a) x= 2, 3...8

a: Ta có: \(36\le6^x\le1296\)

\(\Leftrightarrow2\le x\le4\)

hay \(x\in\left\{2;3;4\right\}\)

b: Ta có: \(100< 5^{2x-1}< 5^6\)

\(\Leftrightarrow2x-1\in\left\{3;5\right\}\)

\(\Leftrightarrow2x\in\left\{4;6\right\}\)

hay \(x\in\left\{2;3\right\}\)

a) 2n + 29 \(⋮\) 2n + 1

\(\Rightarrow\) 2n + 29 - (2n + 1) \(⋮\) 2n + 1

\(\Rightarrow\) 28  \(⋮\) 2n + 1

\(\Rightarrow\) 2n + 1 \(\in\) Ư(28) = {1 ; 2 ; 4 ; 7 ; 14 ; 28} , mà n \(\in\) N

\(\Rightarrow\) n \(\in\) {0 ; 3}

Vậy  n \(\in\) {0 ; 3}

b) 5n + 38 \(⋮\) n + 2

\(\Rightarrow\) 5n + 38 - 5(n + 2) \(⋮\) n + 2

\(\Rightarrow\) 28 \(⋮\) n + 2

\(\Rightarrow\) n + 2 \(\in\) Ư(28) = {1 ; 2 ; 4 ; 7 ; 14 ; 28}, mà n \(\in\) N

\(\Rightarrow\) n \(\in\) {0 ; 2 ; 5 ; 12 ; 26}

Vậy n \(\in\) {0 ; 2 ; 5 ; 12 ; 26}

\(b,A=\left(1+4+4^2\right)+\left(4^3+4^4+4^5\right)+...\left(4^{57}+4^{58}+4^{59}\right)\\ A=\left(1+4+4^2\right)+4^3\left(1+4+4^2\right)+...+4^{57}\left(1+4+4^2\right)\\ A=\left(1+4+4^2\right)\left(1+4^3+...+4^{57}\right)\\ A=21\left(1+4^3+...+4^{57}\right)⋮7\)

a: \(\Leftrightarrow2x+1\in\left\{1;3\right\}\)

hay \(x\in\left\{0;1\right\}\)

1. A

2. D

3. A

4. A

4A

3A

2D

1D

Bài 1:

a: \(\frac{x}{7}-\frac12=\frac{1}{y+1}\)

=>\(\frac{2x-7}{14}=\frac{1}{y+1}\)

=>(2x-7)(y+1)=14

mà 2x-7 lẻ

nên (2x-7;y+1)∈{(1;14);(-1;-14);(7;2);(-7;-2)}

=>(2x;y)∈{(8;13);(6;-15);(14;1);(0;-3)}

=>(x;y)∈{(4;13);(3;-15);(7;1);(0;-3)}

b: \(\frac{5}{x}-\frac{y}{4}=\frac18\)

=>\(\frac{5}{x}=\frac{y}{4}+\frac18=\frac{2y+1}{8}\)

=>x(2y+1)=40

mà 2y+1 lẻ

nên (x;2y+1)∈{(40;1);(-40;-1);(8;5);(-8;-5)}

=>(x;2y)∈{(40;0);(-40;-2);(8;4);(-8;-6)}

=>(x;y)∈{(40;0);(-40;-1);(8;2);(-8;-3)}

c: \(\frac{2}{y}-\frac{1}{x}=\frac{8}{xy}+1\)

=>\(\frac{2x-y}{xy}=\frac{8+xy}{xy}\)

=>xy+8=2x-y

=>xy-2x+y+8=0

=>x(y-2)+y-2+10=0

=>(x+1)(y-2)=-10

=>(x+1;y-2)∈{(1;-10);(-10;1);(-1;10);(10;-1);(2;-5);(-5;2);(-2;5);(5;-2)}

=>(x;y)∈{(0;-8);(-11;3);(-2;12);(9;1);(1;-3);(-6;4);(-3;7);(4;0)}

mà y<>0 và x<>0

nên (x;y)∈{(-11;3);(-2;12);(9;1);(1;-3);(-6;4);(-3;7)}

a: \(\Leftrightarrow2n+1\in\left\{1;3;9\right\}\)

hay \(n\in\left\{0;1;4\right\}\)

\(a,\Leftrightarrow10n+14⋮2n+1\\ \Leftrightarrow5\left(2n+1\right)+9⋮2n+1\\ \Leftrightarrow2n+1\inƯ\left(9\right)=\left\{1;3;9\right\}\\ \Leftrightarrow n\in\left\{0;1;4\right\}\)

Các cậu giải nhanh giúp mình với nhé!

điên à ai tính được

(C=\frac{38}{25}+\frac{9}{10}-\frac{11}{15}+\frac{13}{21}-\frac{15}{28}+\frac{17}{36}-...+\frac{197}{4851}-\frac{199}{4950})

bỏ tạm 38/25 ra 

=(9/10-11/15) +(13/21-15/28) +(197/4851-199/4950)

=1/6+1/12+1/20+...+1/49*50=2450

=3-2/2x3+4-3/3x4+5-4/4x5+....50-49/49x50

=1/2-1/3+1/3-...-1/50

1/2-1/50=12/25

=38/25+12/25=50/25

=2

kiểm tra lại đề giúp

cc