TopStudy

Đáp án và lời giải

Đề cương ôn tập giữa kì 2 Toán 6 Chân trời sáng tạo cấu trúc mới (Tự luận) có đáp án - Phần 1

Về trang đề

Câu 1

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a) 45+−65=4+(−6)5=−25.\frac{4}{5} + \frac{{ - 6}}{5} = \frac{{4 + \left( { - 6} \right)}}{5} = \frac{{ - 2}}{5}.

Câu 2

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b) −18+−53\frac{{ - 1}}{8} + \frac{{ - 5}}{3}=−324+−4024= \frac{{ - 3}}{{24}} + \frac{{ - 40}}{{24}}=−4324= \frac{{ - 43}}{{24}}.

Câu 3

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c) −635⋅−4954\frac{{ - 6}}{{35}} \cdot \frac{{ - 49}}{{54}}

=−6⋅(−49)35⋅54= \frac{{ - 6 \cdot \left( { - 49} \right)}}{{35 \cdot 54}}

=745= \frac{7}{{45}}.

Câu 4

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d) −45:34\frac{{ - 4}}{5}:\frac{3}{4}

=−45.43= \frac{{ - 4}}{5}.\frac{4}{3}

=−1615= \frac{{ - 16}}{{15}}.

Câu 5

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e) −813−(37+513)\frac{{ - 8}}{{13}} - \left( {\frac{3}{7} + \frac{5}{{13}}} \right)

=−813−37−513= \frac{{ - 8}}{{13}} - \frac{3}{7} - \frac{5}{{13}}

=−813−513−37= \frac{{ - 8}}{{13}} - \frac{5}{{13}} - \frac{3}{7}=−1−37= - 1 - \frac{3}{7}

=−7−37= \frac{{ - 7 - 3}}{7}=−107= \frac{{ - 10}}{7}.

Câu 6

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f) 12+−38+59\frac{1}{2} + \frac{{ - 3}}{8} + \frac{5}{9}

=3672+−2772+4072= \frac{{36}}{{72}} + \frac{{ - 27}}{{72}} + \frac{{40}}{{72}}

=4972.= \frac{{49}}{{72}}.

Câu 7

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g) 625−(237+325)6\frac{2}{5} - \left( {2\frac{3}{7} + 3\frac{2}{5}} \right)

=325−177−175= \frac{{32}}{5} - \frac{{17}}{7} - \frac{{17}}{5}

=325−175−177= \frac{{32}}{5} - \frac{{17}}{5} - \frac{{17}}{7}

=3−177= 3 - \frac{{17}}{7}=21−177= \frac{{21 - 17}}{7}=47= \frac{4}{7}.

Câu 8

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h) 23−53⋅925\frac{2}{3} - \frac{5}{3} \cdot \frac{9}{{25}}

=23−5⋅93⋅25=23−35= \frac{2}{3} - \frac{{5 \cdot 9}}{{3 \cdot 25}} = \frac{2}{3} - \frac{3}{5}

=1015−915=115.= \frac{{10}}{{15}} - \frac{9}{{15}} = \frac{1}{{15}}.

Câu 9

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i) 45+915:115\frac{4}{5} + \frac{9}{{15}}:\frac{1}{{15}}

=45+915⋅15= \frac{4}{5} + \frac{9}{{15}} \cdot 15

=45+9= \frac{4}{5} + 9

=45+455= \frac{4}{5} + \frac{{45}}{5}

=495.= \frac{{49}}{5}.

Câu 10

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j) −1021⋅[915+(35)2]- \frac{{10}}{{21}} \cdot \left[ {\frac{9}{{15}} + {{\left( {\frac{3}{5}} \right)}^2}} \right]

=−1021⋅[35+925]= - \frac{{10}}{{21}} \cdot \left[ {\frac{3}{5} + \frac{9}{{25}}} \right]

=−1021⋅[1525+925]= \frac{{ - 10}}{{21}} \cdot \left[ {\frac{{15}}{{25}} + \frac{9}{{25}}} \right]

=−1021⋅2425= \frac{{ - 10}}{{21}} \cdot \frac{{24}}{{25}}

=−10⋅2421⋅25=−1635.= \frac{{ - 10 \cdot 24}}{{21 \cdot 25}} = \frac{{ - 16}}{{35}}.

Câu 11

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k) (14+−513)+(211+−813+34)\left( {\frac{1}{4} + \frac{{ - 5}}{{13}}} \right) + \left( {\frac{2}{{11}} + \frac{{ - 8}}{{13}} + \frac{3}{4}} \right)

=14+−513+211+−813+34= \frac{1}{4} + \frac{{ - 5}}{{13}} + \frac{2}{{11}} + \frac{{ - 8}}{{13}} + \frac{3}{4}

=(14+34)+(−513+−813)+211= \left( {\frac{1}{4} + \frac{3}{4}} \right) + \left( {\frac{{ - 5}}{{13}} + \frac{{ - 8}}{{13}}} \right) + \frac{2}{{11}}

=44+−1313+211= \frac{4}{4} + \frac{{ - 13}}{{13}} + \frac{2}{{11}}

=1+(−1)+211= 1 + \left( { - 1} \right) + \frac{2}{{11}}=211.= \frac{2}{{11}}.

Câu 12

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l) −313+−717+−519+−1013+2417+−1419\frac{{ - 3}}{{13}} + \frac{{ - 7}}{{17}} + \frac{{ - 5}}{{19}} + \frac{{ - 10}}{{13}} + \frac{{24}}{{17}} + \frac{{ - 14}}{{19}}

=(−313+−1013)+(−717+2417)+(−519+−1419)= \left( {\frac{{ - 3}}{{13}} + \frac{{ - 10}}{{13}}} \right) + \left( {\frac{{ - 7}}{{17}} + \frac{{24}}{{17}}} \right) + \left( {\frac{{ - 5}}{{19}} + \frac{{ - 14}}{{19}}} \right)

=−1313+1717+−1919= \frac{{ - 13}}{{13}} + \frac{{17}}{{17}} + \frac{{ - 19}}{{19}}

=(−1)+1+(−1)= \left( { - 1} \right) + 1 + \left( { - 1} \right)

=−1.= - 1.

Câu 13

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m) (517−833)−(2533−1217)−20210\left( {\frac{5}{{17}} - \frac{8}{{33}}} \right) - \left( {\frac{{25}}{{33}} - \frac{{12}}{{17}}} \right) - {2021^0}

=517−833−2533+1217−1= \frac{5}{{17}} - \frac{8}{{33}} - \frac{{25}}{{33}} + \frac{{12}}{{17}} - 1

=(517+1217)−(833+2533)−1= \left( {\frac{5}{{17}} + \frac{{12}}{{17}}} \right) - \left( {\frac{8}{{33}} + \frac{{25}}{{33}}} \right) - 1

=1717−3333−1= \frac{{17}}{{17}} - \frac{{33}}{{33}} - 1

=1−1−1=−1.= 1 - 1 - 1 = - 1.

Câu 14

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n) −412+1845+−69+−2135+630.\frac{-4}{12}+\frac{18}{45}+\frac{-6}{9}+\frac{-21}{35}+\frac{6}{30}.

n) −412+1845+−69+−2135+630\frac{{ - 4}}{{12}} + \frac{{18}}{{45}} + \frac{{ - 6}}{9} + \frac{{ - 21}}{{35}} + \frac{6}{{30}}

=−13+25+−23+−35+15= \frac{{ - 1}}{3} + \frac{2}{5} + \frac{{ - 2}}{3} + \frac{{ - 3}}{5} + \frac{1}{5}=(−13+−23)+(25+−35+15)= \left( {\frac{{ - 1}}{3} + \frac{{ - 2}}{3}} \right) + \left( {\frac{2}{5} + \frac{{ - 3}}{5} + \frac{1}{5}} \right)

=−33+05=−1.= \frac{{ - 3}}{3} + \frac{0}{5} = - 1.

Câu 15

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a) −57⋅211+−57⋅911+57\frac{{ - 5}}{7} \cdot \frac{2}{{11}} + \frac{{ - 5}}{7} \cdot \frac{9}{{11}} + \frac{5}{7}

=57⋅−211+57⋅−911+57⋅1= \frac{5}{7} \cdot \frac{{ - 2}}{{11}} + \frac{5}{7} \cdot \frac{{ - 9}}{{11}} + \frac{5}{7} \cdot 1

=57⋅(−211+−911+1)= \frac{5}{7} \cdot \left( {\frac{{ - 2}}{{11}} + \frac{{ - 9}}{{11}} + 1} \right)

=57⋅(−1111+1)=57⋅(−1+1)= \frac{5}{7} \cdot \left( {\frac{{ - 11}}{{11}} + 1} \right) = \frac{5}{7} \cdot \left( { - 1 + 1} \right)

=57⋅0=0.= \frac{5}{7} \cdot 0 = 0.

Câu 16

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b) 59⋅−47−59⋅(−187)\frac{5}{9} \cdot \frac{{ - 4}}{7} - \frac{5}{9} \cdot \left( {\frac{{ - 18}}{7}} \right)

=59⋅−47+59⋅187= \frac{5}{9} \cdot \frac{{ - 4}}{7} + \frac{5}{9} \cdot \frac{{18}}{7}

=59⋅(−47+187)= \frac{5}{9} \cdot \left( {\frac{{ - 4}}{7} + \frac{{18}}{7}} \right)

=59⋅147= \frac{5}{9} \cdot \frac{{14}}{7}

=59⋅2=109.= \frac{5}{9} \cdot 2 = \frac{{10}}{9}.

Câu 17

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c) 1112⋅715+1112⋅815−2312\frac{{11}}{{12}} \cdot \frac{7}{{15}} + \frac{{11}}{{12}} \cdot \frac{8}{{15}} - \frac{{23}}{{12}}

=1112⋅(715+815)−2312= \frac{{11}}{{12}} \cdot \left( {\frac{7}{{15}} + \frac{8}{{15}}} \right) - \frac{{23}}{{12}}

=1112⋅1515−2312= \frac{{11}}{{12}} \cdot \frac{{15}}{{15}} - \frac{{23}}{{12}}

=1112−2312= \frac{{11}}{{12}} - \frac{{23}}{{12}}

=−1212=−1.= \frac{{ - 12}}{{12}} = - 1.

Câu 18

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d) −37⋅1213+37⋅(−113)+514- \frac{3}{7} \cdot \frac{{12}}{{13}} + \frac{3}{7} \cdot \left( { - \frac{1}{{13}}} \right) + \frac{5}{{14}}

=37⋅−1213+37⋅(−113)+514= \frac{3}{7} \cdot \frac{{ - 12}}{{13}} + \frac{3}{7} \cdot \left( { - \frac{1}{{13}}} \right) + \frac{5}{{14}}

=37⋅[−1213+(−113)]+514= \frac{3}{7} \cdot \left[ {\frac{{ - 12}}{{13}} + \left( { - \frac{1}{{13}}} \right)} \right] + \frac{5}{{14}}

=37⋅−1313+514= \frac{3}{7} \cdot \frac{{ - 13}}{{13}} + \frac{5}{{14}}

=37⋅(−1)+514= \frac{3}{7} \cdot \left( { - 1} \right) + \frac{5}{{14}}=−37+514= \frac{{ - 3}}{7} + \frac{5}{{14}}

=−614+514=−114.= \frac{{ - 6}}{{14}} + \frac{5}{{14}} = \frac{{ - 1}}{{14}}.

Câu 19

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e) −94:72+94⋅167−23\frac{{ - 9}}{4}:\frac{7}{2} + \frac{9}{4} \cdot \frac{{16}}{7} - \frac{2}{3}

=−94⋅27+94⋅167−23= \frac{{ - 9}}{4} \cdot \frac{2}{7} + \frac{9}{4} \cdot \frac{{16}}{7} - \frac{2}{3}

=94⋅−27+94.167−23= \frac{9}{4} \cdot \frac{{ - 2}}{7} + \frac{9}{4}.\frac{{16}}{7} - \frac{2}{3}

=94⋅(−27+167)−23= \frac{9}{4} \cdot \left( {\frac{{ - 2}}{7} + \frac{{16}}{7}} \right) - \frac{2}{3}

=94⋅147−23=94⋅2−23= \frac{9}{4} \cdot \frac{{14}}{7} - \frac{2}{3} = \frac{9}{4} \cdot 2 - \frac{2}{3}

=92−23=276−46=236.= \frac{9}{2} - \frac{2}{3} = \frac{{27}}{6} - \frac{4}{6} = \frac{{23}}{6}.

Câu 20

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f) 311⋅917−1211:179−1311⋅917\frac{3}{{11}} \cdot \frac{9}{{17}} - \frac{{12}}{{11}}:\frac{{17}}{9} - \frac{{13}}{{11}} \cdot \frac{9}{{17}}

=311⋅917−1211⋅917−1311⋅917= \frac{3}{{11}} \cdot \frac{9}{{17}} - \frac{{12}}{{11}} \cdot \frac{9}{{17}} - \frac{{13}}{{11}} \cdot \frac{9}{{17}}

=917⋅(311−1211−1311)= \frac{9}{{17}} \cdot \left( {\frac{3}{{11}} - \frac{{12}}{{11}} - \frac{{13}}{{11}}} \right)

=917⋅−2211= \frac{9}{{17}} \cdot \frac{{ - 22}}{{11}}

=917⋅(−2)=−1817.= \frac{9}{{17}} \cdot \left( { - 2} \right) = \frac{{ - 18}}{{17}}.

Câu 21

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g) 178:−113+318⋅−311+411\frac{{17}}{8}:\frac{{ - 11}}{3} + \frac{{31}}{8} \cdot \frac{{ - 3}}{{11}} + \frac{4}{{11}}

=178⋅−311+318⋅−311+411= \frac{{17}}{8} \cdot \frac{{ - 3}}{{11}} + \frac{{31}}{8} \cdot \frac{{ - 3}}{{11}} + \frac{4}{{11}}

=−311⋅(178+318)+411= \frac{{ - 3}}{{11}} \cdot \left( {\frac{{17}}{8} + \frac{{31}}{8}} \right) + \frac{4}{{11}}

=−311⋅488+411= \frac{{ - 3}}{{11}} \cdot \frac{{48}}{8} + \frac{4}{{11}}

=−311⋅6+411= \frac{{ - 3}}{{11}} \cdot 6 + \frac{4}{{11}}

=−1811+411=−1411.= \frac{{ - 18}}{{11}} + \frac{4}{{11}} = \frac{{ - 14}}{{11}}.

Câu 22

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h) 713⋅11431−3731:137+713\frac{7}{{13}} \cdot 1\frac{{14}}{{31}} - \frac{{37}}{{31}}:\frac{{13}}{7} + \frac{7}{{13}}

=713⋅4531−3731⋅713+713= \frac{7}{{13}} \cdot \frac{{45}}{{31}} - \frac{{37}}{{31}} \cdot \frac{7}{{13}} + \frac{7}{{13}}

=713⋅(4531−3731+1)= \frac{7}{{13}} \cdot \left( {\frac{{45}}{{31}} - \frac{{37}}{{31}} + 1} \right)

=713⋅3931=2131.= \frac{7}{{13}} \cdot \frac{{39}}{{31}} = \frac{{21}}{{31}}.

Câu 23

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i) 45:12+15:12−78\frac{4}{5}:\frac{1}{2} + \frac{1}{5}:\frac{1}{2} - \frac{7}{8}

=(45+15):12−78= \left( {\frac{4}{5} + \frac{1}{5}} \right):\frac{1}{2} - \frac{7}{8}

=1.2−78= 1.2 - \frac{7}{8}

=168−78= \frac{{16}}{8} - \frac{7}{8}=98.= \frac{9}{8}.

Câu 24

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j) 118:−92+278:−92+591\frac{1}{8}:\frac{{ - 9}}{2} + 2\frac{7}{8}:\frac{{ - 9}}{2} + \frac{5}{9}

=98⋅−29+238⋅−29+59= \frac{9}{8} \cdot \frac{{ - 2}}{9} + \frac{{23}}{8} \cdot \frac{{ - 2}}{9} + \frac{5}{9}

=−29⋅(98+238)+59= \frac{{ - 2}}{9} \cdot \left( {\frac{9}{8} + \frac{{23}}{8}} \right) + \frac{5}{9}

=−29⋅328+59= \frac{{ - 2}}{9} \cdot \frac{{32}}{8} + \frac{5}{9}

=−29⋅4+59= \frac{{ - 2}}{9} \cdot 4 + \frac{5}{9}

=−89+59=−39=−13.= \frac{{ - 8}}{9} + \frac{5}{9} = \frac{{ - 3}}{9} = \frac{{ - 1}}{3}.

Câu 25

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k) (45+−97):20242025+(−57−−65):20242025\left( {\frac{4}{5} + \frac{{ - 9}}{7}} \right):\frac{{2024}}{{2025}} + \left( {\frac{{ - 5}}{7} - \frac{{ - 6}}{5}} \right):\frac{{2024}}{{2025}}

=(45+−97+−57−−65):20242025= \left( {\frac{4}{5} + \frac{{ - 9}}{7} + \frac{{ - 5}}{7} - \frac{{ - 6}}{5}} \right):\frac{{2024}}{{2025}}

=[(45−−65)+(−97+−57)]:20242025= \left[ {\left( {\frac{4}{5} - \frac{{ - 6}}{5}} \right) + \left( {\frac{{ - 9}}{7} + \frac{{ - 5}}{7}} \right)} \right]:\frac{{2024}}{{2025}}

=[105+−147]:20242025= \left[ {\frac{{10}}{5} + \frac{{ - 14}}{7}} \right]:\frac{{2024}}{{2025}}

=[2+(−2)]:20242025= \left[ {2 + \left( { - 2} \right)} \right]:\frac{{2024}}{{2025}}

=0:20242025=0.= 0:\frac{{2024}}{{2025}} = 0.

Câu 26

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a) 34+35+37−31164+65+67−611=3⋅(14+14+17−111)6⋅(14+14+17−111)=36=12.\frac{{\frac{3}{4} + \frac{3}{5} + \frac{3}{7} - \frac{3}{{11}}}}{{\frac{6}{4} + \frac{6}{5} + \frac{6}{7} - \frac{6}{{11}}}} = \frac{{3 \cdot \left( {\frac{1}{4} + \frac{1}{4} + \frac{1}{7} - \frac{1}{{11}}} \right)}}{{6 \cdot \left( {\frac{1}{4} + \frac{1}{4} + \frac{1}{7} - \frac{1}{{11}}} \right)}} = \frac{3}{6} = \frac{1}{2}.

Câu 27

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b) 13−15+11063−65+35+56\frac{{\frac{1}{3} - \frac{1}{5} + \frac{1}{{10}}}}{{\frac{6}{3} - \frac{6}{5} + \frac{3}{5}}} + \frac{5}{6}=13−15+1106⋅(13−15+110)+56=16+56=66=1.= \frac{{\frac{1}{3} - \frac{1}{5} + \frac{1}{{10}}}}{{6 \cdot \left( {\frac{1}{3} - \frac{1}{5} + \frac{1}{{10}}} \right)}} + \frac{5}{6} = \frac{1}{6} + \frac{5}{6} = \frac{6}{6} = 1.

Câu 28

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c) 12.517−1314.517+15238−2068+2614.517−15119\frac{{\frac{1}{2}.\frac{5}{{17}} - \frac{{13}}{{14}}.\frac{5}{{17}} + \frac{{15}}{{238}}}}{{\frac{{ - 20}}{{68}} + \frac{{26}}{{14}}.\frac{5}{{17}} - \frac{{15}}{{119}}}}=534−6514.17+15238−2⋅(534−6514.17+15238)=−12= \frac{{\frac{5}{{34}} - \frac{{65}}{{14.17}} + \frac{{15}}{{238}}}}{{ - 2 \cdot \left( {\frac{5}{{34}} - \frac{{65}}{{14.17}} + \frac{{15}}{{238}}} \right)}} = - \frac{1}{2}

Câu 29

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d) 129⋅32−2611⋅323+9238−329+1311⋅323−9119\frac{{\frac{1}{{29}} \cdot \frac{3}{2} - \frac{{26}}{{11}} \cdot \frac{3}{{23}} + \frac{9}{{238}}}}{{\frac{{ - 3}}{{29}} + \frac{{13}}{{11}} \cdot \frac{3}{{23}} - \frac{9}{{119}}}}=−2⋅(358−3911.23+9238)358−3911.23+9238=−2= \frac{{ - 2 \cdot \left( {\frac{3}{{58}} - \frac{{39}}{{11.23}} + \frac{9}{{238}}} \right)}}{{\frac{3}{{58}} - \frac{{39}}{{11.23}} + \frac{9}{{238}}}} = - 2

Câu 30

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e) 12⋅3+13⋅4+14⋅5+15⋅6+16⋅7+17⋅8\frac{1}{{2 \cdot 3}} + \frac{1}{{3 \cdot 4}} + \frac{1}{{4 \cdot 5}} + \frac{1}{{5 \cdot 6}} + \frac{1}{{6 \cdot 7}} + \frac{1}{{7 \cdot 8}}

=12−13+13−14+14−15+15−16+16−17+17−18= \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \frac{1}{4} - \frac{1}{5} + \frac{1}{5} - \frac{1}{6} + \frac{1}{6} - \frac{1}{7} + \frac{1}{7} - \frac{1}{8}

=12−18=48−18=38.= \frac{1}{2} - \frac{1}{8} = \frac{4}{8} - \frac{1}{8} = \frac{3}{8}.

Câu 31

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f) 23.7+27.11+211.15+...+263.67\frac{2}{{3.7}} + \frac{2}{{7.11}} + \frac{2}{{11.15}} + ... + \frac{2}{{63.67}}

=12⋅(43.7+47.11+411.15+...+463.67)= \frac{1}{2} \cdot \left( {\frac{4}{{3.7}} + \frac{4}{{7.11}} + \frac{4}{{11.15}} + ... + \frac{4}{{63.67}}} \right)

=12⋅(13−17+17−111+111−115+...+163−167)= \frac{1}{2} \cdot \left( {\frac{1}{3} - \frac{1}{7} + \frac{1}{7} - \frac{1}{{11}} + \frac{1}{{11}} - \frac{1}{{15}} + ... + \frac{1}{{63}} - \frac{1}{{67}}} \right)

=12⋅(13−167)=12⋅(67201−3201)= \frac{1}{2} \cdot \left( {\frac{1}{3} - \frac{1}{{67}}} \right) = \frac{1}{2} \cdot \left( {\frac{{67}}{{201}} - \frac{3}{{201}}} \right)

=12⋅64201=12⋅32201= \frac{1}{2} \cdot \frac{{64}}{{201}} = \frac{1}{2} \cdot \frac{{32}}{{201}}

Câu 32

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g) 415+435+463+...+4399\frac{4}{{15}} + \frac{4}{{35}} + \frac{4}{{63}} + ... + \frac{4}{{399}}

=43⋅5+45⋅7+47⋅9+...+419⋅21= \frac{4}{{3 \cdot 5}} + \frac{4}{{5 \cdot 7}} + \frac{4}{{7 \cdot 9}} + ... + \frac{4}{{19 \cdot 21}}

=2⋅(23⋅5+25⋅7+27⋅9+...+219⋅21)= 2 \cdot \left( {\frac{2}{{3 \cdot 5}} + \frac{2}{{5 \cdot 7}} + \frac{2}{{7 \cdot 9}} + ... + \frac{2}{{19 \cdot 21}}} \right)

=2⋅(13−15+15−17+17−19+...+119−121)= 2 \cdot \left( {\frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7} + \frac{1}{7} - \frac{1}{9} + ... + \frac{1}{{19}} - \frac{1}{{21}}} \right)

=2⋅(13−121)= 2 \cdot \left( {\frac{1}{3} - \frac{1}{{21}}} \right)=2⋅(721−121)= 2 \cdot \left( {\frac{7}{{21}} - \frac{1}{{21}}} \right)

=2⋅621=2⋅27=47.= 2 \cdot \frac{6}{{21}} = 2 \cdot \frac{2}{7} = \frac{4}{7}.

Câu 33

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h) 13+115+135+163+199+1143+1195\frac{1}{3} + \frac{1}{{15}} + \frac{1}{{35}} + \frac{1}{{63}} + \frac{1}{{99}} + \frac{1}{{143}} + \frac{1}{{195}}

=11⋅3+13⋅5+15⋅7+17⋅9+19⋅11+111⋅13+113⋅15= \frac{1}{{1 \cdot 3}} + \frac{1}{{3 \cdot 5}} + \frac{1}{{5 \cdot 7}} + \frac{1}{{7 \cdot 9}} + \frac{1}{{9 \cdot 11}} + \frac{1}{{11 \cdot 13}} + \frac{1}{{13 \cdot 15}}

=12⋅(21⋅3+23⋅5+25⋅7+27⋅9+29⋅11+211⋅13+213⋅15)= \frac{1}{2} \cdot \left( {\frac{2}{{1 \cdot 3}} + \frac{2}{{3 \cdot 5}} + \frac{2}{{5 \cdot 7}} + \frac{2}{{7 \cdot 9}} + \frac{2}{{9 \cdot 11}} + \frac{2}{{11 \cdot 13}} + \frac{2}{{13 \cdot 15}}} \right)

=12⋅(11−13+13−15+15−17+17−19+19−111+111−113+113−115)= \frac{1}{2} \cdot \left( {\frac{1}{1} - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7} + \frac{1}{7} - \frac{1}{9} + \frac{1}{9} - \frac{1}{{11}} + \frac{1}{{11}} - \frac{1}{{13}} + \frac{1}{{13}} - \frac{1}{{15}}} \right)

=12⋅(11−115)=12⋅(1515−115)= \frac{1}{2} \cdot \left( {\frac{1}{1} - \frac{1}{{15}}} \right) = \frac{1}{2} \cdot \left( {\frac{{15}}{{15}} - \frac{1}{{15}}} \right)

=12⋅1415=715.= \frac{1}{2} \cdot \frac{{14}}{{15}} = \frac{7}{{15}}.

Câu 34

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i) (1−12)⋅(1−13)⋅(1−14)⋅(1−15)⋅(1−16)⋅(1−17)⋅(1−18)\left( {1 - \frac{1}{2}} \right) \cdot \left( {1 - \frac{1}{3}} \right) \cdot \left( {1 - \frac{1}{4}} \right) \cdot \left( {1 - \frac{1}{5}} \right) \cdot \left( {1 - \frac{1}{6}} \right) \cdot \left( {1 - \frac{1}{7}} \right) \cdot \left( {1 - \frac{1}{8}} \right)

=(22−12)⋅(33−13)⋅(44−14)⋅(55−15)⋅(66−16)⋅(77−17)⋅(88−18)= \left( {\frac{2}{2} - \frac{1}{2}} \right) \cdot \left( {\frac{3}{3} - \frac{1}{3}} \right) \cdot \left( {\frac{4}{4} - \frac{1}{4}} \right) \cdot \left( {\frac{5}{5} - \frac{1}{5}} \right) \cdot \left( {\frac{6}{6} - \frac{1}{6}} \right) \cdot \left( {\frac{7}{7} - \frac{1}{7}} \right) \cdot \left( {\frac{8}{8} - \frac{1}{8}} \right)

=12⋅23⋅34⋅45⋅56⋅67⋅78= \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \frac{4}{5} \cdot \frac{5}{6} \cdot \frac{6}{7} \cdot \frac{7}{8}=18.= \frac{1}{8}.

Câu 35

Hướng dẫn chấm do TopStudy biên soạn, mang tính gợi ý

Hướng dẫn giải:

j) (1+110)⋅(1+111)⋅(1+112)⋅...⋅(1+159)⋅(1+160)\left( {1 + \frac{1}{{10}}} \right) \cdot \left( {1 + \frac{1}{{11}}} \right) \cdot \left( {1 + \frac{1}{{12}}} \right) \cdot ... \cdot \left( {1 + \frac{1}{{59}}} \right) \cdot \left( {1 + \frac{1}{{60}}} \right)

=(1010+110)⋅(1111+111)⋅(1212+112)⋅...⋅(5959+159)⋅(6060+160)= \left( {\frac{{10}}{{10}} + \frac{1}{{10}}} \right) \cdot \left( {\frac{{11}}{{11}} + \frac{1}{{11}}} \right) \cdot \left( {\frac{{12}}{{12}} + \frac{1}{{12}}} \right) \cdot ... \cdot \left( {\frac{{59}}{{59}} + \frac{1}{{59}}} \right) \cdot \left( {\frac{{60}}{{60}} + \frac{1}{{60}}} \right)

=1110⋅1211⋅1312⋅...⋅6059⋅6160= \frac{{11}}{{10}} \cdot \frac{{12}}{{11}} \cdot \frac{{13}}{{12}} \cdot ... \cdot \frac{{60}}{{59}} \cdot \frac{{61}}{{60}}=6110.= \frac{{61}}{{10}}.