Đề cương ôn tập cuối kì 1 Toán 8 Cánh diều cấu trúc mới (Tự luận) có đáp án - Phần 1

a) \({2^3} \cdot {2^2} + {4^4}:{4^4} - {6^0}\)

\( = {2^5} + 1 - 1\)

\( = 32\).

b) \[\left( { - 25} \right) \cdot 8 \cdot \left( { - 125} \right) \cdot 3 \cdot \left( { - 4} \right)\]

\( = \left[ {\left( { - 25} \right) \cdot \left( { - 4} \right)} \right] \cdot \left[ {8 \cdot \left( { - 125} \right)} \right] \cdot 3\)

\[ = 100 \cdot \left( { - 1\,\,000} \right) \cdot 3\]

\( = - 300\,\,000\).

c) \(76 - \left( {4 \cdot {5^2} - 3 \cdot {2^3}} \right)\)

\( = 76 - \left( {4 \cdot 25 - 3 \cdot 8} \right)\)

\( = 76 - \left( {100 - 24} \right)\)

\( = 76 - 100 + 24\)

\[ = \left( {76 + 24} \right) - 100\]

\( = 100 - 100 = 0.\)

d) \({10^2} - \left[ {60:\left( {{5^6}:{5^4} - 3 \cdot 5} \right)} \right]\)

\( = {10^2} - \left[ {60:\left( {{5^2} - 15} \right)} \right]\)

\( = {10^2} - \left[ {60:\left( {25 - 15} \right)} \right]\)

\( = {10^2} - \left[ {60:10} \right]\)

\( = 100 - 6\)

\( = 94.\)

e) \(\left( {5 \cdot {2^2} - 20} \right):5 + {3^2} \cdot 6\)

\( = \left( {5 \cdot 4 - 20} \right):5 + 9 \cdot 6\)

\( = \left( {20 - 20} \right):5 + 54\)

\( = 0:5 + 54\)

\( = 0 + 54 = 54.\)

f) \[ - {4^2} + \left\{ { - 54:\left[ {{{\left( { - 2} \right)}^3} + 7 \cdot 2} \right]} \right\}\]

\( = - 16 + \left\{ { - 54:\left[ { - 8 + 14} \right]} \right\}\)

\( = - 16 + \left\{ { - 54:6} \right\}\)

\( = - 16 + \left( { - 9} \right)\)\( - 25.\)

a) \(173 + 315 + 327 + 185\)

\( = \left( {173 + 327} \right) + \left( {315 + 185} \right)\)

\( = 500 + 500 = 100\).

b) \[\left( {1\,\,342 + 265} \right) - \left( { - 735 + 1\,\,342} \right)\]

\[ = 1\,\,342 + 265 + 735 - 1\,\,342\]

\[ = \left( {1\,\,342 - 1\,\,342} \right) + \left( {265 + 735} \right)\]

\[ = 0 + 1\,\,000 = 1\,\,000.\]

c) \(32 \cdot \left( { - 64} \right) - 64 \cdot 68\)

\( = 32 \cdot \left( { - 64} \right) + \left( { - 64} \right) \cdot 68\)

\( = \left( { - 64} \right) \cdot \left( {32 + 68} \right)\)

\( = \left( { - 64} \right) \cdot 100 = - 6\,\,400\).

d) \(125 \cdot 98 - 125 \cdot 46 - 52 \cdot 25\)

\( = 125 \cdot \left( {98 - 46} \right) - 52 \cdot 25\)

\( = 125 \cdot 52 - 52 \cdot 25\)

\( = 52 \cdot \left( {125 - 25} \right)\)

\( = 52 \cdot 100 = 5\,\,200.\)

e) \( - 138 \cdot 2 + 125 \cdot 3 - \left( { - 138 \cdot 2} \right) + 75 \cdot 3\)

\( = \left[ { - 138 \cdot 2 - \left( { - 138 \cdot 2} \right)} \right] + \left( {125 \cdot 3 + 75 \cdot 3} \right)\)

\( = 0 + 3 \cdot \left( {125 + 75} \right)\)

\( = 0 + 3 \cdot 200\)

\( = 0 + 600\)\( = 600\).

f) \( - 68 \cdot 131 + 51 \cdot 68 - 34 \cdot 2 \cdot 20\)

\( = 68 \cdot \left( { - 131} \right) + 51 \cdot 68 - 68 \cdot 20\)

\( = 68 \cdot \left[ {\left( { - 131} \right) + 51 - 20} \right]\)

\( = 68 \cdot \left( { - 100} \right)\)

\( = 6\,\,800.\)

g) \[26 \cdot \left( {74 - 126} \right) - 74 \cdot \left( {126 + 26} \right)\]

\[ = 26 \cdot 74 - 26 \cdot 126 - 74 \cdot 126 - 74 \cdot 26\]

\[ = \left( {26 \cdot 74 - 74 \cdot 126} \right) - \left( {26 \cdot 126 + 74 \cdot 26} \right)\]

\[ = 74 \cdot \left( {26 - 126} \right) - 26 \cdot \left( {126 + 74} \right)\]

\[ = 74 \cdot \left( { - 100} \right) - 26 \cdot 200\]

\[ = - 7\,\,400 - 5\,\,200 = - 12\,\,600.\]

h) \(21 \cdot {6^2} - 11 \cdot {6^2} + 90 \cdot 36 - 36 \cdot 25 \cdot 8\)

\( = 21 \cdot 36 - 11 \cdot 36 + 90 \cdot 36 - 36 \cdot 25 \cdot 8\)

\( = 36 \cdot \left( {21 - 11 + 90 - 25 \cdot 8} \right)\)

\( = 36 \cdot \left( {100 - 200} \right)\)

\( = 36 \cdot \left( { - 100} \right) = - 3\,\,600\).

a) \(19 - x = - 27\)

 \(x = 19 - \left( { - 27} \right)\)

 \(x = 46\).

Vậy \(x = 46\).

b) \( - 23 + x = - 45\)

\(x = - 45 + 23\)

\(x = - 22\).

Vậy \(x = - 22\).

c) \(28 - \left( {x - 3} \right) = 18\)

 \(x - 3 = 28 - 18\)

 \(x - 3 = 10\)

 \(x = 10 + 3\)

 \(x = 13\)

Vậy \(x = 13.\)

d) \[9 - \left( {x + 5} \right):2 = {\rm{36}}\]

\[\left( {x + 5} \right):2 = 9 - {\rm{36}}\]

\[\left( {x + 5} \right):2 = - 27\]

\[x + 5 = \left( { - 27} \right) \cdot 2\]

\[x + 5 = - 54\]

\[x = - 54 - 5\]

\(x = - 59\).

Vậy \(x = - 59\).

e) \( - 7\left( {x + 9} \right) + \left( { - 21} \right) = 35\)

\( - 7\left( {x + 9} \right) = 35 - \left( { - 21} \right)\)

\( - 7\left( {x + 9} \right) = 56\)

 \(x + 9 = - 8\)

 \(x = - 8 - 9\)

 \(x = - 17\).

Vậy \(x = - 17\).

f) \(51 + 2\left( {3x - 5} \right) = 5\)

\(2\left( {3x - 5} \right) = 5 - 51\)

\(2\left( {3x - 5} \right) = - 46\)

\(3x - 5 = - 23\)

\(3x = - 23 + 5\)

\(3x = - 18\)

 \(x = - 6\).

Vậy \(x = - 6\).

g) \({\left( {2 - x} \right)^3} = - 27\)

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

Suy ra \(2 - x = - 3\)

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

\(x = 5\).

Vậy \(x = 5\).

h) \({\left( {x - 1} \right)^4} = 16\) \({\left( {x - 1} \right)^4} = {2^4} = {\left( { - 2} \right)^4}\)
Trường hợp 1: \(x - 1 = 2\) \(x = 3\) Vậy \(x \in \left\{ {3;\,\, - 1} \right\}.\)	Trường hợp 2: \(x - 1 = - 2\) \(x = - 1\)

i) \({3^{2x + 1}} = {9^7} \cdot {3^3}:{9^5}\)

\({3^{2x + 1}} = {\left( {{3^2}} \right)^7} \cdot {3^3}:{\left( {{3^2}} \right)^5}\)

\[{3^{2x + 1}} = {3^{14}} \cdot {3^3}:{3^{10}}\]

\[{3^{2x + 1}} = {3^7}\]

Suy ra \(2x + 1 = 7\)

\(2x = 6\)

\(x = 3\)

Vậy \(x = 3.\)

j) \[{2^x} + {\rm{ }}{2^{x + 4}} = 544\]

\({2^x} + {2^x} \cdot {2^4} = 544\)

 \({2^x} + {2^x} \cdot 16 = 544\)

\[{2^x} \cdot \left( {1 + 16} \right) = 544\]

\[{2^x} \cdot 17 = 544\]

\[{2^x} = 544:17\]

\[{2^x} = 32\]

\({2^x} = {2^5}\)

Suy ra \(x = 5\).

Vậy \[x = 5.\]

k) \(92 - 2x = 2\left( {x + 4} \right)\)

 \(2x = 92 - 2\left( {x + 4} \right)\)

 \(2x = 92 - \left( {2x + 8} \right)\)

\(2x = 92 - 2x - 8\)

\(2x + 2x = 92 - 8\)

\(4x = 84\)

\(x = 21\).

Vậy \(x = 21\).

l) \({5^2} \cdot \left( {2x + 1} \right) - 9 = 12\left( {3x - 1} \right)\)

\(25 \cdot \left( {2x + 1} \right) - 9 = 36x - 12\)

\(50x + 25 - 9 = 36x - 12\)

\(50x - 36x = - 12 - 25 + 9\)

\(14x = - 28\)

   \(x = - 2\).

Vậy \(x = - 2\).