Bài 1:

\(2^{49}=\left(2^7\right)^7=128^7;5^{21}=\left(5^3\right)^7=125^7\\ Vì:128^7>125^7\Rightarrow2^{49}>5^{21}\)

Bài 2:

\(a,S=1+3+3^2+3^3+...+3^{99}\\ =\left(1+3+3^2+3^3\right)+3^4.\left(1+3+3^2+3^3\right)+...+3^{96}.\left(1+3+3^2+3^3\right)\\ =40+3^4.40+...+3^{96}.40\\ =40.\left(1+3^4+...+3^{96}\right)⋮40\\ b,S=1+4+4^2+4^3+...+4^{62}\\ =\left(1+4+4^2\right)+4^3.\left(1+4+4^2\right)+...+4^{60}.\left(1+4+4^2\right)\\ =21+4^3.21+...+4^{60}.21\\ =21.\left(1+4^3+...+4^{60}\right)⋮21\)

Bài 1 :

\(2^{49}=\left(2^7\right)^7=128^7\)

\(5^{21}=\left(5^3\right)^7=125^7\)

mà \(125^7< 128^7\)

\(\Rightarrow2^{49}>5^{21}\)

Bài 2 :

a) \(S=1+3+3^2+3^3+...3^{99}\)

\(\Rightarrow S=\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)...+3^{96}\left(1+3+3^2+3^3\right)\)

\(\Rightarrow S=40+40.3^4+...+40.3^{96}\)

\(\Rightarrow S=40\left(1+3^4+...+3^{96}\right)⋮40\)

\(\Rightarrow dpcm\)

b) \(S=1+4+4^2+4^3+...4^{62}\)

\(\Rightarrow S=\left(1+4+4^2\right)+4^3\left(1+4+4^2\right)+...4^{60}\left(1+4+4^2\right)\)

\(\Rightarrow S=21+4^3.21+...4^{60}.21\)

\(\Rightarrow S=21\left(1+4^3+...4^{60}\right)⋮21\)

\(\Rightarrow dpcm\)

Bạn ko biết gõ số mũ à gõ thế này bố ai mà hiểu được

`#3107.101107`

\(A = 1 + 3 + 3^2 + 3^3 + ... + 3^{98} + 3^{99}\)

\(A = (1 + 3) + (3^2 + 3^3) + ... + (3^{98} + 3^{99})\)

\(A = (1 + 3) + 3^2(1 + 3) + ... + 3^{98}(1 + 3)\)

\(A = (1 + 3)(1 + 3^2 + ... + 3^{98})\)

\(A = 4(1 + 3^2 + ... + 3^{98})\)

Vì \(4(1 + 3^2 + ... + 3^{98}) \) \(\vdots\) \(4\)

`\Rightarrow A \vdots 4`

Vậy, `A \vdots 4` (đpcm).

A = 1 + 3 + 3² + 3³ + ... + 3⁹⁸ + 3⁹⁹

A = (1 + 3) + (3² + 3³) + ... + (3⁹⁸ + 3⁹⁹)

A = 1. (1 + 3) + 3². (1 + 3) + ... + 3⁹⁸. (1 + 3)

A = 1.4 + 3².4 + ... + 3⁹⁸.4

A = 4. (1 + 3² + ... + 3⁹⁸)

⇒ A ⋮ 4

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

\(\Leftrightarrow M=1+3+\left(3^2+3^3+3^4\right)+\left(3^5+3^6+3^7\right)+.......+\left(3^{98}+3^{99}+3^{100}\right)\)

\(\Leftrightarrow M=4+3^2\left(1+3+3^2\right)+3^5\left(1+3+3^2\right)+......+3^{98}\left(1+3+3^2\right)\)

\(\Leftrightarrow M=4+3^2.13+3^5.13+.........+3^{98}.13\)

\(\Leftrightarrow M=4+13\left(3^2+3^5+..........+3^{98}\right)\)

Mà \(13\left(3^2+3^5+......+3^{98}\right)⋮13\)

\(4:13\left(dư4\right)\)

\(\Leftrightarrow M:13\left(dư4\right)\)

b, tương tự

a: (x-3)(y+1)=15

=>\(\left(x-3\right)\left(y+1\right)=1\cdot15=15\cdot1=\left(-1\right)\cdot\left(-15\right)=\left(-15\right)\cdot\left(-1\right)=3\cdot5=5\cdot3=\left(-3\right)\cdot\left(-5\right)=\left(-5\right)\cdot\left(-3\right)\)

=>(x-3;y+1)\(\in\){(1;15);(15;1);(-1;-15);(-15;-1);(3;5);(5;3);(-3;-5);(-5;-3)}

=>(x,y)\(\in\){(4;14);(18;0);(2;-16);(-12;-2);(6;4);(8;2);(0;-6);(-2;-4)}

b: Sửa đề:\(m=1+3+3^2+3^3+...+3^{99}+3^{100}\)

\(m=1+3+\left(3^2+3^3+3^4\right)+\left(3^5+3^6+3^7\right)+...+\left(3^{98}+3^{99}+3^{100}\right)\)

\(=4+3^2\left(1+3+3^2\right)+3^5\left(1+3+3^2\right)+...+3^{98}\left(1+3+3^2\right)\)

\(=4+13\left(3^2+3^5+...+3^{98}\right)\)

=>m chia 13 dư 4

\(m=1+3+3^2+...+3^{99}+3^{100}\)

\(=1+\left(3+3^2+3^3+3^4\right)+...+\left(3^{97}+3^{98}+3^{99}+3^{100}\right)\)

\(=1+3\left(1+3+3^2+3^3\right)+3^5\left(1+3+3^2+3^3\right)+...+3^{97}\left(1+3+3^2+3^3\right)\)

\(=1+40\left(3+3^5+...+3^{97}\right)\)

=>m chia 40 dư 1

Bài 1 :

a) \(a.b+b.19=713\) \(\left(a;b\inℕ^∗\right)\)

\(\Rightarrow b.\left(a+19\right)=713\)

\(\Rightarrow\left(a+19\right);b\in\left\{1;23;31;713\right\}\)

\(\Rightarrow\left(a;b\right)\in\left\{\left(-18;713\right);\left(4;31\right);\left(12;23\right);\left(694;1\right)\right\}\)

\(\Rightarrow\left(a;b\right)\in\left\{\left(4;31\right);\left(12;23\right);\left(694;1\right)\right\}\left(a;b\inℕ^∗\right)\)

b) \(a.b-10.b=650\)

\(\Rightarrow b.\left(a-10\right)=650\)

\(\Rightarrow\left(a-10\right);b\in\left\{1;5;10;13;25;26;50;65;130;325;650\right\}\)

Bạn lập bảng sẽ tìm ra (a;b)...

Bài 2 :

a) \(3^4+3^5+3^6+3^7=3^4\left(1+3+3^2+3^3\right)=3^4.40\)

b) \(B=1+3+3^2+3^3+...+3^{99}\)

\(\Rightarrow B=\left(1+3+3^2+3^3\right)+3^4.\left(1+3+3^2+3^3\right)...+3^{96}.\left(1+3+3^2+3^3\right)\)

\(\Rightarrow B=40+3^4.40...+3^{96}.40\)

\(\Rightarrow B=40\left(1+3^4...+3^{96}\right)⋮40\)

\(\Rightarrow dpcm\)

Lồn bâm

Gâu gâu 

S = (1 - 3 + 3² - 3³) + 3⁴ . (1 - 3 + 3² - 3³) + .... + 3⁹⁶ . (1 - 3 + 3² - 3³)

S = (-20) + 3⁴ . (-20) +.... + 3⁹⁶ . (-20)

S = (-20) . (1 + 3⁴ +...+ 3⁹⁶) 

\(\Rightarrow\)S \(⋮\) 20 

(Ko bt có đúng ko)

*KO CHÉP MẠNG*