VD: 1/2 là 1 phần 2 đó nha.

a: x/3-1/6=1/5

=>x/3=11/30

hay x=11/90

b: =>1/2x=2

hay x=4

c: =>2/3:x=-7-1/3=-22/3

=>x=-1/11

\(3n+1⋮2n-1\)

\(\Leftrightarrow2n-1\in\left\{1;-1;5;-5\right\}\)

\(\Leftrightarrow2n\in\left\{2;0;6;-4\right\}\)

hay \(n\in\left\{1;0;3;-2\right\}\)

a: (x-2)(y-3)=5

=>\(\left(x-2\right)\cdot\left(y-3\right)=1\cdot5=5\cdot1=\left(-1\right)\cdot\left(-5\right)=\left(-5\right)\cdot\left(-1\right)\)

=>\(\left(x-2;y-3\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(3;8\right);\left(7;4\right);\left(1;-2\right);\left(-3;2\right)\right\}\)

b: (2x-1)*(y-4)=-11

=>\(\left(2x-1\right)\cdot\left(y-4\right)=1\cdot\left(-11\right)=\left(-11\right)\cdot1=\left(-1\right)\cdot11=11\cdot\left(-1\right)\)

=>\(\left(2x-1;y-4\right)\in\left\{\left(1;-11\right);\left(-11;1\right);\left(-1;11\right);\left(11;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(1;-7\right);\left(-5;5\right);\left(0;15\right);\left(6;3\right)\right\}\)

c: xy-2x+y=3

=>\(x\left(y-2\right)+y-2=1\)

=>\(\left(x+1\right)\left(y-2\right)=1\)

=>\(\left(x+1\right)\cdot\left(y-2\right)=1\cdot1=\left(-1\right)\cdot\left(-1\right)\)

=>\(\left(x+1;y-2\right)\in\left\{\left(1;1\right);\left(-1;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(0;3\right);\left(-2;1\right)\right\}\)

\(n+2⋮n-3\)

\(n-3+5⋮n-3\)

\(5⋮n-3\Rightarrow n-3\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

Ta có: \(n+2=n-3+5\)

Để \(n+2⋮n-3\)\(\Rightarrow\)\(n-3+5⋮n-3\)mà  \(n-3⋮n-3\)

\(\Rightarrow\)\(5⋮n-3\)\(\Rightarrow\)\(n-3\inƯ\left(5\right)\in\left\{\pm1;\pm5\right\}\)

\(\Rightarrow\)\(n\in\left\{2;4;-2;8\right\}\)( các giá trị trên đều thoả mãn )

Vậy...........

a,n=0;2;6;12;14;....

b,n=1

c,n=0

d,n=2;4;6;10;12;...

\(2\cdot11^x=\left(3^2+2\right)^3:\left(5^3-2^5:2^3\right)\)

\(\Leftrightarrow11^x\cdot2=1331:121\)

\(\Leftrightarrow11^x\cdot2=11\)

=> Phương trình vô nghiệm

a, \(x,y\in Z\Rightarrow\left\{{}\begin{matrix}x-3,2y-6\in Z\\x-3,2y-6\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\end{matrix}\right.\)

Ta có bảng:

Vậy không có x,y thỏa mãn đề bài 

b, tương tự câu a

 \(c,xy-5x+2y=7\\ \Rightarrow x\left(y-5\right)+2y-10=-3\\ \Rightarrow x\left(y-5\right)+2\left(y-5\right)=-3\\ \Rightarrow\left(x+2\right)\left(y-5\right)=-3\)

Rồi làm tương tự câu a

\(d,xy-3x-4y=5\\ \Rightarrow x\left(y-3\right)-4y+12=17\\ \Rightarrow x\left(y-3\right)-4\left(y-3\right)=17\\ \Rightarrow\left(x-4\right)\left(y-3\right)=17\)

Rồi làm tương tự câu a

Câu b,

= 50x + (1 + 3 + 5 + ..... + 99) = 2550

= 50x + 2500 = 2550

= 50x = 2550 - 2500

= 50x = 50

= x = 50 : 50

= x = 1

refer

A) 2448:[119-[x-6]]=24

<=> 199 - ( x-6 ) = 2448 : 24 = 102

<=> x - 6 = 199 - 102 = 97

<=> x = 97 +6 = 103

B) (x+1)+(x+3)+(x+5)+......+(x+99) = 2550

(x+x+x+....+x)+(1+3+5+....+99) = 2550

50x + 2500 = 2550

50x = 2550 - 2500

50x = 50

x = 50 : 50

x = 1