Tính giá trị của các biểu thức sau (tính hợp lý nếu có thể): 1) ( -

Lời giải của Tự Học 365

Giải chi tiết:

1) \(\left( { - 3,2} \right).\frac{{ - 15}}{{64}} + \left( {0,8 - 2\frac{4}{{15}}} \right):3\frac{2}{3}\)

\(\begin{array}{l} = \frac{{ - 32}}{{10}} \cdot \frac{{ - 15}}{{64}} + \left( {\frac{8}{{10}} - \frac{{34}}{{15}}} \right):\frac{{11}}{3}\\ = \frac{3}{4} + \left( {\frac{{24}}{{30}} - \frac{{68}}{{30}}} \right) \cdot \frac{3}{{11}}\\ = \frac{3}{4} + \frac{{ - 44}}{{30}} \cdot \frac{3}{{11}}\, = \frac{3}{4} + \frac{{ - 2}}{5}\\ = \frac{{15}}{{20}} + \frac{{ - 8}}{{20}} = \frac{7}{{20}}.\end{array}\)

3) \({\left( {\frac{2}{5}} \right)^2} + 5\frac{1}{2} \cdot \left( {4,5 - 2} \right) + \frac{{{2^3}}}{{\left( { - 4} \right)}}\)

\(\begin{array}{l} = \frac{4}{{25}} + \frac{{11}}{2} \cdot 2,5 + \frac{8}{{ - 4}}\\ = \frac{4}{{25}} + \frac{{11}}{2} \cdot \frac{5}{2} + \left( { - 2} \right)\\ = \frac{4}{{25}} + \frac{{55}}{4} + \left( { - 2} \right)\\\, = \frac{{4.4 + 25.55 - 2.25.4}}{{100}}\\ = \frac{{1191}}{{100}}.\end{array}\)

5) \(\frac{5}{{19}} \cdot \frac{7}{{13}} + \frac{5}{9} \cdot \frac{9}{{13}} - \frac{5}{9} \cdot \frac{3}{{13}}\)

\(\begin{array}{l} = \frac{5}{{13}} \cdot \frac{7}{{19}} + \frac{5}{{13}} - \frac{5}{{13}} \cdot \frac{3}{9}\\ = \frac{5}{{13}} \cdot \frac{7}{{19}} + \frac{5}{{13}} - \frac{5}{{13}} \cdot \frac{1}{3}\\ = \frac{5}{{13}} \cdot \left( {\frac{7}{{19}} + 1 - \frac{1}{3}} \right)\end{array}\)

\(\begin{array}{l} = \frac{5}{{13}} \cdot \left( {\frac{{21}}{{57}} + \frac{{57}}{{57}} - \frac{{19}}{{57}}} \right)\\ = \frac{5}{{13}} \cdot \frac{{59}}{{57}} = \frac{{295}}{{741}}\end{array}\)

7) \(\frac{{{5^2}{{.6}^{11}}{{.16}^2} + {6^2}{{.12}^6}{{.15}^2}}}{{{{2.6}^{12}}{{.10}^4} - {{81}^2}{{.960}^3}}}\)

\(\begin{array}{l} = \frac{{{5^2}.{{\left( {2.3} \right)}^{11}}.{{\left( {{2^4}} \right)}^2} + {{\left( {2.3} \right)}^2}.{{\left( {{2^2}.3} \right)}^6}.{{\left( {3.5} \right)}^2}}}{{2.{{\left( {2.3} \right)}^{12}}.{{\left( {2.5} \right)}^4} - {{\left( {{3^4}} \right)}^2}.{{\left( {{2^6}.3.5} \right)}^3}}}\\ = \frac{{{5^2}{{.2}^{11}}{{.3}^{11}}{{.2}^8} + {2^2}{{.3}^2}{{.2}^{12}}{{.3}^6}{{.3}^2}{{.5}^2}}}{{{{2.2}^{12}}{{.3}^{12}}{{.2}^4}{{.5}^4} - {3^8}{{.2}^{18}}{{.3}^3}{{.5}^3}}}\\\, = \frac{{{5^2}{{.2}^{19}}{{.3}^{11}} + {2^{14}}{{.3}^{10}}{{.5}^2}}}{{{2^{17}}{{.3}^{12}}{{.5}^4} - {3^{11}}{{.2}^{18}}{{.5}^3}}}\\ = \frac{{{5^2}{{.2}^{14}}{{.3}^{10}}.\left( {{2^5}.3 + 1} \right)}}{{{2^{17}}{{.3}^{11}}{{.5}^3}.\left( {3.5 - 2} \right)}}\\ = \frac{{{2^5}.3 + 1}}{{{2^3}.3.5.13}}\\ = \frac{{32.3 + 1}}{{120.13}} = \frac{{97}}{{1560}}\end{array}\)

9) \(\frac{3}{4} \cdot \frac{8}{9} \cdot \frac{{15}}{{16}} \ldots \frac{{9999}}{{10000}}\)

\(\begin{array}{l} = \frac{{1.3}}{{2.2}} \cdot \frac{{2.4}}{{3.3}} \cdot \frac{{3.5}}{{4.4}} \ldots \frac{{99.101}}{{100.100}}\\ = \frac{{1.3.2.4.3.5 \ldots 99.101}}{{2.2.3.3.4.4 \ldots 100.100}}\\ = \frac{{\left( {1.2.3 \ldots 99} \right).\left( {3.4.5 \ldots 101} \right)}}{{\left( {2.3.4 \ldots 100} \right).\left( {2.3.4 \ldots 100} \right)}}\\\, = \frac{{1.2.3 \ldots 99}}{{2.3.4 \ldots 100}} \cdot \frac{{3.4.5 \ldots 101}}{{2.3.4 \ldots 100}}\\ = \frac{1}{{100}} \cdot \frac{{101}}{2} = \frac{{101}}{{200}}.\end{array}\)

2) \({\left( { - 2} \right)^3}\left( {\frac{3}{4} - 0,25} \right):\left( {2\frac{1}{4} - 1\frac{1}{6}} \right)\)

\(\begin{array}{l} =  - 8 \cdot \left( {\frac{3}{4} - \frac{1}{4}} \right):\left( {\frac{9}{4} - \frac{7}{6}} \right)\\ =  - 8 \cdot \frac{2}{4}:\left( {\frac{{27}}{{12}} - \frac{{14}}{{12}}} \right)\\ =  - 4:\frac{{13}}{{12}} = \frac{{ - 4.12}}{{13}} =  - \frac{{48}}{{13}}.\end{array}\)

4) \(\,\,\,\frac{5}{4} \cdot {\left( {\frac{{ - 1}}{2}} \right)^2}:\left( {1\frac{5}{{16}} - 1,5} \right) + {2019^0}\)

\(\begin{array}{l}\, = \frac{5}{4} \cdot \frac{1}{4}:\left( {\frac{{21}}{{16}} - \frac{3}{2}} \right) + 1\\ = \frac{5}{4} \cdot \frac{1}{4}:\left( {\frac{{21}}{{16}} - \frac{{24}}{{16}}} \right) + 1\\ = \frac{5}{4} \cdot \frac{1}{4}:\frac{{ - 3}}{{16}} + 1 = \frac{5}{{16}} \cdot \frac{{16}}{{ - 3}} + 1\\\, = \frac{5}{{ - 3}} + 1 = \frac{{ - 5}}{3} + \frac{3}{3} =  - \frac{2}{3}\end{array}\)

6) \(\frac{{2018.2019 + 2020.21 + 1998}}{{2020.2019 - 2018.2019}}\)  

\(\begin{array}{l} = \frac{{2018.2019 + \left( {2019 + 1} \right).21 + 1998}}{{2019.\left( {2020 - 2018} \right)}}\\ = \frac{{2018.2019 + 2019.21 + 1.21 + 1998}}{{2019.2}}\\\, = \frac{{2019.\left( {2018 + 21} \right) + 2019}}{{2019.2}}\\ = \frac{{2019.2039 + 2019}}{{2019.2}}\end{array}\)

\( = \frac{{2019.2040}}{{2019.2}} = 1020.\)

8) \(\frac{1}{{18}} + \frac{1}{{54}} + \frac{1}{{108}} +  \ldots  + \frac{1}{{990}}\)

\(\begin{array}{l} = \frac{1}{{9.2}} + \frac{1}{{9.6}} + \frac{1}{{9.12}} +  \ldots  + \frac{1}{{9.110}}\\ = \frac{1}{9} \cdot \left( {\frac{1}{2} + \frac{1}{6} + \frac{1}{{12}} +  \ldots  + \frac{1}{{110}}} \right)\\\,\, = \frac{1}{9} \cdot \left( {\frac{1}{{1.2}} + \frac{1}{{2.3}} + \frac{1}{{3.4}} +  \ldots  + \frac{1}{{10.11}}} \right)\\ = \frac{1}{9} \cdot \left( {1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} +  \ldots  + \frac{1}{{10}} - \frac{1}{{11}}} \right)\\\, = \frac{1}{9} \cdot \left( {1 - \frac{1}{{11}}} \right) = \frac{1}{9} \cdot \left( {\frac{{11}}{{11}} - \frac{1}{{11}}} \right)\\ = \frac{1}{9} \cdot \frac{{10}}{{11}} = \frac{{10}}{{99}}.\end{array}\)

10) \(\,\,\,\,1\frac{6}{{41}} \cdot \left( {\frac{{12 + \frac{{12}}{{19}} - \frac{{12}}{{37}} - \frac{{12}}{{53}}}}{{3 + \frac{3}{{19}} - \frac{3}{{37}} - \frac{3}{{53}}}}:\frac{{4 + \frac{4}{{17}} + \frac{4}{{19}} + \frac{4}{{2006}}}}{{5 + \frac{5}{{17}} + \frac{5}{{19}} + \frac{5}{{2006}}}}} \right) \cdot \frac{{124242423}}{{237373735}}\)

\(\begin{array}{l} = \frac{{47}}{{41}} \cdot \left( {\frac{{12 \cdot \left( {1 + \frac{1}{{19}} - \frac{1}{{37}} - \frac{1}{{53}}} \right)}}{{3 \cdot \left( {1 + \frac{1}{{19}} - \frac{1}{{37}} - \frac{1}{{53}}} \right)}}:\frac{{4 \cdot \left( {1 + \frac{1}{{17}} + \frac{1}{{19}} + \frac{1}{{2006}}} \right)}}{{5 \cdot \left( {1 + \frac{1}{{17}} + \frac{1}{{19}} + \frac{1}{{2006}}} \right)}}} \right) \cdot \frac{{41.3.1010101}}{{47.5.1010101}}\\ = \frac{{47}}{{41}} \cdot \left( {\frac{{12}}{3} \cdot \frac{5}{4}} \right) \cdot \frac{{41.3}}{{47.5}} = \frac{{47}}{{41}} \cdot \frac{5}{1} \cdot \frac{{41.3}}{{47.5}} = 3\end{array}\)

11) \(\frac{{\frac{1}{{1.300}} + \frac{1}{{2.301}} + \frac{1}{{3.302}} +  \ldots  + \frac{1}{{101.400}}}}{{\frac{1}{{1.102}} + \frac{1}{{2.103}} + \frac{1}{{3.104}} +  \ldots  + \frac{1}{{299.400}}}}\)

\(\begin{array}{l} = \frac{{101}}{{299}} \cdot \frac{{299.\left( {\frac{1}{{1.300}} + \frac{1}{{2.301}} + \frac{1}{{3.302}} +  \ldots  + \frac{1}{{101.400}}} \right)}}{{101.\left( {\frac{1}{{1.102}} + \frac{1}{{2.103}} + \frac{1}{{3.104}} +  \ldots  + \frac{1}{{299.400}}} \right)}}\\ = \frac{{101}}{{299}} \cdot \frac{{\frac{{299}}{{1.300}} + \frac{{299}}{{2.301}} + \frac{{299}}{{3.302}} +  \ldots  + \frac{{299}}{{101.400}}}}{{\frac{{101}}{{1.102}} + \frac{{101}}{{2.103}} + \frac{{101}}{{3.104}} +  \ldots  + \frac{{101}}{{299.400}}}}\\ = \frac{{101}}{{299}} \cdot \frac{{\frac{{300 - 1}}{{1.300}} + \frac{{301 - 2}}{{2.301}} + \frac{{302 - 3}}{{3.302}} +  \ldots  + \frac{{400 - 101}}{{101.400}}}}{{\frac{{102 - 1}}{{1.102}} + \frac{{103 - 2}}{{2.103}} +  \ldots  + \frac{{400 - 299}}{{299.400}}}}\\ = \frac{{101}}{{299}} \cdot \frac{{1 - \frac{1}{{300}} + \frac{1}{2} - \frac{1}{{301}} + \frac{1}{3} - \frac{1}{{302}} +  \ldots  + \frac{1}{{101}} - \frac{1}{{400}}}}{{1 - \frac{1}{{102}} + \frac{1}{2} - \frac{1}{{103}} +  \ldots  + \frac{1}{{299}} - \frac{1}{{400}}}}\\ = \frac{{101}}{{299}} \cdot \frac{{\left( {1 + \frac{1}{2} + \frac{1}{3} +  \ldots  + \frac{1}{{101}}} \right) - \left( {\frac{1}{{300}} + \frac{1}{{301}} + \frac{1}{{302}} +  \ldots  + \frac{1}{{400}}} \right)}}{{\left( {1 + \frac{1}{2} + \frac{1}{3} +  \ldots  + \frac{1}{{299}}} \right) - \left( {\frac{1}{{102}} + \frac{1}{{103}} +  \ldots  + \frac{1}{{400}}} \right)}}\\ = \frac{{101}}{{299}} \cdot \frac{{\left( {1 + \frac{1}{2} + \frac{1}{3} +  \ldots  + \frac{1}{{101}}} \right) - \left( {\frac{1}{{300}} + \frac{1}{{301}} + \frac{1}{{302}} +  \ldots  + \frac{1}{{400}}} \right)}}{{\left( {1 + \frac{1}{2} + \frac{1}{3} +  \ldots  + \frac{1}{{101}}} \right) + \left( {\frac{1}{{102}} + \frac{1}{{103}} +  \ldots  + \frac{1}{{299}}} \right) - \left( {\frac{1}{{102}} + \frac{1}{{103}} +  \ldots  + \frac{1}{{299}}} \right) - \left( {\frac{1}{{300}} + \frac{1}{{301}} + \frac{1}{{302}} +  \ldots  + \frac{1}{{400}}} \right)}}\\ = \frac{{101}}{{299}} \cdot \frac{{\left( {1 + \frac{1}{2} + \frac{1}{3} +  \ldots  + \frac{1}{{101}}} \right) - \left( {\frac{1}{{300}} + \frac{1}{{301}} + \frac{1}{{302}} +  \ldots  + \frac{1}{{400}}} \right)}}{{\left( {1 + \frac{1}{2} + \frac{1}{3} +  \ldots  + \frac{1}{{101}}} \right) - \left( {\frac{1}{{300}} + \frac{1}{{301}} + \frac{1}{{302}} +  \ldots  + \frac{1}{{400}}} \right)}}\\ = \frac{{101}}{{299}} \cdot \end{array}\)

12) \(\frac{{\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + .... + \frac{1}{{200}}}}{{\frac{1}{{199}} + \frac{2}{{198}} + \frac{3}{{197}} + ..... + \frac{{198}}{2} + \frac{{199}}{1}}}\)

Xét mẫu:

\(\begin{array}{l}\frac{1}{{199}} + \frac{2}{{198}} + \frac{3}{{197}} +  \ldots  + \frac{{198}}{2} + \frac{{199}}{1}\\ = \frac{{200 - 199}}{{199}} + \frac{{200 - 198}}{{198}} + \frac{{200 - 197}}{{197}} +  \ldots  + \frac{{200 - 2}}{2} + \frac{{200 - 1}}{1}\\\, = \frac{{200}}{{199}} - 1 + \frac{{200}}{{198}} - 1 + \frac{{200}}{{197}} - 1 +  \ldots  + \frac{{200}}{2} - 1 + \frac{{200}}{1} - 1\\ = \left( {\frac{{200}}{{199}} + \frac{{200}}{{198}} + \frac{{200}}{{197}} +  \ldots  + \frac{{200}}{1}} \right) - \left( {1 + 1 + 1 +  \ldots  + 1} \right)\\ = 200.\left( {\frac{1}{{199}} + \frac{1}{{198}} +  \ldots  + \frac{1}{2} + 1} \right) - 199\\ = 200.\left( {\frac{1}{{199}} + \frac{1}{{198}} +  \ldots  + \frac{1}{2}} \right) + 200 - 199\\ = 200.\left( {\frac{1}{{199}} + \frac{1}{{198}} +  \ldots  + \frac{1}{2}} \right) + 1\\\,\, = 200.\left( {\frac{1}{{199}} + \frac{1}{{198}} +  \ldots  + \frac{1}{2}} \right) + \frac{{200}}{{200}}\\ = 200 \cdot \left( {\frac{1}{{200}} + \frac{1}{{199}} + \frac{1}{{198}} +  \ldots  + \frac{1}{2}} \right)\end{array}\)

Do vậy ta có: \(\,\,\,\frac{{\frac{1}{2} + \frac{1}{3} + \frac{1}{4} +  \ldots  + \frac{1}{{200}}}}{{\frac{1}{{199}} + \frac{2}{{198}} + \frac{3}{{197}} +  \ldots  + \frac{{198}}{2} + \frac{{199}}{1}}} = \frac{{\frac{1}{2} + \frac{1}{3} + \frac{1}{4} +  \ldots  + \frac{1}{{200}}}}{{200 \cdot \left( {\frac{1}{{200}} + \frac{1}{{199}} + \frac{1}{{198}} +  \ldots  + \frac{1}{2}} \right)}} = \frac{1}{{200}}\)