ta có:\(x^2y+2x^2y+3x^2y+...+nx^2y=210x^2y\)

  \(x^2y\left(1+2+3+4+...+n\right)=210x^2y\)

\(1+2+3+...+n=210x^2y:\left(x^2y\right)\)

\(1+2+3+...+n=210\)

\(\frac{\left(n-1\right):1+1}{2}.\left(n+1\right)=210\)

\(n\left(n+1\right):2=210\)

\(n.\left(n+1\right)=420=20.21\)

vậy n=20

1+2+3+...+n = 210

n(n+1):2=210

n(n+1)=420 =20.21

n =20

n=20 nha rồi mình kik hộ cho

\(x^2y+2x^2y+3x^2y+....+nx^2y=210x^2y\)

\(x^2y\left(1+2+3+...+n\right)=210x^2y\)

\(1+2+3+...+n=210\)

\(\frac{n\left(n+1\right)}{2}=210\)

\(n\left(n+1\right)=420\)

\(n\left(n+1\right)=20.21\)

\(\Rightarrow n=20\)

x^2.y+2x^2.y+3x^2.y+...+n.x^2y=210x^2.y

x^2.y(1+2+3+..+n)=210x^2.y

1+2+3+..+n=210

=>(n+1)(n-1+1)/2=210

(n+1)n/2=210

(n+1)n=420=21.20

=>n+1=21

n=20

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

\(=\frac{\left(n+1\right)+\left(n+1\right)+...+\left(n+1\right)}{2}.\left(có.n.nhóm.n+1\right)\)

\(=\frac{n\left(n+1\right)}{2}.\)

tìm n nhak các bạn

x^2y*(1+2+3+...+n)=210x^2y

1+2+3+...+n=210

[(n-1):1+1]:2*(n+1)=210

n*(n+1):2=210

n*(n+1)=420=20*21

suy ra n=20

k mik nhe

ta có:

x^2y(1+2+3+4+...+n)=210x^2y

1+2+3+4+...+n=210x^2y/x^2y

1+2+3+4+...+n=210

(n-1):1+1/2.(n+1)=210

n(n+1)/2=210

n(n+1)=420=20.21

Vậy n=20

ko có yêu cầu đề thì sao làm được ?

bạn coi lại đề đi nhá, không tìm được n

1. Đặt A = 3x + 1

=> 2A = 6x + 2 = 3(2x - 1) + 5

Để A \(⋮\)2x - 1 <=> 2A \(⋮\)2x - 1 

              <=> 3(2x - 1) + 5 \(⋮\) 2x - 1

          <=> 5 \(⋮\)2x - 1 (vì 3(2x - 1) \(⋮\)2x - 1)

         <=> 2x - 1 \(\in\)Ư(5) = {1; 5}

   Với: +) 2x - 1 = 1 => 2x = 2 => x = 1

+) 2x - 1 = 5 => 2x = 6 => x = 3

Vậy ...

Sao b chỉ trả lời có 1 câu thế 

a) \(xy+x+2y=5\\ \Rightarrow y\left(x+2\right)+x+2=5+2\\ \Rightarrow\left(x+2\right)\left(y+1\right)=7\)

Ta xét bảng:

Vậy \(\left(x;y\right)\in\left\{\left(-1;6\right);\left(5;0\right);\left(-3;-8\right);\left(-9;-2\right)\right\}\)

b) \(xy-3x-y=0\\ \Rightarrow x\left(y-3\right)-y+3=3\\ \Rightarrow\left(y-3\right)\left(x-1\right)=3\)

Ta xét bảng:

Vậy \(\left(x;y\right)\in\left\{\left(2;6\right);\left(4;4\right);\left(0;0\right);\left(-2;2\right)\right\}\)

c) \(xy+2x+2y=-16\\ \Rightarrow x\left(y+2\right)+2y+4=-12\\ \Rightarrow\left(y+2\right)\left(x+2\right)=-12\)

Ta xét bảng:

Vậy \(\left(x;y\right)\in\left\{\left(-1;-14\right);\left(0;-8\right);\left(1;-6\right);\left(2;-5\right);\left(4;-4\right);\left(10;-3\right);\left(-3;10\right);\left(-4;4\right);\left(-5;2\right);\left(-6;1\right);\left(-8;0\right);\left(-14;-1\right)\right\}\)