Đề cương ôn tập cuối kì 1 Toán 7 Cánh diều cấu trúc mới (Tự luận) có đáp án - Phần 1

a) \(\frac{{11}}{{24}} - \frac{5}{{41}} + \frac{{13}}{{24}} + 0,5 - \frac{{36}}{{41}}\)

\( = \left( {\frac{{11}}{{24}} + \frac{{13}}{{24}}} \right) + \left( { - \frac{5}{{41}} - \frac{{36}}{{41}}} \right) + 0,5\)

\( = 1 + \left( { - 1} \right) + 0,5\)\( = 0,5\).

b) \(16\frac{2}{7}:\left( {\frac{{ - 3}}{5}} \right) + 28\frac{2}{7}:\frac{3}{5}\)

\( = 16\frac{2}{7} \cdot \left( {\frac{{ - 5}}{3}} \right) + 28\frac{2}{7} \cdot \frac{5}{3}\)

\( = \frac{5}{3}\, \cdot \,\left( { - 16\frac{2}{7} + 28\frac{2}{7}} \right)\)\( = \frac{5}{3}\, \cdot \,12\)\( = 20\).

c) \(5\frac{5}{{27}} + \frac{7}{{23}} + 0,5 - \left| {\frac{{ - 5}}{{27}}} \right| + \frac{{16}}{{23}}\)

\( = 5\frac{5}{{27}} + \frac{7}{{23}} + 0,5 - \frac{5}{{27}} + \frac{{16}}{{23}}\)

\( = \left( {5\frac{5}{{27}} - \frac{5}{{27}}} \right) + \left( {\frac{7}{{23}} + \frac{{16}}{{23}}} \right) + 0,5\)

\( = 5 + 1 + 0,5\)\( = 6,5\).

d) \(\frac{{2\,.\,{6^9} - {2^5}\,.\,{{18}^4}}}{{{2^2}\,.\,{6^8}}}\)

\[ = \frac{{2\,.\,{{\left( {2\,.\,3} \right)}^9} - {2^5}\,.\,{{\left( {2\,.\,{3^2}} \right)}^4}}}{{{2^2}\,.\,{{\left( {2\,.\,3} \right)}^8}}}\]

\[ = \frac{{2\,.\,{2^9}\,.\,{3^9} - {2^5}\,.\,{2^4}\,.\,{3^8}}}{{{2^2}\,.\,{2^8}\,.\,{3^8}}}\]

\[ = \frac{{{2^{10}}\,.\,{3^9} - {2^9}\,.\,{3^8}}}{{{2^{10}}\,.\,{3^8}}}\]\[ = \frac{{{2^9}\,.\,{3^8}\left( {6 - 1} \right)}}{{{2^{10}}\,.\,{3^8}}}\].\[ = \frac{5}{2}\]..

e) \(\frac{{15}}{{37}} \cdot \left( {\frac{{38}}{{41}} - \frac{{74}}{{45}}} \right) - \frac{{38}}{{41}} \cdot \left( {\frac{{15}}{{37}} + \frac{{82}}{{76}}} \right)\)

\( = \frac{{15}}{{37}} \cdot \frac{{38}}{{41}} - \frac{{15}}{{37}} \cdot \frac{{74}}{{45}} - \frac{{38}}{{41}} \cdot \frac{{15}}{{37}} - \frac{{38}}{{41}} \cdot \frac{{82}}{{76}}\)

\[ = \left( {\frac{{15}}{{37}} \cdot \frac{{38}}{{41}} - \frac{{38}}{{41}} \cdot \frac{{15}}{{37}}} \right) + \left( { - \frac{{15}}{{37}} \cdot \frac{{74}}{{45}} - \frac{{38}}{{41}} \cdot \frac{{82}}{{76}}} \right)\]

\[ = 0 + \left( { - \frac{1}{1} \cdot \frac{2}{3} - \frac{{38}}{{41}} \cdot \frac{{41 \cdot 2}}{{38 \cdot 2}}} \right)\]

\[ = \frac{{ - 2}}{3} - 1 = \frac{{ - 2 - 3}}{3} = - \frac{5}{3}\].

f) \(\left( { - \frac{2}{3} + \frac{3}{7}} \right):\frac{4}{5} + \left( { - \frac{1}{3} + \frac{4}{7}} \right):\frac{4}{5}\)

\( = \left( { - \frac{2}{3} + \frac{3}{7}} \right) \cdot \frac{5}{4} + \left( { - \frac{1}{3} + \frac{4}{7}} \right) \cdot \frac{5}{4}\)

\( = \frac{5}{4} \cdot \left( { - \frac{2}{3} + \frac{3}{7} - \frac{1}{3} + \frac{4}{7}} \right)\)

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

\( = \frac{5}{4} \cdot \left[ {\frac{{ - 3}}{3} + \frac{7}{7}} \right] = \frac{5}{4} \cdot \left[ { - 1 + 1} \right] = 0\).

g) \(\frac{{15}}{{34}} + \frac{7}{{21}} + \frac{{19}}{{34}} - \frac{{20}}{{15}} + \frac{3}{7}\)

\( = {\rm{\;}}\left( {\frac{{15}}{{34}} + \frac{{19}}{{34}}} \right) + \left( {\frac{1}{3} - \frac{4}{3}} \right) + \frac{3}{7}\)

\( = 1 - 1 + \frac{3}{7} = \frac{3}{7}\).

h) \(\frac{{11}}{{24}} - \frac{5}{{41}} + \frac{{13}}{{24}} + 0,5 - \frac{{36}}{{41}}\)

\( = \left( {\frac{{11}}{{24}} + \frac{{13}}{{24}}} \right) + 0,5 + \left( { - \frac{{36}}{{41}} - \frac{5}{{41}}} \right)\)

\( = 1 + 0,5 - 1 = 0,5\).

i) \(\frac{{ - {2^3}}}{9} + \frac{4}{9}\left[ { - \left( {2012\frac{2}{7} - 2012\frac{2}{7}} \right):\left| { - \frac{1}{{21}}} \right|} \right]\)

\( = \frac{{ - 8}}{9} + \frac{4}{9}\left[ { - 0:\frac{1}{{21}}} \right] = \frac{{ - 8}}{9}\).

k) \(\sqrt {{{12}^2} - 44} - {\left( {1\frac{3}{4} - 2} \right)^2} + {\left( {1,5} \right)^5}:{\left( {\frac{{ - 3}}{2}} \right)^4}\)

\( = \sqrt {144 - 44} - {\left( {\frac{7}{4} - \frac{8}{4}} \right)^2} + {\left( {\frac{3}{2}} \right)^5}:{\left( {\frac{3}{2}} \right)^4}\)

\( = \sqrt {100} - \frac{1}{{16}} + \frac{3}{2} = 10 + \frac{{23}}{{16}} = 11\frac{7}{{16}} = \frac{{183}}{{16}}.\)

a) \(x + \frac{1}{3} = \frac{2}{5} - \left( { - \frac{1}{4}} \right)\)

\(x + \frac{1}{3} = \frac{2}{5} + \frac{1}{4}\)

\(x + \frac{1}{3} = \frac{{13}}{{20}}\)

\(x = \frac{{13}}{{20}} - \frac{1}{3}\)

\(x = \frac{{19}}{{60}}\).

Vậy \(x = \frac{{19}}{{60}}\).

b) \(\frac{2}{3} - 1\frac{4}{{15}}x = \frac{{ - 3}}{5}\)

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

\(\frac{{19}}{{15}}x = \frac{{19}}{{15}}\)

\(x = \frac{{19}}{{15}}:\frac{{19}}{{15}}\)

\(x = 1\)

Vậy \(x = 1\).

c) \(\frac{7}{4} - \left( {x + \frac{5}{3}} \right) = \frac{{ - 12}}{5}\)

\(x + \frac{5}{3} = \frac{7}{4} - \frac{{ - 12}}{5}\)

\(x + \frac{5}{3} = \frac{{83}}{{20}}\)

\(x = \frac{{83}}{{20}} - \frac{5}{3}\)

\(x = \frac{{149}}{{60}}\)

Vậy \(x = \frac{{149}}{{60}}\).

d) \(\frac{{{{( - 3)}^x}}}{{81}} = - 27\)

\({( - 3)^x} = {\left( { - 3} \right)^3}.{\left( { - 3} \right)^4}\)

\({( - 3)^x} = {\left( { - 3} \right)^{3 + 4}}\)

\({( - 3)^x} = {\left( { - 3} \right)^7}\)

\(x = 7\)

Vậy \(x = 7\).

e) \(\frac{3}{4} - {\left( {x + \frac{1}{2}} \right)^2} = \frac{{11}}{{16}}\)

\({\left( {x + \frac{1}{2}} \right)^2} = \frac{3}{4} - \frac{{11}}{{16}}\)

\({\left( {x + \frac{1}{2}} \right)^2} = \frac{1}{{16}}\)

\({\left( {x + \frac{1}{2}} \right)^2} = {\left( {\frac{1}{4}} \right)^2}\)   hoặc    \({\left( {x + \frac{1}{2}} \right)^2} = {\left( { - \frac{1}{4}} \right)^2}\)

\(x + \frac{1}{2} = \frac{1}{4}\)                           \(x + \frac{1}{2} = - \frac{1}{4}\)

\(x = \frac{1}{4} - \frac{1}{2}\)                         \(x = - \frac{1}{4} - \frac{1}{2}\)

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

Vậy \(x \in \left\{ {\frac{{ - 1}}{4}\;;\;\frac{{ - 3}}{4}} \right\}\)

f) \(\frac{2}{3}{.3^{x + 1}} - {7.3^x} = - 405\)

\({2.3^x} - {7.3^x} = - 405\)

\( - {5.3^x} = - 405\)

\({3^x} = 81\)

\({3^x} = {3^4}\)

\(x = 4\)

Vậy \(x = 4\).

g) \(1\frac{2}{3}:\frac{x}{4} = 6:0,3\)

\(\frac{5}{3}:\frac{x}{4} = 20\)

\(\frac{x}{4} = \frac{5}{3}:20\)

\(x = \frac{1}{{12}}.4\)

\(x = \frac{1}{3}\)

Vậy \(x = \frac{1}{3}\).

h) \({2^{x + 4}} - {12.2^{x - 2}} = 26\)

\({16.2^x} - \frac{{12}}{4}{.2^x} = 26\)

\({2^x}\left( {16 - 3} \right) = 26\)

\({13.2^x} = 26\)

\({2^x} = 2\)

\(x = 1\)

Vậy \(x = 1\).

i) \({\left( {x - \frac{2}{9}} \right)^3} = {\left( {\frac{2}{3}} \right)^6}\)

\({\left( {x - \frac{2}{9}} \right)^3} = {\left( {\frac{4}{9}} \right)^3}\)

\(x - \frac{2}{9} = \frac{4}{9}\)

\(x = \frac{6}{9}\)

\(x = \frac{2}{3}\)

Vậy \(x = \frac{2}{3}\).

k) \[2\left( {x + \frac{1}{2}} \right) - 3\left( {x - \frac{4}{3}} \right) = - x + 1\]

\[2.x + 2.\frac{1}{2} - 3.x - 3.\left( { - \frac{4}{3}} \right) + x - 1 = 0\].

\[2x + 1 - 3x + 4 + x - 1 = 0\].

\[\left( {2x - 3x + x} \right) + \left( {1 + 4 - 1} \right) = 0\].

\[4 = 0\] (vô lí)

Vậy không tìm được \[x\].

a) \[\left| {2x - \frac{1}{3}} \right| + \frac{3}{2} = 2\]

\[\left| {2x - \frac{1}{3}} \right| = 2 - \frac{3}{2}\]

\[\left| {2x - \frac{1}{3}} \right| = \frac{1}{2}\]

TH 1: \[2x - \frac{1}{3} = \frac{1}{2}\]

\[2x = \frac{1}{2} + \frac{1}{3}\]

\[2x = \frac{3}{6} + \frac{2}{6}\]

\[2x = \frac{5}{6}\]

\[x = \frac{5}{{12}}\]

TH 2: \[2x - \frac{1}{3} = - \frac{1}{2}\]

\[2x = - \frac{1}{2} + \frac{1}{3}\]

\[2x = - \frac{3}{6} + \frac{2}{6}\]

\[2x = - \frac{1}{6}\]

\[x = \frac{{ - 1}}{{12}}\]

Vậy \[x \in \left\{ {\frac{5}{{12}};\frac{{ - 1}}{{12}}} \right\}\].

b) \[\left| {2x + \frac{1}{3}} \right| - \sqrt {1\frac{9}{{16}}} = {\left( { - \frac{{\sqrt 3 }}{2}} \right)^2}\]

\[\left| {2x + \frac{1}{3}} \right| - \sqrt {\frac{{25}}{{16}}} = \frac{3}{4}\]

\[\left| {2x + \frac{1}{3}} \right| - \frac{5}{4} = \frac{3}{4}\]

\[\left| {2x + \frac{1}{3}} \right| = 2\]

TH1: \(2x + \frac{1}{3} = 2\)

\(2x = \frac{5}{3}\)

\(x = \frac{5}{6}\)

TH2: \(2x + \frac{1}{3} = - 2\)

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

\(x = \frac{{ - 7}}{6}\)

Vậy \(x \in \left\{ {\frac{5}{6};\frac{{ - 7}}{6}} \right\}\).

c) \[\left| {\frac{1}{3}\sqrt {x - 1} - \frac{2}{9}} \right| - \frac{1}{6} = \frac{1}{9}\]

\[\left| {\frac{1}{3}\sqrt {x - 1} - \frac{2}{9}} \right| = \frac{1}{9} + \frac{1}{6}\]

\[\left| {\frac{1}{3}\sqrt {x - 1} - \frac{2}{9}} \right| = \frac{5}{{18}}\]

TH1: \[\frac{1}{3}\sqrt {x - 1} - \frac{2}{9} = \frac{5}{{18}}\]

\[\frac{1}{3}\sqrt {x - 1} = \frac{1}{2}\]

\[\sqrt {x - 1} = \frac{3}{2}\]

\( \Rightarrow x - 1 = \frac{9}{4}\)

\(x = \frac{{13}}{4}\)

TH2: \[\frac{1}{3}\sqrt {x - 1} - \frac{2}{9} = \frac{{ - 5}}{{18}}\]

 (vô lí vì \[\frac{1}{3}\sqrt {x - 1} \ge 0;\frac{{ - 1}}{{18}} < 0\])

Vậy \(x = \frac{{13}}{4}\).

d) \( - 3,5 - \left| {x - \frac{1}{2}} \right| = 0,75\)

\(\left| {x - \frac{1}{2}} \right| = - 3,5 - 0,75\)

\(\left| {x - \frac{1}{2}} \right| = - 4,25\) (vô lí do \(\left| {x - \frac{1}{2}} \right| \ge 0\)).

Vậy phương trình không có \(x\) th\[\frac{1}{3}\sqrt {x - 1} = \frac{{ - 1}}{{18}}\]ỏa mãn.

e) \(\left| {x + \frac{4}{{15}}} \right| - \left| { - 3,75} \right| = - \left| { - 2,15} \right|\)

\(\left| {x + \frac{4}{{15}}} \right| = - 2,15 + 3,75\)

\(\left| {x + \frac{4}{{15}}} \right| = 1,6\)

\(\left| {x + \frac{4}{{15}}} \right| = \frac{8}{5}\)

TH1: \[x + \frac{4}{{15}} = \frac{8}{5}\]

\[x = \frac{8}{5} - \frac{4}{{15}}\]

\[x = \frac{{24}}{{15}} - \frac{4}{{15}}\]

\[x = \frac{4}{3}\].

TH2: \[x + \frac{4}{{15}} = - \frac{8}{5}\]

\[x = - \frac{8}{5} - \frac{4}{{15}}\]

\[x = - \frac{{24}}{{15}} - \frac{4}{{15}}\]

\[x = - \frac{{28}}{{15}}\]

Vậy \[x \in \left\{ { - \frac{{28}}{{15}};\frac{4}{3}} \right\}\].

f) \(2 - \left| {x - \frac{2}{5}} \right| = - \frac{1}{2}\)

\(\left| {x - \frac{2}{5}} \right| = 2 - \left( { - \frac{1}{2}} \right)\)

\(\left| {x - \frac{2}{5}} \right| = \frac{5}{2}\)

TH1: \[x - \frac{2}{5} = \frac{5}{2}\]

\[x = \frac{5}{2} + \frac{2}{5}\]

\[x = \frac{{29}}{{10}}\].

TH2: \[x - \frac{2}{5} = - \frac{5}{2}\]

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

\[x = - \frac{{21}}{{10}}\].

Vậy \[x \in \left\{ { - \frac{{21}}{{10}};\frac{{29}}{{10}}} \right\}\].

g) \(0,2 + \left| {x - 2,3} \right| = 1,1\)

\(\left| {x - 2,3} \right| = 1,1 - 0,2\)

\(\left| {x - 2,3} \right| = 0,9\)

TH1: \(x - 2,3 = 0,9\)

\(x = 0,9 + 2,3\)

\(x = 3,2\).

TH2: \(x - 2,3 = - 0,9\)

\(x = - 0,9 + 2,3\)

\(x = 1,4\).

\(x \in \left\{ {3,2\,;\,\,1,4} \right\}\).

h) \( - 1 + \left| {x + 4,5} \right| = - 6,2\)

\(\left| {x + 4,5} \right| = - 6,2 + 1 = - 5,2 < 0\)

\(\left| {x + 4,5} \right| = - 5,2\)

Vì \(\left| {x + 4,5} \right| \ge 0\) với mọi \(x\).

Do đó không có giá trị nào của \(x\) để

 \( - 1 + \left| {x + 4,5} \right| = - 6,2\).