a,Tính tổng:S=1+52+54+...+5200

=>5²S=5²+5⁴+5⁶+...+5²⁰²

=>25S-S=24S=5²⁰²-1

=>S=\(\frac{5^{202}-1}{24}\)

b,So sánh 230+330+430 và 3.2410

3.24^10=3^11.4^15 
4^30=4^15.4^15 
hiển nhiên 4^15>3^11 
=>3.24^10<<4^30<<<2^30+3^20+4^30

Ta có: 2³⁰+3³⁰+4³⁰>2³⁰+2³⁰+4³⁰=2³¹+2³⁰.2³⁰

                                                                 =2³¹(1+2²⁹) (1)

Lại có:3.24^10=3^11.2^30 (2)

So sánh (1)và (2): Vì 3^11<4^11=2^22<2^29

                              và 2^30<2^31

=> 3^11.2^30 <(1+2^29)2^31<2^30+3^30+4^30

\(1,\\ \left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\\ \Leftrightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^{10}\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-7\right)^{x+1}=0\\\left(x-7\right)^{10}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x-7=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\end{matrix}\right.\)

\(2,\\ a,\left|2x-3\right|>5\Leftrightarrow\left[{}\begin{matrix}2x-3< -5\\2x-3>5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< -1\\x>4\end{matrix}\right.\\ b,\left|3x-1\right|\le7\Leftrightarrow\left[{}\begin{matrix}3x-1\le7\\1-3x\le7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\le\dfrac{8}{3}\\x\ge-2\end{matrix}\right.\\ c,\cdot x< -\dfrac{3}{2}\\ \Leftrightarrow5-3x+\left(-2x-3\right)=7\Leftrightarrow2-5x=7\Leftrightarrow x=-1\left(ktm\right)\\ \cdot-\dfrac{3}{2}\le x\le\dfrac{5}{3}\\ \Leftrightarrow\left(5-3x\right)+\left(2x+3\right)=7\Leftrightarrow8-x=7\Leftrightarrow x=1\left(tm\right)\\ \cdot x>\dfrac{5}{3}\\ \Leftrightarrow\left(3x-5\right)+\left(2x+3\right)=7\Leftrightarrow5x-2=7\Leftrightarrow x=\dfrac{9}{5}\left(tm\right)\\ \Leftrightarrow S=\left\{1;\dfrac{9}{5}\right\}\)

Mai lam

Đề bài thiếu yêu cầu cụ thể em nhé. em cập nhật lại câu hỏi để được sự hỗ trợ tốt nhất cho tài khoản olm vip

#@₫!%&@^@₫@₫=_++_×%@%@&@@@@=@

A = 5 + 5² + 5³ + ... + 5⁴⁹ + 5⁵⁰

⇒ 5A = 5² + 5³ + 5⁴ + ... + 5⁵⁰ + 5⁵¹

⇒ 4A = 5A - A

= (5² + 5³ + 5⁴ + ... + 5⁵⁰ + 5⁵¹) - (5 + 5² + 5³ + ... + 5⁴⁹ + 5⁵⁰)

= 5⁵¹ - 5

⇒ A = (5⁵¹ - 5) : 4

A = 5 + 5² + 5³ + ... + 5²⁰²³

⇒ 5A = 5² + 5³ + 5⁴ + ... + 5²⁰²⁴

⇒ 4A = 5A - A

= (5² + 5³ + 5⁴ + ... + 5²⁰²⁴) - (5 + 5² + 5³ + ... + 5²⁰²³)

= 5²⁰²⁴ - 5

⇒ A = (5²⁰²⁴ - 5)/4

A = 5 + 5² + 5³ + ... + 5²⁰²³

⇒ 5A = 5² + 5³ + 5⁴ + ... + 5²⁰²⁴

⇒ 4A = 5A - A

= (5² + 5³ + 5⁴ + ... + 5²⁰²⁴) - (5 + 5² + 5³ + ... + 5²⁰²³)

= 5²⁰²⁴ - 5

⇒ A = (5²⁰²⁴ - 5)/4

5A=5+5^2+...+5^2023

=>4A=5^2023-1

=>\(A=\dfrac{5^{2023}-1}{4}\)

\(2B-A=\dfrac{5^{2023}}{4}-\dfrac{5^{2023}-1}{4}=\dfrac{1}{4}\)

a) \(S=5+5^2+...+5^{2006}\)

\(5S=5^2+5^3+...+5^{2007}\)

\(5S-S=5^2+5^3+5^4+...+5^{2007}-5-5^2-5^3-...-5^{2006}\)

\(4S=5^{2007}-5\)

\(S=\dfrac{5^{2007}-5}{4}\)

b) \(S=5+5^2+5^3+...+5^{2006}\)

\(S=\left(5+5^4\right)+\left(5^2+5^5\right)+...+\left(5^{2003}+5^{2006}\right)\)

\(S=5\cdot\left(1+5^3\right)+5^2\cdot\left(1+5^3\right)+...+5^{2003}\cdot\left(1+5^3\right)\)

\(S=\left(1+5^3\right)\cdot\left(5+5^2+...+5^{2003}\right)\)

\(S=126\cdot\left(5+5^2+...+5^{2003}\right)\) ⋮ 126 

a: \(12+2^2+3^2+4^2+5^2\)

\(=12+4+9+16+25\)

\(=16+50=66\)

\(\left(1+2+3+4+5\right)^2=15^2=225\)

=>\(12+2^2+3^2+4^2+5^2< \left(1+2+3+4+5\right)^2\)

b: \(1^3+2^3+3^3+4^3=\left(1+2+3+4\right)^2< \left(1+2+3+4\right)^3\)

c: \(5^{202}=5^2\cdot5^{200}=25\cdot5^{200}>16\cdot5^{200}\)

d: \(18\cdot4^{500}=18\cdot2^{1000}\)

\(2^{1004}=2^4\cdot2^{1000}=16\cdot2^{1000}\)

=>\(18\cdot4^{500}>2^{1004}\)

e: \(2022\cdot2023^{2024}+2023^{2024}=2023^{2024}\left(2022+1\right)\)

\(=2023^{2025}\)

dạ em nhầm ạ phần a) số 12 phải là 1²