\(A=2^1+2^2+2^3+2^4+2^5+2^6+2^7+...+2^{99}\)

    \(=\left(2^1+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+\left(2^7+2^8+2^9\right)+...+\left(2^{97}+2^{98}+2^{99}\right)\)

 \(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+2^7\left(1+2+2^2\right)+...+2^{97}\left(1+2+2^2\right)\)

 \(=2.7+2^4.7+2^7.7+...+2^{97}.7\)

   \(=\left(2+2^4+2^7+...+2^{97}\right).7⋮7\)

\(\Rightarrow A⋮7\)

A = 2¹ +2² +2³ +2⁴ +2⁵  +2⁶ +2⁷ ….+ 2⁹⁹ 

A = (2¹ +2² +2³) +(2⁴ +2⁵  +2⁶) + ….+ (2⁹⁷+2⁹⁸+2⁹⁹)

A = 14 + (2¹.2³ +2².2³  +2³.2³) + ….+ (2¹.2⁹⁶+2².2⁹⁶+2³.2⁹⁶)

A = 14 + 2³(2¹+2²+2³) + ...... + 2⁹⁶(2¹+2²+2³)

A = 14.1 + 2³.14 + ....... + 2⁹⁶.14

A = 14.(1+2³+....+2⁹⁶)

14 \(⋮\) 7

=> A \(⋮\) 7 (đpcm)

a: \(S=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-99}{100}=-\dfrac{1}{100}\)

c: \(5S_3=5^6+5^7+...+5^{101}\)

\(\Leftrightarrow4\cdot S_3=5^{101}-5^5\)

hay \(S_3=\dfrac{5^{101}-5^5}{4}\)

d: \(S_4=7\cdot\left(\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+...+\dfrac{1}{69}-\dfrac{1}{70}\right)\)

\(=7\left(\dfrac{1}{10}-\dfrac{1}{70}\right)=7\cdot\dfrac{6}{70}=\dfrac{6}{10}=\dfrac{3}{5}\)

Câu 1: 

a: =>37-39+x=13-13-17

=>x-2=-17

=>x=-15

b: |x-3|+x=3

=>|x-3|=3-x

=>x-3<=0

=>x<=3

c: (x²+7)(x²-49)=0

=>(x+7)(x-7)=0

=>x=-7 hoặc x=7

Ta có:

3s₁=3+3²+3³+3⁴+...+3⁵⁰

=>3s₁-s₁=3+3²+3³+3⁴+...+3⁵⁰-1-3-3²-3³-...-3⁴⁹

=>2s₁=3⁵⁰-1

=>a₁=(3⁵⁰-1)/2

Tính s₂ tương tự như s₁

ta lấy 4s₂-s₂ đoực kết quả s₂=(4⁵⁰-1)/3

S₁ = 1+3+3²+3³+3⁴+..........+3⁴⁹

3S₁ =  3+3²+3³+3⁴+3⁵+.........+3⁵⁰

3S₁ - S₁ = 3+3²+3³+3⁴+3⁵+.........+3⁵⁰ - (1+3+3²+3³+3⁴+..........+3⁴⁹)

             = 3+3²+3³+3⁴+3⁵+.........+3⁵⁰ - 1 - 3 - 3² - 3³ - 3⁴-..........-3⁴⁹

2S₁ = 3⁵⁰ - 1

S₁ =\(\frac{3^{50}-1}{2}\)

\(A=\frac{1}{2}+\frac{5}{6}+...+\frac{89}{90}=1-\frac{1}{2}+1-\frac{5}{6}+...+1-\frac{1}{90}=\left(1++...+1\right)-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{9\cdot10}\right)\)\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)=9-\left(\frac{10}{10}-\frac{1}{10}\right)=9-\frac{9}{10}=\frac{90}{10}-\frac{9}{10}=\frac{89}{10}\)

\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\)\(=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)

\(=\)\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-...-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=\)\(1-\dfrac{1}{10}\)

\(=\dfrac{9}{10}\)

=1-1/10=9/10

\(\frac{-5}{6}+\frac{8}{3}+\frac{29}{-6}\le x\le\frac{-1}{2}+2+\frac{5}{2}\)

\(\Rightarrow-3\le x\le4\)

\(\Rightarrow x\in\left\{-3;-2;-1;0;1;2;3;4\right\}\)

\(\left(\frac{-2}{3}-\frac{1}{2}\right):\frac{-1}{4}\le x\le\left(\frac{-5}{6}+\frac{9}{4}:\frac{-3}{2}\right)\cdot\frac{-13}{2}\)

\(\Rightarrow\frac{14}{3}\le x\le\frac{91}{6}\)

\(\Rightarrow\frac{28}{6}\le x\le\frac{91}{6}\)

\(\Rightarrow x\in\left\{\frac{28}{6};\frac{29}{6};...;\frac{90}{6};\frac{91}{6}\right\}\)

Câu 4:
a) Ta có: \(\left|-x+8\right|\ge0\)

\(\Rightarrow A=\left|-x+8\right|-21\ge-21\)

Vậy \(MIN_A=-21\) khi x = 8

b) Ta có: \(\left|-x-17\right|+\left|y-36\right|\ge0\)

\(\Rightarrow B=\left|-x-17\right|+\left|y-36\right|+12\ge12\)

Vậy \(MIN_B=12\) khi \(x=-17;y=36\)

c) Ta có: \(-\left|2x-8\right|\le0\)

\(\Rightarrow C=-\left|2x-8\right|-35\le-35\)

Vậy \(MAX_C=-35\) khi \(x=4\)

d) Ta có: \(3\left(3x-12\right)^2\ge0\)

\(\Rightarrow D=3\left(3x-12\right)^2-37\ge-37\)

Vậy \(MIN_D=-37\) khi x = 4

e) Ta có: \(-3\left|2x+50\right|\le0\)

\(\Rightarrow E=-21-3\left|2x+50\right|\le-21\)

Vậy \(MAX_E=-21\) khi x = -25

g) \(\left(x-3\right)^2+\left|x^2-9\right|\ge0\)

\(\Rightarrow G=\left(x-3\right)^2+\left|x^2-9\right|+25\ge25\)

Vậy \(MIN_G=25\) khi x = 3

MIN_{A là j}