《2025年重点中学招生分班五年真题分类卷数学》

16. (2024陕西TYZLG学校)$\frac {19981998}{999}×\frac {121}{11111111}$

答案： 解:原式=$\frac{999×2×10001}{999}×\frac{11×11}{101×11×10001}$
=2×10001×$\frac{11}{101×10001}$
=$\frac{22}{101}$

17. (2023陕西GXYZSY中学)$246×\frac {345345345345}{123123123123}$

答案： 解:原式=246×$\frac{345×1001001001}{123×1001001001}$
=2×345
=690

18. (2024陕西GDFZ)$\frac {341}{275}+2×\frac {3441}{2775}+3×\frac {34441}{27775}+... +8×\frac {344444441}{277777775}+9×\frac {3444444441}{2777777775}$

答案： 解:原式=$\frac{31×11}{25×11}+2×\frac{31×111}{25×111}+3×\frac{31×1111}{25×1111}+\cdots+8×\frac{31×111111111}{25×111111111}+9×\frac{31×1111111111}{25×1111111111}$
=$\frac{31}{25}+2×\frac{31}{25}+3×\frac{31}{25}+\cdots+8×\frac{31}{25}+9×\frac{31}{25}$
=$\frac{31}{25}×(1+2+3+4+5+6+7+8+9)$
=$\frac{279}{5}$

19. (2020陕西SDFZ)$1-(\frac {101010}{202020})^{2}×(\frac {202020}{303030})^{2}×(\frac {333033}{555055})^{2}×(\frac {555555}{777777})^{2}$

答案： 解:原式=1-$(\frac{1}{2}×\frac{101010}{101010})^{2}×(\frac{2}{3}×\frac{101010}{101010})^{2}×(\frac{3}{5}×\frac{111011}{111011})^{2}×(\frac{5}{7}×\frac{111111}{111111})^{2}$
=1-$(\frac{1}{2})^{2}×(\frac{2}{3})^{2}×(\frac{3}{5})^{2}×(\frac{5}{7})^{2}$
=1-$\frac{1}{4}×\frac{4}{9}×\frac{9}{25}×\frac{25}{49}$
=1-$\frac{1}{49}$
=$\frac{48}{49}$

20. (2024陕西GXTN中学)$\frac {1}{21}+\frac {202}{2121}+\frac {50505}{212121}+\frac {13131313}{21212121}$

答案： 解:原式=$\frac{1}{21}+\frac{101×2}{101×21}+\frac{10101×5}{10101×21}+\frac{1010101×13}{1010101×21}$
=$\frac{1}{21}+\frac{2}{21}+\frac{5}{21}+\frac{13}{21}$
=$\frac{21}{21}$
=1