Đề cương ôn tập giữa kì 2 Toán 6 Cánh diều cấu trúc mới (Tự luận) có đáp án - Phần 2

a) \(\frac{{ - 5}}{7} \cdot \frac{2}{{11}} + \frac{{ - 5}}{7} \cdot \frac{9}{{11}} + \frac{5}{7}\)

\[ = \frac{5}{7} \cdot \frac{{ - 2}}{{11}} + \frac{5}{7} \cdot \frac{{ - 9}}{{11}} + \frac{5}{7} \cdot 1\]

\[ = \frac{5}{7} \cdot \left( {\frac{{ - 2}}{{11}} + \frac{{ - 9}}{{11}} + 1} \right)\]

\[ = \frac{5}{7} \cdot \left( {\frac{{ - 11}}{{11}} + 1} \right) = \frac{5}{7} \cdot \left( { - 1 + 1} \right)\]

\( = \frac{5}{7} \cdot 0 = 0.\)

b) \(\frac{5}{9} \cdot \frac{{ - 4}}{7} - \frac{5}{9} \cdot \left( {\frac{{ - 18}}{7}} \right)\)

\( = \frac{5}{9} \cdot \frac{{ - 4}}{7} + \frac{5}{9} \cdot \frac{{18}}{7}\)

\( = \frac{5}{9} \cdot \left( {\frac{{ - 4}}{7} + \frac{{18}}{7}} \right)\)

\( = \frac{5}{9} \cdot \frac{{14}}{7}\)

\( = \frac{5}{9} \cdot 2 = \frac{{10}}{9}.\)

c) \[\frac{{11}}{{12}} \cdot \frac{7}{{15}} + \frac{{11}}{{12}} \cdot \frac{8}{{15}} - \frac{{23}}{{12}}\]

\( = \frac{{11}}{{12}} \cdot \left( {\frac{7}{{15}} + \frac{8}{{15}}} \right) - \frac{{23}}{{12}}\)

\( = \frac{{11}}{{12}} \cdot \frac{{15}}{{15}} - \frac{{23}}{{12}}\)

\( = \frac{{11}}{{12}} - \frac{{23}}{{12}}\)

\( = \frac{{ - 12}}{{12}} = - 1.\)

d) \( - \frac{3}{7} \cdot \frac{{12}}{{13}} + \frac{3}{7} \cdot \left( { - \frac{1}{{13}}} \right) + \frac{5}{{14}}\)

\( = \frac{3}{7} \cdot \frac{{ - 12}}{{13}} + \frac{3}{7} \cdot \left( { - \frac{1}{{13}}} \right) + \frac{5}{{14}}\)

\[ = \frac{3}{7} \cdot \left[ {\frac{{ - 12}}{{13}} + \left( { - \frac{1}{{13}}} \right)} \right] + \frac{5}{{14}}\]

\[ = \frac{3}{7} \cdot \frac{{ - 13}}{{13}} + \frac{5}{{14}}\]

\[ = \frac{3}{7} \cdot \left( { - 1} \right) + \frac{5}{{14}}\]\[ = \frac{{ - 3}}{7} + \frac{5}{{14}}\]

\[ = \frac{{ - 6}}{{14}} + \frac{5}{{14}} = \frac{{ - 1}}{{14}}.\]

e) \(\frac{{ - 9}}{4}:\frac{7}{2} + \frac{9}{4} \cdot \frac{{16}}{7} - \frac{2}{3}\)

\( = \frac{{ - 9}}{4} \cdot \frac{2}{7} + \frac{9}{4} \cdot \frac{{16}}{7} - \frac{2}{3}\)

\( = \frac{9}{4} \cdot \frac{{ - 2}}{7} + \frac{9}{4}.\frac{{16}}{7} - \frac{2}{3}\)

\( = \frac{9}{4} \cdot \left( {\frac{{ - 2}}{7} + \frac{{16}}{7}} \right) - \frac{2}{3}\)

\( = \frac{9}{4} \cdot \frac{{14}}{7} - \frac{2}{3} = \frac{9}{4} \cdot 2 - \frac{2}{3}\)

\( = \frac{9}{2} - \frac{2}{3} = \frac{{27}}{6} - \frac{4}{6} = \frac{{23}}{6}.\)

f) \(\frac{3}{{11}} \cdot \frac{9}{{17}} - \frac{{12}}{{11}}:\frac{{17}}{9} - \frac{{13}}{{11}} \cdot \frac{9}{{17}}\)

\[ = \frac{3}{{11}} \cdot \frac{9}{{17}} - \frac{{12}}{{11}} \cdot \frac{9}{{17}} - \frac{{13}}{{11}} \cdot \frac{9}{{17}}\]

\( = \frac{9}{{17}} \cdot \left( {\frac{3}{{11}} - \frac{{12}}{{11}} - \frac{{13}}{{11}}} \right)\)

\( = \frac{9}{{17}} \cdot \frac{{ - 22}}{{11}}\)

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

g) \(\frac{{17}}{8}:\frac{{ - 11}}{3} + \frac{{31}}{8} \cdot \frac{{ - 3}}{{11}} + \frac{4}{{11}}\)

\( = \frac{{17}}{8} \cdot \frac{{ - 3}}{{11}} + \frac{{31}}{8} \cdot \frac{{ - 3}}{{11}} + \frac{4}{{11}}\)

\( = \frac{{ - 3}}{{11}} \cdot \left( {\frac{{17}}{8} + \frac{{31}}{8}} \right) + \frac{4}{{11}}\)

\( = \frac{{ - 3}}{{11}} \cdot \frac{{48}}{8} + \frac{4}{{11}}\)

\( = \frac{{ - 3}}{{11}} \cdot 6 + \frac{4}{{11}}\)

\( = \frac{{ - 18}}{{11}} + \frac{4}{{11}} = \frac{{ - 14}}{{11}}.\)

h) \[\frac{7}{{13}} \cdot 1\frac{{14}}{{31}} - \frac{{37}}{{31}}:\frac{{13}}{7} + \frac{7}{{13}}\]

\( = \frac{7}{{13}} \cdot \frac{{45}}{{31}} - \frac{{37}}{{31}} \cdot \frac{7}{{13}} + \frac{7}{{13}}\)

\[ = \frac{7}{{13}} \cdot \left( {\frac{{45}}{{31}} - \frac{{37}}{{31}} + 1} \right)\]

\[ = \frac{7}{{13}} \cdot \frac{{39}}{{31}} = \frac{{21}}{{31}}.\]

i) \(\frac{4}{5}:\frac{1}{2} + \frac{1}{5}:\frac{1}{2} - \frac{7}{8}\)

\[ = \left( {\frac{4}{5} + \frac{1}{5}} \right):\frac{1}{2} - \frac{7}{8}\]

\[ = 1.2 - \frac{7}{8}\]

\[ = \frac{{16}}{8} - \frac{7}{8}\]\[ = \frac{9}{8}.\]

j) \[1\frac{1}{8}:\frac{{ - 9}}{2} + 2\frac{7}{8}:\frac{{ - 9}}{2} + \frac{5}{9}\]

\[ = \frac{9}{8} \cdot \frac{{ - 2}}{9} + \frac{{23}}{8} \cdot \frac{{ - 2}}{9} + \frac{5}{9}\]

\[ = \frac{{ - 2}}{9} \cdot \left( {\frac{9}{8} + \frac{{23}}{8}} \right) + \frac{5}{9}\]

\[ = \frac{{ - 2}}{9} \cdot \frac{{32}}{8} + \frac{5}{9}\]

\[ = \frac{{ - 2}}{9} \cdot 4 + \frac{5}{9}\]

\[ = \frac{{ - 8}}{9} + \frac{5}{9} = \frac{{ - 3}}{9} = \frac{{ - 1}}{3}.\]

k) \(\left( {\frac{4}{5} + \frac{{ - 9}}{7}} \right):\frac{{2024}}{{2025}} + \left( {\frac{{ - 5}}{7} - \frac{{ - 6}}{5}} \right):\frac{{2024}}{{2025}}\)

\[ = \left( {\frac{4}{5} + \frac{{ - 9}}{7} + \frac{{ - 5}}{7} - \frac{{ - 6}}{5}} \right):\frac{{2024}}{{2025}}\]

\[ = \left[ {\left( {\frac{4}{5} - \frac{{ - 6}}{5}} \right) + \left( {\frac{{ - 9}}{7} + \frac{{ - 5}}{7}} \right)} \right]:\frac{{2024}}{{2025}}\]

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

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

\[ = 0:\frac{{2024}}{{2025}} = 0.\]

a) \[\frac{{\frac{3}{4} + \frac{3}{5} + \frac{3}{7} - \frac{3}{{11}}}}{{\frac{6}{4} + \frac{6}{5} + \frac{6}{7} - \frac{6}{{11}}}} = \frac{{3 \cdot \left( {\frac{1}{4} + \frac{1}{4} + \frac{1}{7} - \frac{1}{{11}}} \right)}}{{6 \cdot \left( {\frac{1}{4} + \frac{1}{4} + \frac{1}{7} - \frac{1}{{11}}} \right)}} = \frac{3}{6} = \frac{1}{2}.\]

b) \[\frac{{\frac{1}{3} - \frac{1}{5} + \frac{1}{{10}}}}{{\frac{6}{3} - \frac{6}{5} + \frac{3}{5}}} + \frac{5}{6}\]\[ = \frac{{\frac{1}{3} - \frac{1}{5} + \frac{1}{{10}}}}{{6 \cdot \left( {\frac{1}{3} - \frac{1}{5} + \frac{1}{{10}}} \right)}} + \frac{5}{6} = \frac{1}{6} + \frac{5}{6} = \frac{6}{6} = 1.\]

c) \[\frac{{\frac{1}{2}.\frac{5}{{17}} - \frac{{13}}{{14}}.\frac{5}{{17}} + \frac{{15}}{{238}}}}{{\frac{{ - 20}}{{68}} + \frac{{26}}{{14}}.\frac{5}{{17}} - \frac{{15}}{{119}}}}\]\[ = \frac{{\frac{5}{{34}} - \frac{{65}}{{14.17}} + \frac{{15}}{{238}}}}{{ - 2 \cdot \left( {\frac{5}{{34}} - \frac{{65}}{{14.17}} + \frac{{15}}{{238}}} \right)}} =  - \frac{1}{2}\]

d) \(\frac{{\frac{1}{{29}} \cdot \frac{3}{2} - \frac{{26}}{{11}} \cdot \frac{3}{{23}} + \frac{9}{{238}}}}{{\frac{{ - 3}}{{29}} + \frac{{13}}{{11}} \cdot \frac{3}{{23}} - \frac{9}{{119}}}}\)\[ = \frac{{ - 2 \cdot \left( {\frac{3}{{58}} - \frac{{39}}{{11.23}} + \frac{9}{{238}}} \right)}}{{\frac{3}{{58}} - \frac{{39}}{{11.23}} + \frac{9}{{238}}}} =  - 2\]

e) \[\frac{1}{{2 \cdot 3}} + \frac{1}{{3 \cdot 4}} + \frac{1}{{4 \cdot 5}} + \frac{1}{{5 \cdot 6}} + \frac{1}{{6 \cdot 7}} + \frac{1}{{7 \cdot 8}}\]

\( = \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \frac{1}{4} - \frac{1}{5} + \frac{1}{5} - \frac{1}{6} + \frac{1}{6} - \frac{1}{7} + \frac{1}{7} - \frac{1}{8}\)

\( = \frac{1}{2} - \frac{1}{8} = \frac{4}{8} - \frac{1}{8} = \frac{3}{8}.\)

f) \(\frac{2}{{3.7}} + \frac{2}{{7.11}} + \frac{2}{{11.15}} + ... + \frac{2}{{63.67}}\)

\( = \frac{1}{2} \cdot \left( {\frac{4}{{3.7}} + \frac{4}{{7.11}} + \frac{4}{{11.15}} + ... + \frac{4}{{63.67}}} \right)\)

\( = \frac{1}{2} \cdot \left( {\frac{1}{3} - \frac{1}{7} + \frac{1}{7} - \frac{1}{{11}} + \frac{1}{{11}} - \frac{1}{{15}} + ... + \frac{1}{{63}} - \frac{1}{{67}}} \right)\)

\( = \frac{1}{2} \cdot \left( {\frac{1}{3} - \frac{1}{{67}}} \right) = \frac{1}{2} \cdot \left( {\frac{{67}}{{201}} - \frac{3}{{201}}} \right)\)

\[ = \frac{1}{2} \cdot \frac{{64}}{{201}} = \frac{1}{2} \cdot \frac{{32}}{{201}}\]

g) \(\frac{4}{{15}} + \frac{4}{{35}} + \frac{4}{{63}} + ... + \frac{4}{{399}}\)

\[ = \frac{4}{{3 \cdot 5}} + \frac{4}{{5 \cdot 7}} + \frac{4}{{7 \cdot 9}} + ... + \frac{4}{{19 \cdot 21}}\]

\[ = 2 \cdot \left( {\frac{2}{{3 \cdot 5}} + \frac{2}{{5 \cdot 7}} + \frac{2}{{7 \cdot 9}} + ... + \frac{2}{{19 \cdot 21}}} \right)\]

\[ = 2 \cdot \left( {\frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7} + \frac{1}{7} - \frac{1}{9} + ... + \frac{1}{{19}} - \frac{1}{{21}}} \right)\]

\[ = 2 \cdot \left( {\frac{1}{3} - \frac{1}{{21}}} \right)\]\[ = 2 \cdot \left( {\frac{7}{{21}} - \frac{1}{{21}}} \right)\]               

\[ = 2 \cdot \frac{6}{{21}} = 2 \cdot \frac{2}{7} = \frac{4}{7}.\]

h) \[\frac{1}{3} + \frac{1}{{15}} + \frac{1}{{35}} + \frac{1}{{63}} + \frac{1}{{99}} + \frac{1}{{143}} + \frac{1}{{195}}\]

\( = \frac{1}{{1 \cdot 3}} + \frac{1}{{3 \cdot 5}} + \frac{1}{{5 \cdot 7}} + \frac{1}{{7 \cdot 9}} + \frac{1}{{9 \cdot 11}} + \frac{1}{{11 \cdot 13}} + \frac{1}{{13 \cdot 15}}\)

\( = \frac{1}{2} \cdot \left( {\frac{2}{{1 \cdot 3}} + \frac{2}{{3 \cdot 5}} + \frac{2}{{5 \cdot 7}} + \frac{2}{{7 \cdot 9}} + \frac{2}{{9 \cdot 11}} + \frac{2}{{11 \cdot 13}} + \frac{2}{{13 \cdot 15}}} \right)\)

\[ = \frac{1}{2} \cdot \left( {\frac{1}{1} - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7} + \frac{1}{7} - \frac{1}{9} + \frac{1}{9} - \frac{1}{{11}} + \frac{1}{{11}} - \frac{1}{{13}} + \frac{1}{{13}} - \frac{1}{{15}}} \right)\]

\[ = \frac{1}{2} \cdot \left( {\frac{1}{1} - \frac{1}{{15}}} \right) = \frac{1}{2} \cdot \left( {\frac{{15}}{{15}} - \frac{1}{{15}}} \right)\]

\[ = \frac{1}{2} \cdot \frac{{14}}{{15}} = \frac{7}{{15}}.\]

i) \[\left( {1 - \frac{1}{2}} \right) \cdot \left( {1 - \frac{1}{3}} \right) \cdot \left( {1 - \frac{1}{4}} \right) \cdot \left( {1 - \frac{1}{5}} \right) \cdot \left( {1 - \frac{1}{6}} \right) \cdot \left( {1 - \frac{1}{7}} \right) \cdot \left( {1 - \frac{1}{8}} \right)\]

\[ = \left( {\frac{2}{2} - \frac{1}{2}} \right) \cdot \left( {\frac{3}{3} - \frac{1}{3}} \right) \cdot \left( {\frac{4}{4} - \frac{1}{4}} \right) \cdot \left( {\frac{5}{5} - \frac{1}{5}} \right) \cdot \left( {\frac{6}{6} - \frac{1}{6}} \right) \cdot \left( {\frac{7}{7} - \frac{1}{7}} \right) \cdot \left( {\frac{8}{8} - \frac{1}{8}} \right)\]

\[ = \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \frac{4}{5} \cdot \frac{5}{6} \cdot \frac{6}{7} \cdot \frac{7}{8}\]\[ = \frac{1}{8}.\]

Hướng dẫn giải:

j) \(\left( {1 + \frac{1}{{10}}} \right) \cdot \left( {1 + \frac{1}{{11}}} \right) \cdot \left( {1 + \frac{1}{{12}}} \right) \cdot ... \cdot \left( {1 + \frac{1}{{59}}} \right) \cdot \left( {1 + \frac{1}{{60}}} \right)\)

\[ = \left( {\frac{{10}}{{10}} + \frac{1}{{10}}} \right) \cdot \left( {\frac{{11}}{{11}} + \frac{1}{{11}}} \right) \cdot \left( {\frac{{12}}{{12}} + \frac{1}{{12}}} \right) \cdot ... \cdot \left( {\frac{{59}}{{59}} + \frac{1}{{59}}} \right) \cdot \left( {\frac{{60}}{{60}} + \frac{1}{{60}}} \right)\]

\[ = \frac{{11}}{{10}} \cdot \frac{{12}}{{11}} \cdot \frac{{13}}{{12}} \cdot ... \cdot \frac{{60}}{{59}} \cdot \frac{{61}}{{60}}\]\[ = \frac{{61}}{{10}}.\]                                                        

a) \(\frac{x}{{ - 5}} = \frac{6}{{ - 10}}\)

Ta có \(\frac{x}{{ - 5}} = \frac{6}{{ - 10}} = \frac{{6:2}}{{\left( { - 10} \right):2}} = \frac{3}{{ - 5}}\)

Vậy \(x = 3.\)

b) \[\frac{{ - 5}}{x} = \frac{{20}}{{28}}\]\[\left( {x \ne 0} \right)\].

Ta có: \[\frac{{ - 5}}{x} = \frac{{20}}{{28}} = \frac{{20:\left( { - 4} \right)}}{{28:\left( { - 4} \right)}} = \frac{{ - 5}}{{ - 7}}\].

Vậy \[x = - 7\].

c) \[\frac{{ - 5}}{2} = \frac{{x + 2}}{8}\]

Ta có \[\frac{{x + 2}}{8} = \frac{{ - 5}}{2} = \frac{{ - 5 \cdot 4}}{{2 \cdot 4}} = \frac{{ - 20}}{8}\]

Suy ra \(x + 2 =  - 20\)

\(x =  - 22\)

Vậy \(x =  - 22.\)

d) \(\frac{{x - 2}}{{ - 5}} = \frac{3}{{15}}\)

Suy ra \(15 \cdot \left( {x - 2} \right) = 3 \cdot \left( { - 5} \right)\)

  \(15\left( {x - 2} \right) =  - 15\)

\(x - 2 =  - 1\)

\(x = 1\)

Vậy \(x = 1.\)

e) \(\frac{{ - 3}}{7} = \frac{{ - 9}}{{2x - 5}}\)

Ta có \[\frac{{ - 9}}{{2x - 5}} = \frac{{ - 3}}{7} = \frac{{\left( { - 3} \right) \cdot 3}}{{7 \cdot 3}} = \frac{{ - 9}}{{21}}\]

Suy ra \(2x - 5 = 21\)

\(2x = 26\)

\(x = 13\)

Vậy \(x = 13.\)

f) \(\frac{x}{{ - 6}} = \frac{{ - 6}}{x}\)

Suy ra \(x \cdot x = \left( { - 6} \right) \cdot \left( { - 6} \right)\)

\({x^2} = 36 = {6^2} = {\left( { - 6} \right)^2}\)

Do đó \(x = 6\) hoặc \(x =  - 6.\)

Vậy \(x \in \left\{ {6; - 6} \right\}.\)

g) \(\frac{{x + 1}}{6} = \frac{2}{x}\) Suy ra \(x\left( {x + 1} \right) = 2 \cdot 6\) \(x\left( {x + 1} \right) = 12\) Vì \(x \in \mathbb{Z}\) nên \(\left( {x + 1} \right) \in \mathbb{Z}\) do đó \(\left( {x + 1} \right) \in \)Ư\(\left( {12} \right) = \left\{ {1; - 1;2; - 2;3; - 3;4; - 4;6; - 6;12; - 12} \right\}.\) Ta có bảng sau: \(x + 1\) \(1\) \( - 1\) \(2\) \( - 2\) \(3\) \( - 3\) \(4\) \( - 4\) \(6\) \( - 6\) \(12\) \( - 12\) \(x\) \(0\) \( - 2\) \(1\) \( - 3\) \(2\) \( - 4\) \(3\) \( - 5\) \(5\) \( - 7\) \(11\) \( - 13\) Mà \(x \in \)Ư\(\left( {12} \right)\) nên từ bảng trên ta có \(x \in \left\{ { - 2;1; - 3;2; - 4;3} \right\}.\)

h) \[\frac{{x - 2}}{{27}} = \frac{3}{{x - 2}}\] Suy ra \(\left( {x - 2} \right)\left( {x - 2} \right) = 27 \cdot 3\) \({\left( {x - 2} \right)^2} = 81 = {9^2} = {\left( { - 9} \right)^2}\)
Trường hợp 1: \(x - 2 = 9\) \(x = 11\) Vậy \(x \in \left\{ { - 7;11} \right\}.\)	Trường hợp 2: \(x - 2 =  - 9\) \(x =  - 7\)

i) \(\frac{1}{3} < \frac{x}{{12}} < \frac{1}{2}\)

Suy ra \(\frac{{1 \cdot 4}}{{3 \cdot 4}} < \frac{x}{{12}} < \frac{{1 \cdot 6}}{{2 \cdot 6}}\)

Hay \[\frac{4}{{12}} < \frac{x}{{12}} < \frac{6}{{12}}\]

Do đó \(4 < x < 6\)

Mà \(x \in \mathbb{Z}\) nên \(x = 5.\)

Vậy \(x = 5.\)

j) \(5 \cdot \frac{1}{6} + \frac{1}{6} \le x \le \frac{1}{4}:\frac{1}{{13}} + \frac{7}{4}\)

\[\left( {5 + 1} \right) \cdot \frac{1}{6} \le x \le \frac{1}{4} \cdot \frac{{13}}{1} + \frac{7}{4}\]

\[\frac{1}{6} \cdot 6 \le x \le \frac{{13}}{4} + \frac{7}{4}\]

\[1 \le x \le \frac{{20}}{4}\]

\[1 \le x \le 5\]

Vì \[x \in \mathbb{Z}\] nên \[x \in \left\{ {1;\,\,2;\,\,3;\,\,4;\,\,5} \right\}\].

Vậy \[x \in \left\{ {1;\,\,2;\,\,3;\,\,4;\,\,5} \right\}\].

k\(\frac{{ - 5}}{3} \cdot \frac{1}{2} + \frac{8}{3} + \frac{{29}}{{ - 3}} \cdot \frac{1}{2} \le x \le \frac{{ - 1}}{2} + 2 + \frac{5}{2}.\)

\[\frac{1}{2} \cdot \left( {\frac{{ - 5}}{3} + \frac{{ - 29}}{3}} \right) + \frac{8}{3} \le x \le \frac{{ - 1}}{2} + \frac{4}{2} + \frac{5}{2}\]

\[\frac{1}{2} \cdot \frac{{ - 34}}{3} + \frac{8}{3} \le x \le \frac{8}{2}\]

\[\frac{{ - 17}}{3} + \frac{8}{3} \le x \le 4\]

\[ - 3 \le x \le 4\]

\[ - 3 \le x \le 4\]

Vì \[x \in \mathbb{Z}\] nên \[x \in \left\{ { - 3;\,\, - 2;\,\, - 1;\,\,0;\,\,1;\,\,2;\,\,3;\,\,4} \right\}\].

Vậy \[x \in \left\{ { - 3;\,\, - 2;\,\, - 1;\,\,0;\,\,1;\,\,2;\,\,3;\,\,4} \right\}\].