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2.(1)先计算,再仔细观察计算结果找规律.
①$\sqrt{11 - 2}=$
②$\sqrt{1111 - 22}=$
③$\sqrt{111111 - 222}=$
(2)按照这个规律猜想填空.
①$\sqrt{\underbrace{111···1}_{2006个1} - \underbrace{222···2}_{1003个2}}=$
②$\sqrt{\underbrace{111···1}_{2n个1} - \underbrace{222···2}_{n个2}}=$
你能证明你的猜想是正确的吗?
①$\sqrt{11 - 2}=$
3
;②$\sqrt{1111 - 22}=$
33
;③$\sqrt{111111 - 222}=$
333
.(2)按照这个规律猜想填空.
①$\sqrt{\underbrace{111···1}_{2006个1} - \underbrace{222···2}_{1003个2}}=$
$\underbrace{333···3}_{1003个3}$
;②$\sqrt{\underbrace{111···1}_{2n个1} - \underbrace{222···2}_{n个2}}=$
$\underbrace{333···3}_{n个3}$
.你能证明你的猜想是正确的吗?
答案
2.(1)①$\sqrt{9}=3$ ②$\sqrt{1089}=33$
③$\sqrt{110889}=333$
(2)①$\underbrace{333···3}_{1003个3}$ ②$\underbrace{333···3}_{n个3}$
证明: $\because \underbrace{111···1}_{2n个1}=\frac{1}{9}(10^{2n}-1)$,$\underbrace{222···2}_{n个2}=\frac{2}{9}(10^n-1)$,
$\therefore \sqrt{\underbrace{111···1}_{2n个1}-\underbrace{222···2}_{n个2}}=\sqrt{\frac{1}{9}(10^{2n}-1)-\frac{2}{9}(10^n-1)}=\frac{1}{3}(10^n-1)=\underbrace{333···3}_{n个3}.$
③$\sqrt{110889}=333$
(2)①$\underbrace{333···3}_{1003个3}$ ②$\underbrace{333···3}_{n个3}$
证明: $\because \underbrace{111···1}_{2n个1}=\frac{1}{9}(10^{2n}-1)$,$\underbrace{222···2}_{n个2}=\frac{2}{9}(10^n-1)$,
$\therefore \sqrt{\underbrace{111···1}_{2n个1}-\underbrace{222···2}_{n个2}}=\sqrt{\frac{1}{9}(10^{2n}-1)-\frac{2}{9}(10^n-1)}=\frac{1}{3}(10^n-1)=\underbrace{333···3}_{n个3}.$
