Ta có: \(xy+3x-7y=21\)

\(\Rightarrow xy+3x-7y-21=0\)

\(\Rightarrow\left(xy+3x\right)-\left(7y+21\right)=0\)

\(\Rightarrow x\left(y+3\right)-7\left(y+3\right)=0\)

\(\Rightarrow\left(y+3\right)\left(x-7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}y+3=0\\x-7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=-3\left(tm\right)\\x=7\left(tm\right)\end{matrix}\right.\)

Vậy: \(x=7;y=-3\).

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

áp dụng tính chất của DTSBN, ta được

\(\dfrac{x}{21}=\dfrac{y}{14}=\dfrac{z}{10}=\dfrac{3x-7y+5z}{3\cdot21-7\cdot14+5\cdot10}=\dfrac{-30}{15}=-2\)

=>x=-42; y=-28; z=-20

Ta có: `x/21=y/14=z/10 -> (3x)/63=(7y)/98=(5z)/50`

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

`(3x)/63=(7y)/98=(5z)/50=(3x-7y+5z)/(63-98+50)=-30/15=-2`

`-> x/21=y/14=z/10=-2`

`-> x=21*(-2)=-42, y=14*(-2)=-28, z=10*(-2)=-20`

x.y + 3x - 2y = 11

=> x.y + 3x - 2y - 11 = 0 

=> x.y + 3x - 2y - 6 - 5 = 0 

=> x.(y+3) -2(y+3)- 5 = 0 

=> (x-2)(y+3) = 5 

tự tính nhé đến đây dễ rồi

x.y + 3x - 7y = 21 

=> x.y + 3x -7y - 21 = 0 

=> x( y+3) - 7(y+3) = 0 

=> (x-7)(y+3) = 0 

=>\(\orbr{\begin{cases}x-7=0\\y+3=0\end{cases}}\)

=> x= 7 hoặc y = -3

a, (x;y) = (2;-3)

b, (x;y) = (7;-3)

Ez game

a, xy + 3x - 7y = 21

    x( y + 3) - 7y - 21 = 0

    x(y + 3)- 7( y+ 3) = 0

   ( x - 7)( y + 3)= 0 

=> x - 7  = 0 hoặc y + 3 = 0

=> x = 7 hoặc y = -3

lâlalakakkssksk

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\(\dfrac{x}{-3}=\dfrac{y}{5}\)⇒\(\dfrac{x}{-6}=\dfrac{y}{10}\)

\(\dfrac{y}{2}=\dfrac{z}{7}\)⇒\(\dfrac{y}{10}=\dfrac{z}{35}\)

⇒\(\dfrac{x}{-6}=\dfrac{y}{10}=\dfrac{z}{35}\)

⇒\(\dfrac{2x}{-12}=\dfrac{3y}{30}=\dfrac{z}{35}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{2x}{-12}=\dfrac{3y}{30}=\dfrac{z}{35}=\dfrac{2x-3y+z}{-12-30+35}=\dfrac{42}{-7}=-6\)

⇒\(\left\{{}\begin{matrix}x=-6.-6=36\\y=-6.10=-60\\z=-6.35=-210\end{matrix}\right.\)

\(a,\dfrac{x}{-3}=\dfrac{y}{5}\Rightarrow\dfrac{x}{-6}=\dfrac{y}{10};\dfrac{y}{2}=\dfrac{z}{7}\Rightarrow\dfrac{y}{10}=\dfrac{z}{35}\\ \Rightarrow\dfrac{x}{-6}=\dfrac{y}{10}=\dfrac{z}{35}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{-6}=\dfrac{y}{10}=\dfrac{z}{35}=\dfrac{2x}{-12}=\dfrac{3y}{30}=\dfrac{2x-3y+z}{-12-30+35}=\dfrac{42}{-7}=-6\\ \Rightarrow\left\{{}\begin{matrix}x=36\\y=-60\\z=-210\end{matrix}\right.\)

\(b,6x=4y=z\Rightarrow\dfrac{6x}{12}=\dfrac{4y}{12}=\dfrac{z}{12}\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{12}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{12}=\dfrac{2x}{4}=\dfrac{3y}{9}=\dfrac{2x-3y+z}{4-9+12}=\dfrac{42}{7}=6\\ \Rightarrow\left\{{}\begin{matrix}x=12\\y=18\\z=72\end{matrix}\right.\)

\(c,x=-2y\Rightarrow\dfrac{x}{-2}=y\Rightarrow\dfrac{x}{-4}=\dfrac{y}{2}\\ 7y=2z\Rightarrow\dfrac{y}{2}=\dfrac{z}{7}\\ \Rightarrow\dfrac{x}{-4}=\dfrac{y}{2}=\dfrac{z}{7}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{-4}=\dfrac{y}{2}=\dfrac{z}{7}=\dfrac{2x}{-8}=\dfrac{3y}{6}=\dfrac{2x-3y+z}{-8+6+7}=\dfrac{42}{5}\\ \Rightarrow\left\{{}\begin{matrix}x=-\dfrac{168}{5}\\y=\dfrac{84}{5}\\z=\dfrac{294}{5}\end{matrix}\right.\)

Ta có 

xy + 3x - 7y = xy + 21 - 21 + 3x - 7y = xy + 3x + 21 - 21 - 7y

                                                      = x ( y + 3 ) + 21 - 7 ( 3 + y )

                                                      = x ( y + 3 ) - 7 ( 3 + y ) + 21

                                                      = ( x - 7 ) ( y + 3 ) + 21 = 21

                                                      = ( x - 7 ) ( y + 3 ) = 0

=> Nếu x - 7 = 0 => x = 7 ; y \(\in\) Z

=> Nếu y + 3 = 0 => y = -3 ; x \(\in\) Z

=> Nếu x - 7 = 0 và y + 3 = 0 thì x = 7 ; y = -3

xy + 3x - 7y = 21

=> x.(y + 3) - 7y - 21 = 0

=> x.(y + 3) - 7.(y + 3) = 0

=> (y + 3).(x - 7) = 0

=> y + 3 = 0; x - 7 = 0

=> y = -3; x = 7

tim x, y thuoc Z thoa xy = x + y