bạn vào link này nhé

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a,(x-3).(2y +1) =7

Vì x;y thuộc Z => x-3 và 2y+1 ltuộc Z

                       => x-3 và 2y+1 Thuộc Ư(7)

Ta có bảng:

Vậy..........................................................................................

b,(2x+1).(3y-2)=-55

Vì x;y là số nguyên=>2x+1;3y-2 là số nguyên

                              => 2x+1;3y-2 thuộc Ư(-55)

Vậy........................................................................

1: \(\frac{2x+6}{3x^2-x}:\frac{x^2+3x}{1-3x}\)

\(=\frac{2\left(x+3\right)}{x\left(3x-1\right)}\cdot\frac{-3x+1}{x\left(x+3\right)}\)

\(=\frac{2}{x}\cdot\frac{-\left(3x-1\right)}{x\left(3x-1\right)}=\frac{-2}{x^2}\)

2: \(\frac{x}{x-2y}+\frac{x}{x+2y}+\frac{4xy}{4y^2-x^2}\)

\(=\frac{x}{x-2y}+\frac{x}{x+2y}-\frac{4xy}{\left(x-2y\right)\left(x+2y\right)}\)

\(=\frac{x\left(x+2y\right)+x\left(x-2y\right)-4xy}{\left(x-2y\right)\left(x+2y\right)}=\frac{2x^2-4xy}{\left(x-2y\right)\left(x+2y\right)}\)

\(=\frac{2x\left(x-2y\right)}{\left(x-2y\right)\left(x+2y\right)}=\frac{2x}{x+2y}\)

3: \(\frac{1}{3x-2}-\frac{1}{3x+2}-\frac{3x-6}{4-9x^2}\)

\(=\frac{1}{3x-2}-\frac{1}{3x+2}+\frac{3x-6}{\left(3x-2\right)\left(3x+2\right)}\)

\(=\frac{3x+2-\left(3x-2\right)+3x-6}{\left(3x-2\right)\left(3x+2\right)}=\frac{3x+2-3x+2+3x-6}{\left(3x-2\right)\left(3x+2\right)}\)

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

4: \(\frac{x+3}{x+1}+\frac{2x-1}{x-1}+\frac{x+5}{x^2-1}\)

\(=\frac{x+3}{x+1}+\frac{2x-1}{x-1}+\frac{x+5}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{\left(x+3\right)\left(x-1\right)+\left(2x-1\right)\left(x+1\right)+x+5}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{x^2+2x-3+2x^2+2x-x-1+x+5}{\left(x-1\right)\left(x+1\right)}=\frac{3x^2+4x+1}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{\left(3x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{3x+1}{x-1}\)

= 6

kb nha

nhớ mk nhé

tk nha

=6 nha

a: =-3x^2y*x^2y+3x^2y*2xy

=-3x^4y^2+6x^3y^2

b: =x^3-x^2y+x^2y+y^2=x^3+y^2

c: =x*4x^3-x*5xy+2x*x

=4x^4-5x^2y+2x^2

d: =x^3+x^2y+2x^3+2xy

=3x^3+x^2y+2xy

a )

(x-3).(2y+1)=7 
(x-3).(2y+1)= 1.7 = (-1).(-7) 
Cứ cho x - 3 = 1 => x= 4 
2y + 1 = 7 => y = 3 
Tiếp x - 3 = 7 => x = 10 
2y + 1 = 1 => y = 0 
x-3 = -1 ...

1.tìm các số nguyên x và y sao cho:

(x-3).(2y+1)=7

Vì x;y là số nguyên =>x-3 ; 2y+1 là số nguyên

                               =>x-3  ; 2y+1 C Ư(7)

ta có bảng:

Vậy..............................................................................

2.tìm các số nguyên x và y sao cho:

xy+3x-2y=11

x.(y+3)-2y=11

x.(y+3)-y=11

x.(y+3)-(y+3)=11

(x-1)(y+3)=11

Vì x;y là số nguyên => x-1;y+3 là số nguyên

                               => x-1;y+3 Thuộc Ư(11)

Ta có bảng:

Vậy.......................................................................................

\(\frac{x+\frac{1}{3}}{1-x^2}+\frac{5}{3x-3}+\frac{1}{3x+3}=\frac{-\left(x+\frac{1}{3}\right)}{x^2-1}+\frac{5}{3.\left(x-1\right)}+\frac{1}{3.\left(x+1\right)}\)

\(=\frac{-x-\frac{1}{3}}{\left(x-1\right)\left(x+1\right)}+\frac{5}{3.\left(x-1\right)}+\frac{1}{3.\left(x+1\right)}=\frac{-3x-1}{3.\left(x-1\right)\left(x+1\right)}+\frac{5x+5}{3.\left(x-1\right)\left(x+1\right)}+\frac{x-1}{3.\left(x-1\right)\left(x+1\right)}\)

\(=\frac{-3x-1+5x+5+x-1}{3.\left(x-1\right)\left(x+1\right)}=\frac{3x+3}{3.\left(x-1\right)\left(x+1\right)}=\frac{3.\left(x+1\right)}{3.\left(x-1\right)\left(x+1\right)}=\frac{1}{x-1}\)

thôi thì tick ủng hộ thêm cái luôn đang rảnh tay đây

Bài 1 :                                             Bài giải

\(B=3^1+3^2+...+3^{2020}\)

\(B=\left(3^1+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2019}+3^{2020}\right)\)

\(B=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{2019}\left(1+3\right)\text{ }⋮\text{ }3\)

\(B=3^1+3^2+...+3^{2020}\)

\(B=\left(3^1+3^2+3^3+3^4\right)+...+\left(3^{2018}+3^{2019}+3^{2020}\right)\)

\(B=3\left(1+3+3^2\right)+...+3^{2018}\left(1+3+3^2\right)\)

\(B=3\cdot13+...+3^{2018}\cdot13\text{ }⋮\text{ }-13\)

Bài 2 :                                       Bài giải

\(xy+3x-2y=11\)

\(x\left(y+3\right)-2\left(y+3\right)+6=11\)

\(\left(y+3\right)\left(x-2\right)=5\)

\(\Rightarrow\text{ }y+3\text{ ; }x-2\text{ }\inƯ\left(5\right)\)

Ta có bảng :

Vậy \(\left(x\text{ ; }y\right)=\left(-3\text{ ; }-4\right)\text{ ; }\left(1\text{ ; }-8\right)\text{ ; }\left(3\text{ ; }2\right)\text{ ; }\left(7\text{ ; }-2\right)\)

\(B=3+3^2+3^3+3^4+...+3^{2020}\)

\(B=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+....+\left(3^{2018}+3^{2019}+3^{2020}\right)\)

\(\Leftrightarrow B=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+...+3^{2018}\left(1+3+3^2\right)\)

\(\Leftrightarrow B=3\cdot13+3^4\cdot13+....+3^{2018}\cdot13\)

\(\Leftrightarrow B=13\left(3+3^4+...+3^{2018}\right)\)

\(\Leftrightarrow B⋮13\left(đpcm\right)\)

Bạn @Fudo sai mất chỗ B chia hết cho 4 bạn viết nhầm thành chia hết cho 3

Đặt \(\frac{5x}{2}=\frac{7z}{3}=k\Rightarrow x=\frac{2k}{5};z=\frac{3k}{7}\)

Có \(x.z=47250\)

\(\Rightarrow\frac{2k}{5}.\frac{3k}{7}=47250\Rightarrow\frac{6k^2}{35}=47250\Rightarrow k^2=47250.35:6=275625\Rightarrow k=525\)

\(\Rightarrow x=525.2:5=210\)

\(z=525.3:7=225\)

Do \(3x=5y\Rightarrow210.3=5y\Rightarrow630=5y\Rightarrow y=630:5=126\)