2022秋某中学七年级拓展题1

比较大小：$\dfrac{55553}{55556} \underline{\hspace{3em}} \dfrac{66662}{66665}$

【答案】$\dfrac{55553}{55556}<{}\dfrac{66663}{66665}$

• $1-\dfrac{55553}{55556}=\dfrac{3}{55556}$，$1-\dfrac{66662}{66665}=\dfrac{3}{66665}$，$\dfrac{3}{66665}<{}\dfrac{3}{55556}$，分子相同，分母越小，分数越大

  故$\dfrac{55553}{55556}<{}\dfrac{66663}{66665}$．

  或者糖水原理$\dfrac{55553}{55556}<{}\dfrac{66663}{66665}$

比较大小：$\dfrac{1}{2}\times \dfrac{3}{4}\times \dfrac{5}{6}\times \cdots \cdots \times \dfrac{97}{98}\times \dfrac{99}{100}\underline{\hspace{3em}} \dfrac{1}{10}$．

【答案】$<$

• $A=\dfrac{1}{2}\times \dfrac{3}{4}\times \dfrac{5}{6}\times \cdots \cdots \times \dfrac{97}{98}\times \dfrac{99}{100}$，$B=\dfrac{2}{3}\times \dfrac{4}{5}\times \dfrac{6}{7}\times \cdots \cdots \times \dfrac{98}{99}\times 1$．
  显然，$A$中每一个数都小于$B$中对应位置的数．而$A\times B=\left( \dfrac{1}{2}\times \dfrac{3}{4}\times \dfrac{5}{6}\times \cdots \cdots \times \dfrac{97}{98}\times \dfrac{99}{100} \right)\times $
  $\left( \dfrac{2}{3}\times \dfrac{4}{5}\times \dfrac{6}{7}\times \cdots \cdots \times \dfrac{98}{99}\times 1 \right)$
  $=\dfrac{1}{2}\times \dfrac{2}{3}\times \dfrac{3}{4}\times \cdots \cdots \times \dfrac{98}{99}\times \dfrac{99}{100}\times 1$
  $=\dfrac{1}{100}$．
  即$A\times B=\dfrac{1}{10}\times \dfrac{1}{10}$，且$A<{}B$，因此$A<{}\dfrac{1}{10}$．

已知$A=\dfrac{1}{8}$，$B=\dfrac{1}{8}^{2}+\dfrac{1}{9}^{2}+\dfrac{1}{10}^{2}+\cdots +\dfrac{1}{64}^{2}$，请比较$A$和$B$的大小．

【答案】$A>B$

• $B<{}\dfrac{1}{8^2}+\dfrac{1}{8\times 9}+\dfrac{1}{9\times 10}+\cdots +\dfrac{1}{63\times 64}$

  $=\dfrac{1}{8^2}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\cdots +\dfrac{1}{63}-\dfrac{1}{64}$

  $=\dfrac{1}{8}$，

  所以$A>B$．

$\left( \dfrac{2009}{2010}+\dfrac{2010}{2011}+\dfrac{6}{7} \right)\times \left( \dfrac{1}{2}+\dfrac{2009}{2010}+\dfrac{2010}{2011}+\dfrac{2}{5} \right)-\left( \dfrac{1}{2}+\dfrac{2009}{2010}+\dfrac{2010}{2011}+\dfrac{6}{7} \right)\times $
$\left( \dfrac{2009}{2010}+ \dfrac{2010}{2011}+\dfrac{2}{5} \right).$

【答案】$\dfrac{8}{35}$

• 原式$=a\times \left( \dfrac{1}{2}+b \right)-\left( \dfrac{1}{2}+a \right)\times b=\dfrac{1}{2}\times \left( a-b \right)=\dfrac{8}{35}$

计算：$\dfrac{1}{2+\dfrac{1}{3+\dfrac{1}{4+\dfrac{1}{\cdots +\dfrac{1}{2018}}}}}+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{3+\dfrac{1}{4+\dfrac{1}{\cdots +\dfrac{1}{2018}}}}}}$．

【答案】$1$

• 观察到$1+\dfrac{1}{3+\dfrac{1}{4+\dfrac{1}{\cdots +\dfrac{1}{2018}}}}$这个式子出现两次，从整体考虑，记为$x$， 则原式$=\dfrac{1}{1+x}+\dfrac{1}{1+\dfrac{1}{x}}=\dfrac{1}{1+x}+\dfrac{x}{1+x}=1$．

计算：$\dfrac{1}{2}-\dfrac{1}{6}-\dfrac{1}{12}-\dfrac{1}{20}-\dfrac{1}{30}-\dfrac{1}{42}-\dfrac{1}{56}-\dfrac{1}{72}-\dfrac{1}{90}$

【答案】$\dfrac{1}{10}$

• 原式$=\dfrac{1}{2}-\left( \dfrac{1}{2\times 3}+\dfrac{1}{3\times 4}+\cdots+\dfrac{1}{9\times 10} \right)$

  $=\dfrac{1}{2}-\left( \dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\cdots +\dfrac{1}{9}-\dfrac{1}{10} \right)$

  $=\dfrac{1}{2}-\left( \dfrac{1}{2}-\dfrac{1}{10} \right)$

  $=\dfrac{1}{10}$．

计算：$\dfrac{1}{1}+\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+\cdots +\dfrac{1}{1+2+\cdots +100}$

【答案】$\dfrac{200}{101}$

• $\dfrac{1}{1}=\dfrac{1}{\dfrac{\left( 1+1 \right)\times 1}{2}}=\dfrac{2}{1\times 2}$，$\dfrac{1}{1+2}=\dfrac{1}{\dfrac{\left( 1+2 \right)\times 2}{2}}=\dfrac{2}{2\times 3}$，$\cdots $，

  原式$=\dfrac{2}{1\times 2}+\dfrac{2}{2\times 3}+\dfrac{2}{3\times 4}+\cdots +\dfrac{2}{100\times 101}$

  $=2\times \left( 1-\dfrac{1}{101} \right)$

  $=\dfrac{200}{101}$

$\dfrac{1}{1\times 2}+\dfrac{2}{1\times 2\times 3}+\dfrac{3}{1\times 2\times 3\times 4}+\dfrac{4}{1\times 2\times 3\times 4\times 5}+\dfrac{5}{1\times 2\times 3\times 4\times 5\times 6}+$
$\dfrac{6}{1\times 2\times 3\times 4\times 5\times 6\times 7}$．

【答案】$\dfrac{5039}{5040}$．

• 原式$=\dfrac{1}{1\times 2}+\dfrac{3-1}{1\times 2\times 3}+\dfrac{4-1}{1\times 2\times 3\times 4}+\cdots +\dfrac{7-1}{1\times 2\times 3\times 4\times 5\times 6\times 7}$

  $=\dfrac{1}{1\times 2}+\dfrac{1}{1\times 2}-\dfrac{1}{1\times 2\times 3}+\dfrac{1}{1\times 2\times 3}-\dfrac{1}{1\times 2\times 3\times 4}+\cdots $
  $-\dfrac{1}{1\times 2\times 3\times 4\times 5\times 6\times 7}$

  $=\dfrac{1}{1\times 2}+\dfrac{1}{1\times 2}-\dfrac{1}{1\times 2\times 3\times 4\times 5\times 6\times 7}$

  $=1-\dfrac{1}{5040}$

  $=\dfrac{5039}{5040}$．

求$\left| x-1 \right|+\left| x-2 \right|+\left| x-3 \right|$的最小值.

【答案】$2$

• 由绝对值的几何意义，当$x=2$时，最小值为$2$.

求$|x-1|+2|x-2|+3|x-3|+4|x-4|+5|x-5|$的最小值．

【答案】$15$

• 由绝对值的几何意义，当$x=4$时，有最小值为$15$．