\(2A=\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{101}}\)

\(2A-A=\frac{1}{2^{101}}-\frac{1}{2}\)

\(\Rightarrow A=\frac{1}{2^{101}}-\frac{1}{2}\)

\(\Rightarrow A>0\) ( đpcm )

Bài này phải làm như thế này nha lần trước tui làm nhầm sorry

Study well 

uk cám ơn bn nhiều

A = 1/2 + 1/2² + 1/2³ + 1/2⁴ + ... + 1/2¹⁰⁰

2A = 1 + 1/2 + 1/2² + 1/2³ + ... + 1/2⁹⁹

2A - A = (1 + 1/2 + 1/2² + 1/2³ + ... + 1/2⁹⁹) - (1/2 + 1/2² + 1/2³ + 1/2⁴ + ... + 1/2¹⁰⁰)

A = 1 - 1/2¹⁰⁰ < 1

Do 1 > 1/2¹⁰⁰ => A > 0

=> 0 < A < 1

=> đpcm

Bài 15 :

a) Đặt \(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2019\cdot2020}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2019}-\frac{1}{2020}\)

\(A=1-\frac{1}{2020}=\frac{2019}{2020}< \frac{2020}{2020}=1\)

b) Ta có : \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{1000}}\)

\(2A=\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{1001}}\)

\(2A-A=\left(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{1001}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{1000}}\right)\)

\(A=\frac{1}{2^{1001}}-\frac{1}{2}\)

Tới đây là so sánh đi nhé

Cái này mình làm hôm qua rồi mà '-'

a) Đặt \(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2019\cdot2020}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)

\(A=\frac{1}{1}-\frac{1}{2020}=\frac{2019}{2020}\)

\(\Rightarrow A< 1\)

b) \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{1000}}\)

\(2A=2\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{1000}}\right)\)

\(2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{999}}\)

\(2A-A=A\)

\(=\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{999}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{1000}}\right)\)

\(=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^{999}}-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{1000}}\)

\(=1-\frac{1}{2^{1000}}\)

\(\Rightarrow A=1-\frac{1}{2^{1000}}< 1\left(đpcm\right)\)

1. A = 75(4²⁰⁰⁴ + 4²⁰⁰³ +...+ 4² + 4 + 1) + 25

    A = 25 . [3 . (4²⁰⁰⁴ + 4²⁰⁰³ +...+ 4² + 4 + 1) + 1]

    A = 25 . (3 . 4²⁰⁰⁴ + 3 . 4²⁰⁰³ +...+ 3 . 4² + 3 . 4 + 3 + 1)

    A = 25 . (3 . 4²⁰⁰⁴ + 3 . 4²⁰⁰³ +...+ 3 . 4² + 3 . 4 + 4)

    A = 25 . 4 . (3 . 4²⁰⁰³ + 3 . 4²⁰⁰² +...+ 3 . 4 + 3 + 1)

    A =100 . (3 . 4²⁰⁰³ + 3 . 4²⁰⁰² +...+ 3 . 4 + 3 + 1) \(⋮\) 100

3a) |x| = 1/2 

=> x = 1/2 hoặc x = -1/2

với x = 1/2:

A = \(3.\left(\frac{1}{2}\right)^2-2.\frac{1}{2}+1\)

\(A=\frac{3}{4}-1+1=\frac{3}{4}\)

với x = -1/2

A = \(3.\left(-\frac{1}{2}\right)^2-2\left(-\frac{1}{2}\right)+1\)

\(A=\frac{3}{4}+1+1=\frac{3}{4}+2=\frac{11}{4}\)

Chào bạn, bạn hãy theo dõi bài giải của mình nhé!

Ta có : 

\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)

\(=>2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)

\(=>2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)

\(=>A=1-\frac{1}{2^{100}}\)

Ta có : \(1>\frac{1}{2^{100}}=>A>1-1=0\)

\(\frac{1}{2^{100}}>0=>1-\frac{1}{2^{100}}< 1-0=1\)

\(=>0< A< 1\)

Chúc bạn học tốt!

Dễ thấy A>0(vì 1/2>0;1/2^2>0;...;1/2^100>0 =>1/2+1/2^2+1/2^3+...+1/2^100>0)

2A=1+2/2^2+2/2^3+...+2/2^100(rút gọn 1 bước)

2A=1+1/2+1/2^2+...+1/2^99

2A-A=(1+1/2+1/2^2+...+1/2^99)-(1/2+1/2^2+1/2^3+...+1/2^99+1/2^100)

A=1-1/2^100<1

Vậy A<1

Cậu tự KL nhé

\(a)\) Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\) ta có : 

\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(A< 1-\frac{1}{100}=\frac{99}{100}< 1\)

Vậy \(A< 1\)

Chúc bạn học tốt ~ 

a) Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)=> \(2.A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\)

=> \(2.A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\right)\)

\(A=1-\frac{1}{2^{10}}\)=> \(1-A=1-\left(1-\frac{1}{2^{10}}\right)=\frac{1}{2^{10}}>\frac{1}{2^{11}}\)=> đpcm

b) Đặt B = \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)

Vì \(\frac{1}{2^2}