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}
→ \(B=\left\{-3;-2;-1;0\right\}\)

\(a,=703-140:70-17^2=703-2-289=412\\ b,=36.5.11-37.5.11+11.11.5=11.5(36-37+11)=55.10=550\\ c,=1152-374-1152-65+374=-65\\ d,=53-76+76-53=0\)

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.

$275-(113+x)+63=158$

$338-(113+x)=158$

$113+x=338-158=180$

$x=180-113=67$

\(a,=2^3\left(17-14\right)=2^3\cdot3=24\\ b,=80-\left(130-8^2\right)=80-\left(130-64\right)=80-66=14\)

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

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:

\(\dfrac{\left(100+2\right)\cdot50}{2}=2550\)

a)A=(50+1).50:2=51.50:2=1275

b)B=(100+2).((100-2):2+1):2=98.50:2=2450

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: \(=1-\dfrac{1}{2009}-1-\dfrac{1}{2008}+\dfrac{1}{2009}+1-\dfrac{1}{2008}\)

\(=1-\dfrac{2}{2008}=1-\dfrac{1}{1004}=\dfrac{1003}{1004}\)

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: \(1350+360+650+40=2000+400=2400\)

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 \(A=\frac{1023}{1024}\)