Đề cương ôn tập giữa kì 2 Toán 6 Kết nối tri thức 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 \(x \in \left\{ {6; - 6} \right\}.\)

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 \[x = \frac{5}{{24}}.\]