Lời giải:

$A=\frac{1}{7^2}+\frac{2}{7^3}+\frac{3}{7^4}+....+\frac{69}{7^{70}}$

$7A=\frac{1}{7}+\frac{2}{7^2}+\frac{3}{7^3}+...+\frac{69}{7^{69}}$

$\Rightarrow 6A=7A-A=\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{69}}-\frac{69}{7^{70}}$

$42A=1+\frac{1}{7}+\frac{1}{7^2}+...+\frac{1}{7^{68}}-\frac{69}{7^{69}}$

$\Rightarrow 36A=42A-6A=1-\frac{69}{7^{69}}+\frac{69}{7^{70}}<1$

$\Rightarrow A< \frac{1}{36}$

$A=\genfrac{}{}{0ex}{}{1}{7^2}+\genfrac{}{}{0ex}{}{2}{7^3}+\genfrac{}{}{0ex}{}{3}{7^4}+....+\genfrac{}{}{0ex}{}{69}{7^{70}}$

$7A=\genfrac{}{}{0ex}{}{1}{7}+\genfrac{}{}{0ex}{}{2}{7^2}+\genfrac{}{}{0ex}{}{3}{7^3}+...+\genfrac{}{}{0ex}{}{69}{7^{69}}$

$\Rightarrow 6A=7A-A=\genfrac{}{}{0ex}{}{1}{7}+\genfrac{}{}{0ex}{}{1}{7^2}+\genfrac{}{}{0ex}{}{1}{7^3}+...+\genfrac{}{}{0ex}{}{1}{7^{69}}-\genfrac{}{}{0ex}{}{69}{7^{70}}$

$42A=1+\genfrac{}{}{0ex}{}{1}{7}+\genfrac{}{}{0ex}{}{1}{7^2}+...+\genfrac{}{}{0ex}{}{1}{7^{68}}-\genfrac{}{}{0ex}{}{69}{7^{69}}$

$\Rightarrow 36A=42A-6A=1-\genfrac{}{}{0ex}{}{69}{7^{69}}+\genfrac{}{}{0ex}{}{69}{7^{70}}<1$

$\Rightarrow A<\genfrac{}{}{0ex}{}{1}{36}$
 

280 - x.9 = 450

x.9 = 280 - 450

x.9 = -170

x= -170/9

1. \(x⋮15\Rightarrow x\in B\left(15\right)=\left\{0;15;30;45;60;75;90;105;120;135;150;...\right\}\)

mà \(45< x< 136\)

\(\Rightarrow x\in\left\{60;75;90;105;120;135\right\}\)

2.

\(18⋮x\Rightarrow x\in U\left(18\right)=\left\{1;2;3;6;18\right\}\)

mà \(x>7\Rightarrow\Rightarrow x\in\left\{18\right\}\)

1. A={0;7;14;21;28}

2. B={0;10;20;30;40}

1.

5 phần tử là 5 số nhỏ nhất là bội của 7

\(B\left(7\right)=\left\{0;7;14;21;28\right\}\)

2.

5 phần tử là 5 số nhỏ nhất là bội của 10

\(B\left(10\right)=\left\{0;10;20;30;40\right\}\)

1

\(4^{2x-1}:4=4^4\\ \Leftrightarrow4^{2x-2}=4^4\\ \Leftrightarrow2x-2=4\\ \Leftrightarrow2x=6\\ \Leftrightarrow x=3\)

210 . 13 + 210 . 6528 .104

= 210 . ( 13 + 6528 ) . 104

= 210 . 6541 .104

= 142855440

\(X\in\left\{0;12;24;36;48\right\}\)

 Bằng 60 dc ko c nhỉ, đây là tự luận hay gì ạ. E ko bit giải, chỉ bit đáp án thôi, chị cần ko

2^1, 2^3, 2^4 ,3^3, 3^4, 3^1

\(2A=2+\dfrac{1}{2}.6+\dfrac{1}{3}.12+\dfrac{1}{4}.20+...+\dfrac{1}{200}.40200=\)

\(=2+\dfrac{1}{2}.2.3+\dfrac{1}{3}.3.4+\dfrac{1}{4}.4.5+...+\dfrac{1}{200}.200.201=\)

\(=2+3+4+5+...+201=\dfrac{200\left(2+201\right)}{2}\)

\(=20300\Rightarrow A=\dfrac{20300}{2}=10150\)

\(=2023-1^{2020}+1=2023\)