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

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

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

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

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

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

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:

Mà $x \in$Ư$\left( {12} \right)$ nên từ bảng trên ta có $x \in \left\{ { - 2;1; - 3;2; - 4;3} \right\}.$

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