一级目录

二级目录

三级目录

一、题目

俄罗斯竞赛题
笔算开平方
$$\sqrt{446224}$$

二、解题方式1：设元配方

此方法需要有良好的数感，建议多刷题来提升数感！
$$\begin{array}{c}\begin{array}{rl}\sqrt{446224}& =\sqrt{4\times 111556}\\ & =2\sqrt{111556}\end{array}\end{array}$$
令：$111=m$
$$\begin{array}{c}\begin{array}{rl}\because 111556& =111\times 1000+555+1\\ & =111\times (999+1)+5\ast 111+1\\ & =111\times (9\ast 111+1)+5\ast 111+1\\ & =m\times (9m+1)+5m+1\\ & =9m^2+6m+1\\ & =(3m+1{)}^2\end{array}\end{array}$$
$$\begin{array}{c}\begin{array}{rl}\therefore \sqrt{446224}& =2\sqrt{(3m+1{)}^2}\\ & =6m+2\\ & =6\times 111+2\\ & =666+2\\ & =668\end{array}\end{array}$$

三、解题方式2：逐位逼近

①因式分解

$$\begin{gathered} 2\underline{|446224} \\ 2\underline{|223112} \\ 2\underline{|111556} \\ 2\underline{|55778} \\ 27889 \end{gathered}$$

②数字范围分析

$\because 100^2=10000$且$200^2=40000$ ，即$100^2<27889<200^2$
$\therefore$ 27889是一个三位数的平方，且最高位是1
$\because$
$$\{\begin{array}{ll}{0}^2& \text{个位 }0\\ 1^2& \text{个位 }1\\ 2^2& \text{个位 }4\\ 3^2& \text{个位 }9\\ 4^2& \text{个位 }6\\ 5^2& \text{个位 }5\\ 6^2& \text{个位 }6\\ 7^2& \text{个位 }9\\ 8^2& \text{个位 }4\\ 9^2& \text{个位 }1\end{array}$$
$\therefore$ 三位数的个位数是$3$或$7$

③求十位

设十位为$m$,因为个位是$3$或$7$不为$0$
$$⟹\{\begin{array}{l}(100+10m+0{)}^2<27889\\ 27889<(100+10(m+1)+0{)}^2\end{array}$$
化简:
$$⟹\{\begin{array}{l}(100+10m{)}^2<27889\\ 27889<(110+10m{)}^2\end{array}$$
两边同时除以$10^2$:
$$⟹\{\begin{array}{l}(10+m{)}^2<278.89\\ 278.89<(11+m{)}^2\end{array}$$
设定数值范围：
$$⟹\{\begin{array}{l}16^2=256<278.89\\ 278.89<289=17^2\end{array}$$
$\therefore m=6$

③求个位

设个位n,则
$$\begin{array}{c}\begin{array}{rl}(160+n{)}^2& =27889\\ 25600+320n+n^2& =27889\\ n^2+320n& =2289\\ 320n& =2289-n^2\end{array}\end{array}$$

$\because n=3或7$且为整数
$$\begin{gathered} ⟹2289-7^2\le 320n\le 2289-3^2 \\ ⟹2240\le 320n\le 2280 \\ ⟹224\le 32n\le 228 \\ ⟹112\le 16n\le 114 \\ ⟹56\le 8n\le 57<64 \\ ⟹7\le n<8 \end{gathered}$$
$\therefore n=7$

④整理

$27889$是百位为$1$，十位为$6$，个位为$7$的平方。
$$\begin{array}{c}\begin{array}{rl}\sqrt{446224}& =\sqrt{2\times 2\times 2\times 2\times 27889}\\ & =\sqrt{4^2\times 167^2}\\ & =4\times 167\\ & =668\end{array}\end{array}$$

四、解题方式3：逐位逼近2

①因式分解

$$\begin{gathered} 2\underline{|446224} \\ 2\underline{|223112} \\ 2\underline{|111556} \\ 2\underline{|55778} \\ 27889 \end{gathered}$$

②数字范围分析

$\because 100^2=10000$且$200^2=40000$ ，即$100^2<27889<200^2$
$\therefore$ 27889是一个三位数的平方，且最高位是1
$\because$
$$\{\begin{array}{ll}{0}^2& \text{个位 }0\\ 1^2& \text{个位 }1\\ 2^2& \text{个位 }4\\ 3^2& \text{个位 }9\\ 4^2& \text{个位 }6\\ 5^2& \text{个位 }5\\ 6^2& \text{个位 }6\\ 7^2& \text{个位 }9\\ 8^2& \text{个位 }4\\ 9^2& \text{个位 }1\end{array}$$
$\therefore$ 三位数的个位数是$3$或$7$

③假设个位是3，设十位m,则

$$\begin{array}{c}\begin{array}{rl}(100+10m+3{)}^2& =27889\\ (103+10m{)}^2& =27889\\ 103^2+2\times 103\times 10m+(10m{)}^2& =27889\\ 10609+100m^2+2060m& =27889\\ 100m^2+2060m+10609-27889& =0\\ 100m^2+2060m-17280& =0\\ m^2+20.6m& =172.8\\ m^2+20m+100& =272.8-0.6m\\ (m+10{)}^2& =272.8-0.6m\end{array}\end{array}$$
$\because 0\le m<10$

$$⟹\{\begin{array}{l}272.8-0.6\times 10<272.8-0.6m\\ 272.8-0.6m\le 272.8-0.6\times 0\end{array}$$

$$⟹\{\begin{array}{l}266.8<272.8-0.6m\\ 272.8-0.6m\le 272.8\end{array}$$

$$⟹\{\begin{array}{l}16^2=256<272.8-0.6m\\ 272.8-0.6m<289=17^2\end{array}$$

$16$与$17$之间没有整数，所以个位为$3$不成立；

③假设个位是7，设十位m,则

$$\begin{array}{c}\begin{array}{rl}(100+10m+7{)}^2& =27889\\ (107+10m{)}^2& =27889\\ 107^2+2\times 107\times 10m+(10m{)}^2& =27889\\ 11449+100m^2+2140m& =27889\\ 100m^2+2140m+11449-27889& =0\\ 100m^2+2140m-16440& =0\\ m^2+21.4m& =164.4\\ m^2+20m+100& =264.4-1.6m\\ (m+10{)}^2& =264.40-1.6m\end{array}\end{array}$$
$\because 0\le m<10$

$$⟹\{\begin{array}{l}264.4-1.6\times 10<264.4-0.6m\\ 264.4-1.6m\le 264.4-1.6\times 0\end{array}$$

$$⟹\{\begin{array}{l}248.4<264.4-0.6m\\ 264.4-0.6m\le 264.4\end{array}$$

$$⟹\{\begin{array}{l}15^2=225<272.8-0.6m\\ 272.8-0.6m<289=17^2\end{array}$$

15与17之间的整数为16，所以 $m=6$，代入$m^2+21.4m=164.4$
$6^2+20.6\times 6=36+129.6=164.4$

④整理

$27889$是百位为$1$，十位为$6$，个位为$7$的平方。
$$\begin{array}{c}\begin{array}{rl}\sqrt{446224}& =\sqrt{2\times 2\times 2\times 2\times 27889}\\ & =\sqrt{4^2\times 167^2}\\ & =4\times 167\\ & =668\end{array}\end{array}$$