%86%85%E5%AE%B9)
证明 根号(11...1[2n个1] - 22...2[n个2]) = 33...3[n个3]清晰点,最好有[ ]内的内容
证明 根号(11...1[2n个1] - 22...2[n个2]) = 33...3[n个3]
清晰点,最好有[ ]内的内容
证明 根号(11...1[2n个1] - 22...2[n个2]) = 33...3[n个3]清晰点,最好有[ ]内的内容
√(11...1[2n个1]-22...2[n个2])=33...3[n个3]
当n=1时,
左边=√(11...1[2个1]-22...2[1个2])
=√(11-2)
=3
=右边
原式成立;
设n=k时成立,即√(11...1[2k个1]-22...2[k个2])=33...3[k个3]
11...1[2k个1]=33...3[k个3]^2+22...2[k个2]
当n=k+1时,
√(11...1[2k+2个1]-22...2[k+1个2])
=√[(11...1[2k个1]*100+11)-(22...2[k个2]*10+2)]
=√{[(33...3[k个3]^2+22...2[k个2])*100+11]-(22...2[k个2]*10+2)}
=√{10*33...3[k个3]^2+(11...1[k个1]*10-11...1[k个1])*10*2+9}
=√{10*33...3[k个3]^2+2*10*11...1[k个1]*(10-1)+9}
=√{10*33...3[k个3]^2+2*10*9*11...1[k个1]+9}
=√{10*33...3[k个3]^2+2*3*10*33...3[k个3]+9}
=√{10*33...3[k个3]^2+2*3*(10*33...3[k个3])+3^2}
=√{(10*33...3[k个3]+3)^2]
=10*33...3[k个3]+3
=33...3[k+1个3]
所以
√(11...1[2n个1]-22...2[n个2])=33...3[n个3]成立.
证毕.