Các Bài Toán Rút Gọn Nâng Cao Lớp 6 Có Lời Giải - Học Toán 123

Đây là bài thứ 10 of 27 trong series Toán nâng cao lớp 6

Bài tập rút gọn hay dạng bài thực hiện phép tính là một trong những dạng toán nâng cao trong chương trình lớp 6 dành cho học sinh khá giỏi.

Dưới đây là 21 bài toán rút gọn nâng cao cho học sinh lớp 6 mà Học Toán 123 muốn gửi tới thầy cô và các em học sinh.

Bài 1: Thực hiện phép tính:

a, $ \displaystyle \dfrac{{2^{{12}}{.3}^{5}-4^{6}{.9}^{2}}}{{{(2^{2}.3)}^{6}}}-\dfrac{{5^{{10}}{.7}^{3}-{25}^{5}{.49}^{2}}}{{{(125.7)}^{3}+5^{9}{.14}^{3}}}$

b, $ \displaystyle \dfrac{{2^{{18}}{.18}^{7}{.3}^{3}+3^{{15}}{.2}^{{15}}}}{{2^{{10}}{.6}^{{15}}+3^{{14}}{.15.4}^{{13}}}}$

c, $ \displaystyle \dfrac{{4^{6}{.9}^{5}+6^{9}.120}}{{8^{4}{.3}^{{12}}-6^{{11}}}}$

Hướng dẫn giải:

a, Ta có:

$ \displaystyle \dfrac{{2^{{12}}{.3}^{5}-4^{6}{.9}^{2}}}{{{(2^{2}.3)}^{6}}}-\dfrac{{5^{{10}}{.7}^{3}-{25}^{5}{.49}^{2}}}{{{(125.7)}^{3}+5^{9}{.14}^{3}}}$$ \displaystyle =\dfrac{{2^{{12}}{.3}^{5}-{\left( {2^{2}} \right)}^{6}.{\left( {3^{2}} \right)}^{2}}}{{2^{{12}}{.3}^{6}}}-\dfrac{{5^{{10}}{.7}^{3}-{\left( {5^{2}} \right)}^{5}.{\left( {7^{2}} \right)}^{2}}}{{{\left( {5^{3}} \right)}^{3}{.7}^{3}+5^{9}{.2}^{3}{.7}^{3}}}$

$ \displaystyle =\dfrac{{2^{{12}}{.3}^{5}-2^{{12}}{.3}^{4}}}{{2^{{12}}{.3}^{6}}}-\dfrac{{5^{{10}}{.7}^{3}-5^{{10}}{.7}^{4}}}{{5^{9}{.7}^{3}+5^{9}{.2}^{3}{.7}^{3}}}=\dfrac{{2^{{12}}{.3}^{4}\left( {3-1} \right)}}{{2^{{12}}{.3}^{6}}}-\dfrac{{5^{{10}}{.7}^{3}\left( {1-7} \right)}}{{5^{9}{.7}^{3}\left( {1+8} \right)}}$$ \displaystyle =\dfrac{2}{{3^{2}}}-\dfrac{{5.6}}{9}=\dfrac{{-28}}{9}$

b, Ta có: $ \displaystyle \dfrac{{2^{{18}}{.18}^{7}{.3}^{3}+3^{{15}}{.2}^{{15}}}}{{2^{{10}}{.6}^{{15}}+3^{{14}}{.15.4}^{{13}}}}$

$ \displaystyle =\dfrac{{2^{{18}}{.2}^{7}{.3}^{{14}}{.3}^{3}+3^{{15}}{.2}^{{15}}}}{{2^{{10}}{.2}^{{15}}{.3}^{{15}}+3^{{14}}{.3.5.2}^{{28}}}}=\dfrac{{2^{{25}}{.3}^{{17}}+3^{{15}}{.2}^{{15}}}}{{2^{{25}}{.3}^{{15}}+3^{{15}}{.2}^{{28}}.5}}$

$ \displaystyle =\dfrac{{2^{{15}}{.3}^{{15}}\left( {2^{{10}}{.3}^{2}+1} \right)}}{{2^{{25}}{.3}^{{15}}\left( {1+2^{3}.5} \right)}}=\dfrac{{\left( {2^{{10}}{.3}^{2}+1} \right)}}{{2^{{10}}41}}$

c, Ta có:

$ \displaystyle \dfrac{{4^{6}{.9}^{5}+6^{9}.120}}{{8^{4}{.3}^{{12}}-6^{{11}}}}$ =$ \displaystyle \dfrac{{{\left( {2^{2}} \right)}^{6}.{\left( {3^{2}} \right)}^{5}+2^{9}{.3}^{9}{.2}^{3}.3.5}}{{{\left( {2^{3}} \right)}^{4}{.3}^{{12}}-2^{{11}}{.3}^{{11}}}}=\dfrac{{2^{{12}}{.3}^{{10}}+2^{{12}}{.3}^{{10}}.5}}{{2^{{12}}{.3}^{{12}}-2^{{11}}{.3}^{{11}}}}$

$ \displaystyle =\dfrac{{2^{{12}}{.3}^{{10}}\left( {1+5} \right)}}{{2^{{11}}{.3}^{{11}}\left( {2.3-1} \right)}}=\dfrac{{2.6}}{{3.5}}=\dfrac{4}{5}$

Bài 2: Thực hiện phép tính:

a, $ \displaystyle \dfrac{{{5.4}^{{15}}{.9}^{9}-{4.3}^{{20}}{.8}^{9}}}{{{5.2}^{{29}}{.9}^{{16}}-{7.2}^{{29}}{.27}^{6}}}$

b, $ \displaystyle \dfrac{{2^{4}{.5}^{2}{.11}^{2}.7}}{{2^{3}{.5}^{3}{.7}^{2}.11}}$

c, $ \displaystyle \dfrac{{5^{{11}}{.7}^{{12}}+5^{{11}}{.7}^{{11}}}}{{5^{{12}}{.7}^{{11}}+{9.5}^{{11}}{.7}^{{11}}}}$

Hướng dẫn giải:

a, Ta có:

$ \displaystyle \dfrac{{{5.4}^{{15}}{.9}^{9}-{4.3}^{{20}}{.8}^{9}}}{{{5.2}^{{29}}{.3}^{{16}}-{7.2}^{{29}}{.27}^{6}}}$ =$ \displaystyle \dfrac{{{5.2}^{{30}}{.3}^{{18}}-2^{{29}}{.3}^{{20}}}}{{{5.2}^{{29}}{.3}^{{16}}-{7.2}^{{29}}{.3}^{{18}}}}=\dfrac{{2^{{29}}{.3}^{{18}}\left( {5.2-3^{2}} \right)}}{{2^{{29}}{.3}^{{16}}\left( {5-{7.3}^{2}} \right)}}$=$ \displaystyle \dfrac{{3^{2}}}{{-58}}=\dfrac{{-9}}{{58}}$

b, Ta có:

$ \displaystyle \dfrac{{2^{4}{.5}^{2}{.11}^{2}.7}}{{2^{3}{.5}^{3}{.7}^{2}.11}}$ =$ \displaystyle \dfrac{{2.11}}{{5.7}}=\dfrac{{22}}{{35}}$

c, Ta có:  $ \displaystyle \dfrac{{5^{{11}}{.7}^{{12}}+5^{{11}}{.7}^{{11}}}}{{5^{{12}}{.7}^{{11}}+{9.5}^{{11}}{.7}^{{11}}}}$ =$ \displaystyle \dfrac{{5^{{11}}{.7}^{{11}}\left( {7+1} \right)}}{{5^{{11}}{.7}^{{11}}\left( {5+9} \right)}}=\dfrac{8}{{14}}=\dfrac{4}{7}$

Bài 3: Thực hiện phép tính:

a, $ \displaystyle \dfrac{{{11.3}^{{22}}{.3}^{7}-9^{{15}}}}{{{({2.3}^{{14}})}^{2}}}$

b, $ \displaystyle \dfrac{{2^{{10}}{.3}^{{10}}-2^{{10}}{.3}^{9}}}{{2^{9}{.3}^{{10}}}}$

c, $ \displaystyle \dfrac{{4^{5}{.9}^{4}-{2.6}^{9}}}{{2^{{10}}{.3}^{8}+6^{8}.20}}$

Hướng dẫn giải:

a, Ta có:

$ \displaystyle \dfrac{{{11.3}^{{22}}{.3}^{7}-9^{{15}}}}{{{({2.3}^{{14}})}^{2}}}$ =$ \displaystyle \dfrac{{{11.3}^{{29}}-3^{{30}}}}{{2^{2}{.3}^{{28}}}}=\dfrac{{3^{{29}}.\left( {11-3} \right)}}{{2^{2}{.3}^{{28}}}}=\dfrac{{3.8}}{4}=6$

b, Ta có:  $ \displaystyle \dfrac{{2^{{10}}{.3}^{{10}}-2^{{10}}{.3}^{9}}}{{2^{9}{.3}^{{10}}}}$$ \displaystyle =\dfrac{{2^{{10}}{.3}^{9}\left( {3-1} \right)}}{{2^{9}{.3}^{{10}}}}=\dfrac{{2.2}}{3}=\dfrac{4}{3}$

c, Ta có:  $ \displaystyle \dfrac{{4^{5}{.9}^{4}-{2.6}^{9}}}{{2^{{10}}{.3}^{8}+6^{8}.20}}$$ \displaystyle =\dfrac{{2^{{10}}{.3}^{8}-2^{{10}}{.3}^{9}}}{{2^{{10}}{.3}^{8}+2^{{10}}{.3}^{8}.5}}=\dfrac{{2^{{10}}{.3}^{8}\left( {1-3} \right)}}{{2^{{10}}{.3}^{8}\left( {1+5} \right)}}=\dfrac{{-2}}{6}=\dfrac{{-1}}{3}$

Bài 4: Thực hiện phép tính:

a, $ \displaystyle \dfrac{{2^{{12}}{.3}^{5}-4^{6}{.9}^{2}}}{{{(2^{2}.3)}^{6}+8^{4}{.3}^{5}}}-\dfrac{{5^{{10}}{.7}^{3}-{25}^{5}{.49}^{2}}}{{{(125.7)}^{3}+5^{9}{.14}^{3}}}$

b, $ \displaystyle \dfrac{{{5.4}^{{15}}{.9}^{9}-{4.3}^{{20}}{.8}^{9}}}{{{5.2}^{9}{.6}^{{19}}-{7.2}^{{29}}{.27}^{6}}}$

c, $ \displaystyle \dfrac{{4^{5}{.9}^{4}-{2.6}^{9}}}{{2^{{10}}{.3}^{8}+6^{8}.20}}$

HD:

a, Ta có :

$ \displaystyle \dfrac{{2^{{12}}{.3}^{5}-4^{6}{.9}^{2}}}{{{(2^{2}.3)}^{6}+8^{4}{.3}^{5}}}-\dfrac{{5^{{10}}{.7}^{3}-{25}^{5}{.49}^{2}}}{{{(125.7)}^{3}+5^{9}{.14}^{3}}}$

=$ \displaystyle \dfrac{{2^{{12}}{.3}^{5}-2^{{12}}{.3}^{4}}}{{2^{{12}}{.3}^{6}+2^{{12}}{.3}^{5}}}-\dfrac{{5^{{10}}{.7}^{3}-5^{{10}}{.7}^{4}}}{{5^{9}{.7}^{3}+5^{9}{.7}^{3}{.2}^{3}}}=\dfrac{{2^{{12}}{.3}^{4}\left( {3-1} \right)}}{{2^{{12}}{.3}^{5}\left( {3+1} \right)}}-\dfrac{{5^{{10}}{.7}^{3}\left( {1-7} \right)}}{{5^{9}{.7}^{3}\left( {1+8} \right)}}=\dfrac{{5.\left( {-6} \right)}}{9}=\dfrac{{-10}}{3}$

b, Ta có : $ \displaystyle \dfrac{{{5.4}^{{15}}{.9}^{9}-{4.3}^{{20}}{.8}^{9}}}{{{5.2}^{9}{.6}^{{19}}-{7.2}^{{29}}{.27}^{6}}}$ =$ \displaystyle \dfrac{{{5.2}^{{30}}{.3}^{{18}}-3^{{20}}{.2}^{{29}}}}{{{5.2}^{{28}}{.3}^{{19}}-{7.2}^{{29}}{.3}^{{18}}}}=\dfrac{{2^{{29}}{.3}^{{18}}\left( {5.2-3^{2}} \right)}}{{2^{{28}}{.3}^{{18}}\left( {5.3-7.2} \right)}}=\dfrac{2}{1}=2$

c, Ta có:

$ \displaystyle \dfrac{{4^{5}{.9}^{4}-{2.6}^{9}}}{{2^{{10}}{.3}^{8}+6^{8}.20}}$ =$ \displaystyle \dfrac{{2^{{10}}{.3}^{8}-2^{{10}}{.3}^{9}}}{{2^{{10}}{.3}^{8}+2^{{10}}{.3}^{8}.5}}=\dfrac{{2^{{10}}{.3}^{8}\left( {1-3} \right)}}{{2^{{10}}{.3}^{8}\left( {1+5} \right)}}=\dfrac{{-2}}{6}=\dfrac{{-1}}{3}$

Bài 5: Thực hiện phép tính:

a, $ \displaystyle \dfrac{{{15.4}^{{12}}{.9}^{7}-{4.3}^{{15}}{.8}^{8}}}{{{19.2}^{{24}}{.3}^{{14}}-{6.4}^{{12}}{.27}^{5}}}$

b, $ \displaystyle \dfrac{{3^{{15}}{.2}^{{22}}+6^{{16}}{.4}^{4}}}{{{2.9}^{9}{.8}^{7}-{7.27}^{5}{.2}^{{23}}}}$

c, $ \displaystyle \dfrac{{{16}^{3}{.3}^{{10}}+{120.6}^{9}}}{{4^{6}{.3}^{{12}}+6^{{11}}}}$

Hướng dẫn giải:

a, Ta có:

$ \displaystyle \dfrac{{{15.4}^{{12}}{.9}^{7}-{4.3}^{{15}}{.8}^{8}}}{{{19.2}^{{24}}{.3}^{{14}}-{6.4}^{{12}}{.27}^{5}}}$ =$ \displaystyle \dfrac{{{5.2}^{{24}}{.3}^{{15}}-2^{{26}}{.3}^{{15}}}}{{{19.2}^{{24}}{.3}^{{14}}-2^{{25}}{.3}^{{16}}}}=\dfrac{{2^{{24}}{.3}^{{15}}\left( {5-2^{2}} \right)}}{{2^{{24}}{.3}^{{24}}\left( {19-{2.3}^{2}} \right)}}=\dfrac{3}{1}=3$

b, Ta có:

$ \displaystyle \dfrac{{3^{{15}}{.2}^{{22}}+6^{{16}}{.4}^{4}}}{{{2.9}^{9}{.8}^{7}-{7.27}^{5}{.2}^{{23}}}}$ =$ \displaystyle \dfrac{{3^{{15}}{.2}^{{22}}+2^{{24}}{.3}^{{16}}}}{{2^{{22}}{.3}^{{18}}-{7.3}^{{15}}{.2}^{{23}}}}=\dfrac{{2^{{22}}{.3}^{{15}}\left( {1+2^{2}.3} \right)}}{{2^{{22}}{.3}^{{15}}\left( {3^{3}-7.2} \right)}}=\dfrac{{13}}{{-5}}=\dfrac{{-13}}{5}$

c, Ta có:

$ \displaystyle \dfrac{{{\left( {2^{4}} \right)}^{3}{.3}^{{10}}+2^{3}.3.5.{\left( {2.3} \right)}^{9}}}{{{\left( {2^{2}} \right)}^{6}{.3}^{{12}}+{\left( {2.3} \right)}^{{11}}}}=\dfrac{{2^{{12}}{.3}^{{10}}+2^{{12}}{.3}^{{10}}.5}}{{2^{{12}}{.3}^{{12}}+2^{{11}}{.3}^{{11}}}}=\dfrac{{2^{{12}}{.3}^{{10}}\left( {1+5} \right)}}{{2^{{11}}{.3}^{{11}}\left( {2.3+1} \right)}}=\dfrac{{2.6}}{{3.7}}=\dfrac{{12}}{{21}}$

Bài 6: Thực hiện phép tính :

a, $ \displaystyle A=\dfrac{{2^{{12}}{.3}^{5}-4^{6}{.9}^{2}}}{{{\left( {2^{2}.3} \right)}^{6}+8^{4}{.3}^{5}}}-\dfrac{{5^{{10}}{.7}^{3}-{25}^{5}{.49}^{2}}}{{{\left( {125.7} \right)}^{3}+5^{9}{.14}^{3}}}$

b, $ \displaystyle \dfrac{{{5.4}^{{15}}{.9}^{9}-{4.3}^{{20}}{.8}^{9}}}{{{5.2}^{{10}}{.6}^{{12}}-{7.2}^{{29}}{.27}^{6}}}$

Bài 7: Thực hiện phép tính:

a, $ \displaystyle A=\dfrac{{2^{{12}}{.3}^{5}-4^{6}{.9}^{2}}}{{{\left( {2^{2}.3} \right)}^{6}+8^{4}{.3}^{5}}}$

b, $ \displaystyle B=\dfrac{{4^{5}{.9}^{4}-{2.6}^{9}}}{{2^{{10}}{.3}^{8}+6^{8}.20}}$

Bài 8: Thực hiện phép tính :

a, $ \dfrac{{3^{{10}}.11+3^{{10}}.5}}{{3^{9}{.2}^{4}}}$

b, $ \dfrac{{2^{{10}}.13+2^{{10}}.65}}{{2^{8}.104}}$

Bài 9: Thực hiện phép tính:

a, $ \dfrac{{2^{{30}}{.5}^{7}+2^{{13}}{.5}^{{27}}}}{{2^{{27}}{.5}^{7}+2^{{10}}{.5}^{{27}}}}$

b, $ \dfrac{{{\left( {-3} \right)}^{6}{.15}^{5}+9^{3}.{\left( {-15} \right)}^{6}}}{{{\left( {-3} \right)}^{{10}}{.5}^{5}{.2}^{3}}}$

Bài 10: Thực hiện phép tính:

a, $ \dfrac{{5^{2}{.6}^{{11}}{.16}^{2}+6^{2}{.12}^{6}{.15}^{2}}}{{{2.6}^{{12}}{.10}^{4}-{81}^{2}{.960}^{3}}}$

b, $ A=\dfrac{{2^{{19}}{.27}^{3}.5-15.{\left( {-4} \right)}^{9}{.9}^{4}}}{{6^{9}{.2}^{{10}}-{\left( {-12} \right)}^{{10}}}}$

Bài 11: Thực hiện phép tính:

a, $ \left[ {\dfrac{{{\left( {0,8} \right)}^{5}}}{{{\left( {0,4} \right)}^{6}}}+\dfrac{{2^{{15}}{.9}^{4}}}{{6^{6}{.8}^{3}}}} \right]:\dfrac{{{45}^{{10}}{.5}^{{20}}}}{{{75}^{{15}}}}$

b, $ A=\dfrac{{{2.5}^{{22}}-{9.5}^{{21}}}}{{{25}^{{10}}}}:\dfrac{{5\left( {{3.7}^{{15}}-{19.7}^{{14}}} \right)}}{{7^{{16}}+{3.7}^{{15}}}}$

Bài 12: Tính giá trị của biểu thức: $ \displaystyle A=\dfrac{{{\left( {\dfrac{2}{5}} \right)}^{7}{.5}^{7}+{\left( {\dfrac{9}{4}} \right)}^{3}:{\left( {\dfrac{3}{{16}}} \right)}^{3}}}{{2^{7}{.5}^{7}+512}}$

Bài 13: Tính biểu thức:$ B=\sqrt{{2\dfrac{{14}}{{25}}}}-\sqrt{{1,21}}+\dfrac{{0,6-\dfrac{3}{7}-\dfrac{3}{{13}}}}{{1,2-\dfrac{6}{7}-\dfrac{6}{{13}}}}:\dfrac{{-1\dfrac{1}{6}+0,875-0,7}}{{\dfrac{1}{3}-0,25+0,2}}$

Bài 14: Tính biểu thức: $ A=-84\left( {\dfrac{{-1}}{3}+\dfrac{1}{4}-\dfrac{1}{7}} \right)+51.\left( {-37} \right)-51.\left( {-137} \right)+\dfrac{{3^{3}{.12}^{6}}}{{{\left( {{27.4}^{2}} \right)}^{3}}}$

Bài 15: Thực hiện phép tính:

a, 1024: $ \displaystyle (17.2^{5}+15.2^{5})$

b, $ \displaystyle 5^{3}.2+(23+4^{0}):2^{3}$

c, $ \displaystyle (5.3^{5}+17.3^{4}):6^{2}$

Hướng dẫn giải:

a, Ta có: 1024: $ \displaystyle (17.2^{5}+15.2^{5})$

$ \displaystyle =2^{{10}}:\left[ {2^{5}\left( {17+15} \right)} \right]=2^{{10}}:\left( {2^{5}{.2}^{5}} \right)=1$

b, Ta có: $ \displaystyle 5^{3}.2+(23+4^{0}):2^{3}$$ \displaystyle =5^{3}.2+24:2^{3}=250+3=253$

c, Ta có: $ \displaystyle (5.3^{5}+17.3^{4}):6^{2}$$ \displaystyle \left[ {3^{4}\left( {3.5+17} \right)} \right]:3^{2}.2^{2}=\left( {3^{4}.32} \right):3^{2}.2^{2}=\dfrac{{3^{4}{.2}^{5}}}{{3^{2}{.2}^{2}}}=9.8=72$

Bài 16: Thực hiện phép tính:

a, $ \displaystyle (10^{2}+11^{2}+12^{2}):(13^{2}+14^{2})$

b, $ \displaystyle (2^{3}.9^{4}+9^{3}.45):(9^{2}.10-9^{2})$

Hướng dẫn giải:

a, Ta có:

$ \displaystyle (10^{2}+11^{2}+12^{2}):(13^{2}+14^{2})$$ \displaystyle =\left( {100+121+144} \right):\left( {169+196} \right)=365:365=1$

c Ta có:

$ \displaystyle (2^{3}.9^{4}+9^{3}.45):(9^{2}.10-9^{2})$ =$ \displaystyle \left( {2^{3}{.3}^{8}+3^{{11}}.5} \right):\left( {3^{2}.10+3^{2}} \right)=\dfrac{{3^{8}\left( {8+3^{3}.5} \right)}}{{3^{2}.11}}=\dfrac{{3^{6}.143}}{{11}}=13.3^{6}$

Bài 17: Thực hiện phép tính:

a, $ \displaystyle \left[ {(3^{{14}}.69+3^{{14}}.12):3^{{16}}-7} \right]:2^{4}$

b, $ \displaystyle 24^{4}:3^{4}-32^{{12}}:16^{{12}}$

Hướng dẫn giải:

a, Ta có:

$ \displaystyle \left[ {(3^{{14}}.69+3^{{14}}.12):3^{{16}}-7} \right]:2^{4}$$ \displaystyle =\left[ {\left( {3^{{14}}.3.23+3^{{14}}{.3.2}^{2}} \right):3^{{16}}-7} \right]:2^{4}=\left[ {\left( {3^{{15}}.23+3^{{15}}.4} \right):3^{{16}}-7} \right]:2^{4}$

$ \displaystyle =\left[ {3^{{15}}.27:3^{{16}}-7} \right]:2^{4}=\left( {9-7} \right):2^{4}=\dfrac{1}{{2^{3}}}$

b, Ta có:

$ \displaystyle 24^{4}:3^{4}-32^{{12}}:16^{{12}}$ =$ \displaystyle \left( {24:3} \right)^{4}-\left( {32:16} \right)^{{12}}=8^{4}-2^{{12}}=2^{{12}}-2^{{12}}=0$

Bài 18: Thực hiện phép tính :

a, $ \displaystyle 2010^{{2010}}\left( {7^{{10}}:7^{8}-{3.2}^{4}-2^{{2010}}:2^{{2010}}} \right)$

b, $ \left( {2^{{100}}+2^{{101}}+2^{{102}}} \right):\left( {2^{{97}}+2^{{98}}+2^{{99}}} \right)$

Hướng dẫn giải:

a, Ta có : $ \displaystyle 2010^{{2010}}\left( {7^{{10}}:7^{8}-{3.2}^{4}-2^{{2010}}:2^{{2010}}} \right)=2010^{{2010}}\left( {49-3.16-1} \right)=0$

Bài 19: Tính:

$ \displaystyle A=\dfrac{{\dfrac{{-11}}{2}+\dfrac{{\dfrac{{-5}}{3}}}{{1-\dfrac{4}{3}}}}}{{\dfrac{3}{5}-\dfrac{{\dfrac{{-2}}{5}}}{{\dfrac{4}{5}-\dfrac{2}{3}}}}}$

$ \displaystyle B=\dfrac{{1-\dfrac{1}{{1+\dfrac{4}{3}}}}}{{2+\dfrac{1}{3}-\dfrac{3}{7}}}$

Bài 20: Thực hiện phép tính : $ \displaystyle \dfrac{{45}}{{19}}-\left( {\dfrac{1}{2}+{\left( {\dfrac{1}{3}+{\left( {\dfrac{1}{4}} \right)}^{{-1}}} \right)}^{{-1}}} \right)^{{-1}}$

Hướng dẫn giải:

$ \displaystyle =\dfrac{{45}}{{19}}-\dfrac{1}{{\dfrac{1}{2}+\dfrac{1}{{\dfrac{1}{3}+4}}}}=\dfrac{{45}}{{19}}-\dfrac{{26}}{{19}}=1$

Bài 21: Rút gọn biểu thức: $ \displaystyle A=\left( {\dfrac{3}{2}-\dfrac{2}{5}+\dfrac{1}{{10}}} \right):\left( {\dfrac{3}{2}-\dfrac{2}{3}+\dfrac{1}{{12}}} \right)$

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