一道高中数学题设a1≈√2,令a
解:a1≈√2
(1)证明√2介于a1,a2之间;
(a1-√2)(a2-√2)=(a1-√2)(1+1/(1+a1)-√2)
=(a1-√2)[(1+a1)+1-√2(1+a1)]/(1+a1)
=(a1-√2)[a1+2-√2-√2a1]/(1+a1)
=(a1-√2)[a1(1-√2)+2-√2]/(1+a1)
=(a1-√2)(a1-√2)(1-√2)/(1+a1)
=(a1-√2)^(1-√2)/(1+a1)
∵(a1-√2)^＞0,(1-√2)＜0,∵a1≈√2,∴1+a1＞0
∴(a1-√2)(a2-√2)＜0,也就是√2介于a1,a2之间;
(2)求a1,a2中那一个更接近...全部
解:a1≈√2
(1)证明√2介于a1,a2之间;
(a1-√2)(a2-√2)=(a1-√2)(1+1/(1+a1)-√2)
=(a1-√2)[(1+a1)+1-√2(1+a1)]/(1+a1)
=(a1-√2)[a1+2-√2-√2a1]/(1+a1)
=(a1-√2)[a1(1-√2)+2-√2]/(1+a1)
=(a1-√2)(a1-√2)(1-√2)/(1+a1)
=(a1-√2)^(1-√2)/(1+a1)
∵(a1-√2)^＞0,(1-√2)＜0,∵a1≈√2,∴1+a1＞0
∴(a1-√2)(a2-√2)＜0,也就是√2介于a1,a2之间;
(2)求a1,a2中那一个更接近于√2;??????
需要比较|a1-√2|与|a2-√2|也就是(a1-√2)^与(a2-√2)^的大小。
∵(a1-√2)^-(a2-√2)^=(a1-a2)(a1+a2-2√2)
=[a1-1-1/(1+a1)][a1+1+1/(1+a1)-2√2]
=[a1^-2][(a1+1)^+1-2√2(1+a1)]/(1+a1)^
=[a1^-2][(a1+1)^-2√2(1+a1)+2-1]/(1+a1)^
=[a1^-2]{[(a1+1)-√2]^-1}/(1+a1)^
=[(a1-√2)(a1+√2)]{[a1+2-√2][a1-√2]}/(1+a1)^
=(a1-√2)^(a1+√2)(a1+2-√2)/(1+a1)^
∵(a1-√2)^＞0，
∵a1≈√2，∴(a1+√2)＞0，(a1+2-√2)＞0，
∵(1+a1)^＞0
∴(a1-√2)^-(a2-√2)^＞0就是|a1-√2|＞|a2-√2|也就是a2比a1更接近于√2。
(3)你能设计一个比a2更接近于√2的一个a3吗?并说明理由。
能：a3=1+1/(a2+1)=1+1/[2+1/(1+a1)]
。收起
