Bộ 5 đề thi cuối kì 1 Toán 6 Cánh diều (2023-2024) có đáp án - Đề 1 (Mã 2)

$\left( { - 80} \right) + 25 + \left( { - 20} \right) + 75$

$\begin{array}{l} = \,\,\,\left[ {\left( { - 80} \right) + \left( { - 20} \right)} \right] + \left( {25 + 75} \right)\\ = \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 100\,\,\,\,\,\,\,\,\,\,\, + \,\,\,\,\,\,\,100\\ = \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\end{array}$

$2 \cdot \left[ {\left( { - 3} \right) + 1} \right] + 10$

$$\begin{array}{l} = \,\,\,2.\,\,\,\,\left( { - 2} \right)\,\,\,\,\,\,\,\,\,\, + 10\,\,\\ = \,\,\,\,\,\, - 4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 10\\ = \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,6\,\,\end{array}$$

$2024 - \left[ {10 \cdot \left( {{4^2} - 12} \right):8 + {2^3}} \right] \cdot {2023^0}$

$$\begin{array}{l} = \,2024 - \left[ {10.\left( {16 - 12} \right):8 + 8} \right].\,\,1\\ = \,2024 - \left[ {10.\,\,4:8 + 8} \right].\,\,1\\ = \,2024 - \left[ {40:8 + 8} \right].\,\,1\\ = 2024 - \left[ {5 + 8} \right].\,\,1\\ = 2024 - 13.\,\,1\\ = 2024 - 13\\ = \,\,\,\,\,\,\,\,\,2011\end{array}$$

$$x + 10 = 2$$

$$\begin{array}{l}\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\, = \,2\, - 10\\\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\, = \,\,\, - 8\end{array}$$

Vậy $x = - 8$.

$2 - \left( {x - 7} \right) = 6 - 15$

$$\begin{array}{l}\,\,\,\,2 - \left( {x - 7} \right) = \,\,\,\, - 9\\\,\,\,\,\,\,\,\,\,\,\,\,x - 7\,\,\,\, = \,2 - \left( { - 9} \right)\,\,\\\,\,\,\,\,\,\,\,\,\,\,\,x - 7\,\,\,\, = \,\,\,\,\,\,11\\\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\, = \,\,\,11 + 7\\\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\, = \,\,\,\,\,\,\,18\end{array}$$

Vậy $x = 18$.

$$2n + 7\,\,\; \vdots \,\,\,n + 2$$

Ta có $2n + 7 = 2\left( {n + 2} \right) + 3$

Để $2n + 7\,\,\; \vdots \,\,\,n + 2$

$$2\left( {n + 2} \right) + 3\,\,\; \vdots \,\,\,n + 2$$

Khi đó $3\,\,\; \vdots \,\,\,n + 2$ nên $n + 2 \in$Ư$\left( 3 \right) = \left\{ {1\,;\,\,3} \right\}$.

Ta có bảng giá trị:

Vậy $n = 1$ thì $2n + 7\,\,\; \vdots \,\,\,n + 2$.