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

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

a) $x + \frac{1}{{ - 5}} = \frac{1}{{10}}$

$x = \frac{1}{{10}} - \frac{1}{{ - 5}}$

$x = \frac{1}{{10}} + \frac{2}{{10}}$

$x = \frac{3}{{10}}.$

Vậy $x = \frac{3}{{10}}.$

b) $x - \frac{4}{5} = \frac{2}{3}$.

$$x = \frac{2}{3} + \frac{4}{5}$$

$$x = \frac{{10}}{{15}} + \frac{{12}}{{15}}$$

$$x = \frac{{22}}{{15}}$$

Vậy $x = \frac{{22}}{5}.$

c) $\frac{{ - 8}}{9} - x = \frac{{ - 11}}{{18}}$.

$$x = \frac{{ - 8}}{9} - \frac{{ - 11}}{{18}}$$

$$x = \frac{{ - 8}}{9} + \frac{{11}}{{18}}$$

$$x = \frac{{ - 16}}{{18}} + \frac{{11}}{{18}}$$

$x = \frac{{ - 5}}{{18}}$.

Vậy $x = \frac{{ - 5}}{{18}}.$

d) $x - \frac{2}{3} = \frac{{ - 5}}{{12}}$.

$$x = \frac{{ - 5}}{{12}} + \frac{2}{3}$$

$$x = \frac{{ - 5}}{{12}} + \frac{8}{{12}}$$

$x = \frac{3}{{12}} = \frac{1}{4}$.

Vậy $x = \frac{1}{4}.$

e) $\frac{{ - 5}}{6} - x = \frac{7}{{12}} + \frac{{ - 1}}{3}$

$$\frac{{ - 5}}{6} - x = \frac{7}{{12}} + \frac{{ - 4}}{{12}}$$

$$\frac{{ - 5}}{6} - x = \frac{3}{{12}}$$

$$x = \frac{{ - 5}}{6} - \frac{3}{{12}}$$

$$x = \frac{{ - 10}}{{12}} - \frac{3}{{12}}$$

$x = \frac{{ - 13}}{{12}}$.

Vậy $x = \frac{{ - 13}}{{12}}.$

f) $x - \frac{3}{7} \cdot \frac{{14}}{9} = - \frac{7}{3}$

$x - \frac{2}{3} = - \frac{7}{3}$

$x = - \frac{7}{3} + \frac{2}{3}$

$x = - \frac{5}{3}$

Vậy $x = - \frac{5}{3}.$

g) $\frac{x}{3} = \frac{2}{3} + \frac{{ - 1}}{7}$

$$\frac{x}{3} = \frac{{14}}{{21}} + \frac{{ - 3}}{{21}}$$

$$\frac{x}{3} = \frac{{11}}{{21}}$$

Suy ra $x \cdot 21 = 3 \cdot 11$

$21x = 33$

$$x = \frac{{33}}{{21}} = \frac{{11}}{7}.$$

Vậy $x = \frac{{11}}{7}.$

h) $\frac{4}{{15}}x + \frac{2}{3} = - \frac{1}{5}.$

$\frac{4}{{15}}x = - \frac{1}{5} - \frac{2}{3}$

$$\frac{4}{{15}}x = - \frac{3}{{15}} - \frac{{10}}{{15}}$$

$$\frac{4}{{15}}x = \frac{{ - 13}}{{15}}$$

$$x = \frac{{ - 13}}{{15}}:\frac{4}{{15}}$$

$$x = \frac{{ - 13}}{{15}} \cdot \frac{{15}}{4}$$

$$x = \frac{{ - 13}}{4}.$$

Vậy $x = \frac{{ - 13}}{4}.$

i) $2x - \frac{3}{5} = \frac{4}{7} \cdot \frac{{14}}{6}.$

$$2x - \frac{3}{5} = \frac{{4 \cdot 14}}{{7 \cdot 6}}$$

$$2x - \frac{3}{5} = \frac{4}{3}$$

$$2x = \frac{3}{5} + \frac{4}{3}$$

$$2x = \frac{9}{{15}} + \frac{{20}}{{15}}$$

$$2x = \frac{{29}}{{15}}$$

$$x = \frac{{29}}{{30}}$$

Vậy $x = \frac{{29}}{{30}}$.

j) $\frac{{10}}{{21}} - 3x = \frac{3}{8} \cdot \frac{4}{{15}}.$

$$\frac{{10}}{{21}} - 3x = \frac{{3 \cdot 4}}{{8 \cdot 15}}$$

$$\frac{{10}}{{21}} - 3x = \frac{1}{{10}}$$

$$3x = \frac{{10}}{{21}} - \frac{1}{{10}}$$

$$3x = \frac{{100}}{{210}} - \frac{{21}}{{210}}$$

$$3x = \frac{{79}}{{210}}$$

$$x = \frac{{79}}{{630}}$$

Vậy $x = \frac{{79}}{{630}}.$

k) $\frac{3}{4}:x - \frac{1}{3} = \frac{5}{{14}} \cdot \frac{{ - 7}}{6}$

$\frac{3}{4}:x - \frac{1}{3} = \frac{{5 \cdot \left( { - 7} \right)}}{{14 \cdot 6}}$

$\frac{3}{4}:x - \frac{1}{3} = \frac{{ - 5}}{{12}}$

$\frac{3}{4}:x = \frac{{ - 5}}{{12}} + \frac{1}{3}$

$\frac{3}{4}:x = \frac{{ - 5}}{{12}} + \frac{4}{{12}}$

$\frac{3}{4}:x = \frac{{ - 1}}{{12}}$

$x = \frac{3}{4}:\frac{{ - 1}}{{12}}$

$x = \frac{3}{4} \cdot \left( { - 12} \right)$

$x = - 9.$

Vậy $x = - 9.$

l) $\frac{5}{3} + \frac{1}{6}:x = 1$

$\frac{1}{6}:x = 1 - \frac{5}{3}$

$\frac{1}{6}:x = \frac{{ - 2}}{3}$

$x = \frac{1}{6}:\frac{{ - 2}}{3}$

$x = \frac{1}{6} \cdot \frac{3}{{ - 2}}$

$x = \frac{{ - 1}}{4}.$

Vậy $x = \frac{{ - 1}}{4}.$

m) $\frac{{ - 15}}{{12}} \cdot \left( {x - \frac{1}{2}} \right) = \frac{5}{{24}}$

$x - \frac{1}{2} = \frac{5}{{24}}:\frac{{ - 15}}{{12}}$

$x - \frac{1}{2} = \frac{5}{{24}} \cdot \frac{{12}}{{ - 15}}$

$x - \frac{1}{2} = \frac{{ - 1}}{6}$

$x = \frac{{ - 1}}{6} + \frac{1}{2}$

$x = \frac{{ - 1}}{6} + \frac{3}{6}$

$x = \frac{2}{6} = \frac{1}{3}.$

Vậy $x = \frac{1}{3}.$

n) $\left( {2x + \frac{{ - 3}}{4}} \right):\frac{2}{5} = \frac{{ - 10}}{{12}}$

$$2x + \frac{{ - 3}}{4} = \frac{{ - 10}}{{12}} \cdot \frac{2}{5}$$

$$2x + \frac{{ - 3}}{4} = \frac{{ - 1}}{3}$$

$$2x = \frac{{ - 1}}{3} - \frac{{ - 3}}{4}$$

$$2x = \frac{{ - 1}}{3} + \frac{3}{4}$$

$$2x = \frac{{ - 4}}{{12}} + \frac{9}{{12}}$$

$$2x = \frac{5}{{12}}$$

$$x = \frac{5}{{12}}:2$$

$$x = \frac{5}{{24}}.$$

Vậy $x = \frac{5}{{24}}.$