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![](https://rs.olm.vn/images/avt/0.png?1311)
1) Viết tập hợp sau bằng cách liệt kê các phần tử của nó:
a) 𝑨 = {𝒙 ∈ 𝒁|−𝟓 < 𝒙 ≤ 𝟏}
→ \(A=\left\{-4;-3;-2;-1;0;1\right\}\)
b) 𝑩 = {𝒙 ∈ 𝒁| − 𝟑 ≤ 𝒙 ≤ 𝟎}
→ \(B=\left\{-3;-2;-1;0\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,=703-140:70-17^2=703-2-289=412\\ b,=36.5.11-37.5.11+11.11.5=11.5(36-37+11)=55.10=550\\ c,=1152-374-1152-65+374=-65\\ d,=53-76+76-53=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
a. $7^{2x-6}=49=7^2$
$\Rightarrow 2x-6=2$
$2x=8$
$x=4$
b.
$105-(135-7x):9=97$
$(135-7x):9=105-97=8$
$135-7x=8.9=72$
$7x=135-72=63$
$x=63:7=9$
c.
$3^{x-3}-3^2=2.3^2$
$3^{x-3}=3^2+2.3^2=3^2(1+2)=3^3$
$\Rightarrow x-3=3$
$\Rightarrow x=6$
d.
$275-(113+x)+63=158$
$338-(113+x)=158$
$113+x=338-158=180$
$x=180-113=67$
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,=2^3\left(17-14\right)=2^3\cdot3=24\\ b,=80-\left(130-8^2\right)=80-\left(130-64\right)=80-66=14\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 2.3.12+4.6.42+8.27.3
=2.12.3+4.6.42+8.3.27
=24.3+24.42+24.27
=24.(3+42+27)
=24.72
=1728
=
a) 2 . 3 . 12 + 4 . 6 . 42 + 8 . 27 . 3
= 3 . 24 + 24 . 42 + 24 . 27
= 24 ( 3 + 42 + 27 )
= 24 . 72 = 3024
b) Số số hạng là :(100 – 2) : 2 + 1 = 50 (số hạng)
Tổng là: (100 + 2) x 50 : 2 = 2550
97; 95; ….; 1 có (97 – 1) : 2 + 1 = 49 (số hạng)
Tổng là: (97 + 1) x 49 : 2 = 2401
A=(100+98+96+...+2)-(97+95+...+1) = 2550 – 2401
A = 149
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Số số hạng là:
50-1+1=50(số)
Tổng của dãy là:
\(\dfrac{\left(50+1\right)\cdot50}{2}=1275\)
b: Số số hạng là:
(100-2):2+1=50(số)
Tổng của dãy là:
\(\dfrac{\left(100+2\right)\cdot50}{2}=2550\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\dfrac{2}{5}+\dfrac{4}{5}\times\dfrac{5}{2}\)
\(=\dfrac{2}{5}+\dfrac{4\times5}{5\times2}\)
\(=\dfrac{2}{5}+\dfrac{4}{2}\)
\(=\dfrac{2}{5}+2\)
\(=\dfrac{2}{5}+\dfrac{10}{5}\)
\(=\dfrac{12}{5}\)
b) \(\dfrac{2008}{2009}-\dfrac{2009}{2008}+\dfrac{1}{2009}+\dfrac{2007}{2008}\)
\(=\left(1-\dfrac{1}{2009}\right)-\left(1+\dfrac{1}{2008}\right)+\dfrac{1}{2009}+\left(1-\dfrac{1}{2008}\right)\)
\(=1-\dfrac{1}{2009}-1-\dfrac{1}{2008}+\dfrac{1}{2009}+1-\dfrac{1}{2008}\)
\(=\left(1-1+1\right)-\left(\dfrac{1}{2009}-\dfrac{1}{2009}\right)-\left(\dfrac{1}{2008}+\dfrac{1}{2008}\right)\)
\(=1-\dfrac{2}{2008}\)
\(=\dfrac{2008}{2008}-\dfrac{2}{2008}\)
\(=\dfrac{2006}{2008}\)
\(=\dfrac{1003}{1004}\)
a: =2/5+4/2
=2/5+2
=12/5
b: \(=1-\dfrac{1}{2009}-1-\dfrac{1}{2008}+\dfrac{1}{2009}+1-\dfrac{1}{2008}\)
\(=1-\dfrac{2}{2008}=1-\dfrac{1}{1004}=\dfrac{1003}{1004}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 3:
a: Ta có: \(23\left(42-x\right)=23\)
\(\Leftrightarrow42-x=1\)
hay x=41
b: Ta có: 15(x-3)=30
nên x-3=2
hay x=5
Bài 1:
a: 32+89+68=100+89=189
b: 64+112+236=300+112=412
c: \(1350+360+650+40=2000+400=2400\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\)
\(\Rightarrow2A=1+\frac{1}{2}+...+\frac{1}{2^9}\)
Lấy \(2A-A\), ta có:
\(2A-A=A=\left(1+\frac{1}{2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)
\(=1+\frac{1}{2}+...+\frac{1}{2^9}-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^{10}}\)
\(=\left(1-\frac{1}{2^{10}}\right)+\left(\frac{1}{2}-\frac{1}{2}\right)+...+\left(\frac{1}{2^9}-\frac{1}{2^9}\right)\)
\(=1-\frac{1}{2^{10}}\)
\(=1-\frac{1}{1024}\)
\(=\frac{1023}{1024}\)
Vậy \(A=\frac{1023}{1024}\)
a. C = 12 . (-47) + 12 . (-19) + (-66) . 88
C = 12 . [-47 + (-19)] + (-66) . 88
C = 12 . (-66) + (-66) . 88
C = (-66) . (12 + 88)
C = -66 . 100
C = -6 600
b. D = (-135) . 29 - 13 . 135 + 42 . (-65)
D = 135 . (-29) - 13 . 135 + 42 . (-65)
D = 135 . [(-29) - 13] + 42 . (-65)
D = 135 . (-42) + 42 . (-65)
D = -135 . 42 + 42 . (-65)
D = 42 . [-135 - (-65)]
D = 42 . (-70)
D = -2940