Đề cương ôn tập giữa kì 1 Toán 6 Kết nối tri thức cấu trúc mới (Tự luận) có đáp án - Phần 1

c) \[47 \cdot 29 - 13 \cdot 29 - 24 \cdot 29\]

\[ = 29 \cdot \left( {47 - 13 - 24} \right)\]

\[ = 29 \cdot 10 = 290\].

d) \[2 \cdot 53 \cdot 12 + 4 \cdot 6 \cdot 87 - 3 \cdot 8 \cdot 40\]

\[ = 24 \cdot 53 + 24 \cdot 87 - 24 \cdot 40\]

\[ = 24 \cdot \left( {53 + 87 - 40} \right)\]

\[ = 24 \cdot 100\]

\[ = 2\,\,400\].

e) \[53 \cdot 39 + 47 \cdot 39 - 53 \cdot 21 - 47 \cdot 21\]

\[ = \left( {53 \cdot 39 + 47 \cdot 39} \right) - \left( {53 \cdot 21 + 47 \cdot 21} \right)\]

\[ = 39 \cdot \left( {53 + 47} \right) - 21 \cdot \left( {53 + 47} \right)\]

\[ = 39 \cdot 100 - 21 \cdot 100\]

\[ = 3\,\,900 - 2\,\,100\]

\[ = 1\,\,800\].

f) \[26 \cdot 7 - 17 \cdot 9 + 13 \cdot 26 - 17 \cdot 11\]

\[ = \left( {26 \cdot 7 + 13 \cdot 26} \right) - \left( {17 \cdot 9 + 17 \cdot 11} \right)\]

 \[ = 26 \cdot \left( {7 + 13} \right) - 17 \cdot \left( {9 + 11} \right)\]

 \[ = 26 \cdot 20 - 17 \cdot 20\]

 \[ = 20 \cdot \left( {26 - 17} \right)\]

g) \[1\,\,754:17 - 74:17 + 20:17\]

\[ = \left( {1\,\,754 - 74 + 20} \right):17\]

a) \({2^3} \cdot 17 - {2^3} \cdot 14\)

\( = {2^3} \cdot \left( {17 - 14} \right)\)

\( = 8 \cdot 3 = 24\).

b) \(2 \cdot {5^2} + 3:{71^0} - 54:{3^3}\)

\( = 2 \cdot 25 + 3:1 - 54:27\)

\( = 50 + 3 - 2\)\( = 51\).

c) \[\left( {{6^{2007}} - {6^{2006}}} \right):{6^{2006}}\]            

\( = {6^{2\,007}}:{6^{2\,006}} - {6^{2\,006}}:{6^{2\,006}}\)

\( = 6 - 1 = 5\).

d) \[\left( {{7^5} + {7^9}} \right) \cdot \left( {{5^4} + {5^6}} \right) \cdot \left( {{3^3} \cdot 3 - {9^2}} \right)\]

\( = \left( {{7^5} + {7^9}} \right) \cdot \left( {{5^4} + {5^6}} \right) \cdot \left( {81 - 81} \right)\)

\( = \left( {{7^5} + {7^9}} \right) \cdot \left( {{5^4} + {5^6}} \right) \cdot 0\)

\( = 0\).

e) \(\left[ {\left( {{5^2} \cdot {2^3} - {7^2} \cdot 2} \right):2} \right] \cdot 6 - {3^2} \cdot 17\)

\( = \left[ {\left( {{5^2} \cdot {2^2} - {7^2}} \right) \cdot 2:2} \right] \cdot 6 - {3^2} \cdot 17\)

\( = \left[ {\left( {25 \cdot 4 - 49} \right) \cdot 1} \right] \cdot 6 - {3^2} \cdot 17\)

\( = \left( {100 - 49} \right) \cdot 6 - 3 \cdot 3 \cdot 17\)

\( = 51 \cdot 6 - 3 \cdot 51\)

f) \[\left( {{3^5} \cdot {3^7}} \right):{3^{10}} + 5 \cdot {2^4} - {7^3}:7\]

\[ = {3^{12}}:{3^{10}} + 5 \cdot 16 - {7^2}\]

\[ = {3^2} + 80 - 49\]

\( = 9 + 80 - 49\)

a) \[71 - \left( {33 + x} \right) = 26\]

\[33 + x = 71 - 26\]

\[33 + x = 45\]

\[x = 45 - 33\]

\[x = 12\]

Vậy \[x = 12\].

b) \[97 - \left( {64 - x} \right) = 44\]

\[64 - x = 97 - 44\]

\[64 - x = 53\]

\[x = 64 - 53\]

\[x = 11\]

Vậy \[x = 11\].

c) \(\left( {x - 50} \right) + 12 = 31\)

\(x - 50 = 31 - 12\)

\(x - 50 = 19\)

\(x = 19 + 50\)

\(x = 69\).

Vậy \(x = 69\).

d) \(\left( {x + 5} \right):3 - 11 = 4\)

\(\left( {x + 5} \right):3 = 4 + 11\)

\(\left( {x + 5} \right):3 = 15\)

\(x + 5 = 15 \cdot 3\)

\(x + 5 = 45\)

\(x = 45 - 5\)

\(x = 40\).

Vậy \(x = 40\).

d) \(\left( {x + 5} \right):3 - 11 = 4\)

\(\left( {x + 5} \right):3 = 4 + 11\)

\(\left( {x + 5} \right):3 = 15\)

\(x + 5 = 15 \cdot 3\)

\(x + 5 = 45\)

\(x = 45 - 5\)

\(x = 40\).

Vậy \(x = 40\).

e) \(75:\left( {x - 2} \right) + 4 = 7\)

\(75:\left( {x - 2} \right) = 7 - 4\)

\(75:\left( {x - 2} \right) = 3\)

\(x - 2 = 75:3\)

\(x - 2 = 25\)

\(x = 25 + 2\)

\(x = 27\).

Vậy \(x = 27\).

f) \(125 - 3\left( {x + 3} \right) = 65\)

 \(3\left( {x + 3} \right) = 125 - 65\)

 \(3\left( {x + 3} \right) = 60\)

 \(x + 3 = 60:3\)

 \(x + 3 = 20\)

 \(x = 20 - 3\)

 \(x = 17\).

Vậy \(x = 17\).

g) \[{2^x} + {2^{x + 3}} = 144\]

\[{2^x} + {2^x} \cdot {2^3} = 144\]

\[{2^x} \cdot \left( {1 + {2^3}} \right) = 144\]

\[{2^x} \cdot 9 = 144\]

\[{2^x} = 144:9\]

\[{2^x} = 16\]

\[{2^x} = {2^4}\]

Suy ra \[x = 4\]

Vậy \(x = 4\).

h) \[{\left( {x - 5} \right)^{2\,022}} = {\left( {x - 5} \right)^{2\,021}}\]

\[{\left( {x - 5} \right)^{2\,021}} \cdot \left( {x - 5} \right) - {\left( {x - 5} \right)^{2\,021}} = 0\]

\[{\left( {x - 5} \right)^{2\,021}} \cdot \left( {x - 5 - 1} \right) = 0\]

\[{\left( {x - 5} \right)^{2\,021}} \cdot \left( {x - 6} \right) = 0\]

Suy ra \[{\left( {x - 5} \right)^{2\,021}} = 0\] hoặc \[x - 6 = 0\]

\[x - 5 = 0\] hoặc \[x = 6\]

i) \[{\left( {2x + 1} \right)^3} = 9 \cdot 81\]

\[{\left( {2x + 1} \right)^3} = {9^3}\]

Suy ra \[2x + 1 = 9\]

 \[2x = 9 - 1\]

 \[2x = 8\]

   \[x = 8:2\]

   \[x = 4\]

j) \({\left( {3x - 5} \right)^2} = 16\)

\({\left( {3x - 5} \right)^2} = {4^2}\)

Suy ra \(3x - 5 = 4\)

 \(3x = 4 + 5\)

 \(3x = 9\)

 \(x = 9:3\)

   \(x = 3\).

a) Số \(\overline {7*} \) chia hết cho 2 nên \(* \in \left\{ {0;\,\,2;\,\,4;\,\,6;\,\,8} \right\}\).

b) Số \[\overline {45*} \] chia hết cho 5 nên \(* \in \left\{ {0;\,\,5} \right\}\).

c) Số \(\overline {*75} \) chia hết cho 3 nên \(\left( {* + 7 + 5} \right)\,\, \vdots \,\,3\) hay \(\left( {* + 12} \right)\,\, \vdots \,\,3\)

Suy ra \(* \in \left\{ {0;\,\,3;\,\,6;\,\,9} \right\}\).

Mà số \(\overline {*75} \) là số có ba chữ số nên \(* \in \left\{ {3;\,\,6;\,\,9} \right\}.\)

d) Số \(\overline {745*1} \) chia hết cho 9 nên \(\left( {7 + 4 + 5 + * + 1} \right)\,\, \vdots \,\,9\) hay \(\left( {* + 17} \right)\,\, \vdots \,\,9\)

Suy ra \(* = 1\).

e) Số \(\overline {1373*} \) chia hết cho 2 nên \(* \in \left\{ {0;\,\,2;\,\,4;\,\,6;\,\,8} \right\}\).

Mà số \(\overline {1373*} \) chia hết cho 9 nên \[\left( {1 + 3 + 7 + 3 + *} \right)\,\, \vdots \,\,9\] hay \[\left( {14 + *} \right)\,\, \vdots \,\,9\]

Suy ra \[* = 4\].

f) Số \(\overline {159*} \) chia hết cho 2 nên \(* \in \left\{ {0;\,\,2;\,\,4;\,\,6;\,\,8} \right\}\).

Mà số \(\overline {159*} \) chia hết cho 3 nên \(\left( {1 + 5 + 9 + *} \right)\,\, \vdots \,\,3\) hay \(\left( {15 + *} \right)\,\, \vdots \,\,3\)

Suy ra \(* \in \left\{ {0;\,\,6} \right\}\).