3 mũ 150> 2 mũ 200

3¹⁵⁰ > 2²⁰⁰

^{dung 100%}

cảm ơn nha

nhung lần sau bạn tlời chi tiết 1 chút hộ mình nhé

\(S=1+2+2^2+...+2^9\)

\(\Rightarrow2S=2+2^2+2^3+...+2^{10}\)

\(\Rightarrow S=2^{10}-1\)

Lại có \(5.2^8=\left(2^2+1\right).2^8=2^{10}+2^8\)

Vậy \(S< 5.2^8\)

S=1+2+2^2+2^3+...+2^9​

2S=2+2^2+2^3+...+2^9+2^10

2S-S=(2+2^2+2^3+...+2^9+2^10)-(1+2+2^2+2^3+...+2^9)

S=2^10-1

5.2^8=(2^2+1).2^8=(2^2.2^8)+(1.2^8)=2^10+2^8

Vì 2^10-1<2^10+2^8=> S<5.2^8

Vậy S < 5. 2^8

Ta có: \(S=1+2+2^2+2^3+...+2^9\)

\(\Rightarrow2S=2+2^2+2^3+...+2^{10}\)

\(\Rightarrow2S-S=\left(2+2^2+2^3+...+2^{10}\right)-\left(1+2+2^2+...+2^9\right)\)

\(\Rightarrow S=2^{10}-1\)

Mặt khác: \(5.2^8=\left(1+2^2\right).2^8=2^8+2^2.2^8=2^8+2^{10}\)

Vì \(2^{10}-1< 2^8+2^{10}\Rightarrow S< 5.2^8\)

bạn viết sai đề rồi 2^210=2^2010

\(2A=2.\left(1+2+....+2^{2010}\right)\)

\(2A-A=\left(2+2^2+...+2^{2011}\right)-\left(1+2+...+2^{2010}\right)\)

\(A=2^{2011}-1\)

\(B=2^{2011}-1=>A=B\)

2^x+2^x​⁺¹+2^x+2+2^x​⁺³-480=0

2^x+2^x​.2+2^x.2²+2^x​.2³=0+480

2^x.(1+2+2²+2³)=480

2^x.(1+2+4+8)=480

 2^x.15=480

2^x=480:15

2^x=32=2⁵

Vậy x =5

nếu sai thì thông cảm nha

các anh chị ơi giúp em với ạ

em đang cần gấp

ko bt làm

dng hóng drama

S = 2^0 + 2^2 + 2^4 +...+ 2^100

4S = 2^2 + 2^4 + 2^6 + ... + 2^100 + 2^102

4S - S = 2^2 + 2^4 + 2^6 + ... + 2^100 + 2^102 - ( 2^0 + 2^2 + 2^4 +...+ 2^100 )

3S = 2^102 - 1

S = ( 2^102 - 1 ) / 3 

có ai ko

giúp mk vs

Gọi \(\frac{1}{2^2}\) + \(\frac{1}{2^3}\) + \(\frac{1}{2^4}\) + ... + \(\frac{1}{2^n}\) là A

Ta có :

\(\frac{1}{2^2}\)<\(\frac{1}{1.2}\)

\(\frac{1}{2^3}\)<\(\frac{1}{2.3}\)

\(\frac{1}{2^4}\)<\(\frac{1}{3.4}\)

....

\(\frac{1}{2^n}\)<\(\frac{1}{\text{(n - 1) . n}}\)

❄ Nên :

A < \(\frac{1}{1.2}\) + \(\frac{1}{2.3}\) + \(\frac{1}{3.4}\) + ... + \(\frac{1}{\text{(n - 1) . n}}\)

A < \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\)

A < \(1-\frac{1}{n}\) < 1

Vậy A < 1

\(\frac{1}{2^2}\)\(\frac{1}{2^2}\)