Bài 1. (3 điểm) Tính.
a) 49 + (11 – 25)
c) -8 + 5 . (-9)
e) 40 – (-7)2
f) | -15 + 21| – | 4 – 11|
b) 5 + (-8).3
d) 4 + (-5)2
g) 1 – 2 – 3 + 4 + 5 – 6 – 7 + 8 + … + 801 – 802 – 803 + 804
Bài 2. (1 điểm) Tính tổng các số nguyên x, biết: -3 < x < 2 Bài 3. (3 điểm) Tìm x ∈ Z, biết: a) x + 9 = 2 – 17 c) x - 17 = (-11) . (-5) d) |x – 5| = (-4)2 b) x – 2 = -6 + 17 e) 2x + 5 = x – 1 f) |x – 4| = | -81 Bài 4. (2 điểm) Tìm các số nguyên x sao cho: a) -7 là bội của x + 8 b) x – 2 là ước của 3x – 13 Bài 5. (1 điểm) Tìm x, y ∈ Z, biết: (x – 2)(y + 1) = 23
Đáp án:
$\begin{array}{l}
1)a)49 + \left( {11 – 25} \right)\\
= 49 + 11 – 25\\
= 60 – 25\\
= 35\\
b) – 8 + 5.\left( { – 9} \right)\\
= – 8 – 45\\
= – 53\\
e)40 – {\left( { – 7} \right)^2}\\
= 40 – 49\\
= – 9\\
f)\left| { – 15 + 21} \right| – \left| {4 – 11} \right|\\
= \left| 6 \right| – \left| { – 7} \right|\\
= 6 – 7\\
= – 1\\
b)5 + \left( { – 8} \right).3\\
= 5 – 24\\
= – 19\\
d)4 + {\left( { – 5} \right)^2}\\
= 4 + 25\\
= 29\\
g)1 – 2 – 3 + 4 + 5 – 6 – 7 + 8 + …\\
+ 801 – 802 – 803 + 804\\
= \left( {1 – 2 – 3 + 4} \right) + \left( {5 – 6 – 7 + 8} \right) + … + \\
\left( {801 – 802 – 803 + 804} \right)\\
= 0 + 0 + … + 0\\
= 0\\
B2)\\
– 3 < x < 2\\
\Rightarrow x \in \left\{ { – 2; – 1;0;1} \right\}\\
\Rightarrow S = \left( { – 2} \right) + \left( { – 1} \right) + 0 + 1\\
= – 2\\
B3)a)x + 9 = 2 – 17\\
\Rightarrow x = 2 – 17 – 9\\
\Rightarrow x = – 24\\
Vậy\,x = – 24\\
c)x – 17 = \left( { – 11} \right).\left( { – 5} \right)\\
\Rightarrow x = 17 + 55\\
\Rightarrow x = 72\\
Vậy\,x = 72\\
d)\left| {x – 5} \right| = {\left( { – 4} \right)^2}\\
\Rightarrow \left| {x – 5} \right| = 16\\
\Rightarrow \left[ \begin{array}{l}
x – 5 = 16 \Rightarrow x = 21\\
x – 5 = – 16 \Rightarrow x = – 11
\end{array} \right.\\
Vậy\,x = 21;x = – 11\\
b)x – 2 = – 6 + 17\\
\Rightarrow x = 2 – 6 + 17\\
\Rightarrow x = 13\\
Vậy\,x = 13\\
e)2x + 5 = x – 1\\
\Rightarrow 2x – x = – 1 – 5\\
\Rightarrow x = – 6\\
Vậy\,x = – 6\\
f)\left| {x – 4} \right| = \left| { – 81} \right|\\
\Rightarrow \left| {x – 4} \right| = 81\\
\Rightarrow \left[ \begin{array}{l}
x – 4 = 81 \Rightarrow x = 85\\
x – 4 = – 81 \Rightarrow x = 77
\end{array} \right.\\
Vậy\,x = 85;x = 77\\
B4)a) – 7 \vdots x + 8\\
\Rightarrow \left( {x + 8} \right) \in \left\{ { – 7; – 1;1;7} \right\}\\
\Rightarrow x \in \left\{ { – 15; – 8; – 7; – 1} \right\}\\
b)\left( {3x – 13} \right) \vdots \left( {x – 2} \right)\\
3x – 13 = 3x – 6 – 7 = 3\left( {x – 2} \right) – 7\\
\Rightarrow 7 \vdots \left( {x – 2} \right)\\
\Rightarrow \left( {x – 2} \right) \in \left\{ { – 7; – 1;1;7} \right\}\\
\Rightarrow x \in \left\{ { – 5;1;3;9} \right\}
\end{array}$
$\begin{array}{l}
B5)\left( {x – 2} \right)\left( {y + 1} \right) = 23 = 1.23\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x – 2 = 1\\
y + 1 = 23
\end{array} \right.\\
\left\{ \begin{array}{l}
x – 2 = – 1\\
y + 1 = – 23
\end{array} \right.\\
\left\{ \begin{array}{l}
x – 2 = 23\\
y + 1 = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x – 2 = – 23\\
y + 1 = – 1
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = 3;y = 22\\
x = 1;y = – 24\\
x = 25;y = 0\\
x = – 21;y = – 2
\end{array} \right.\\
Vậy\,\left( {x;y} \right) = \left\{ {\left( {3;22} \right);\left( {1; – 2} \right);\left( {25;0} \right);\left( { – 21; – 2} \right)} \right\}
\end{array}$