\(6x-y+3xy=15\)

\(\Leftrightarrow\left(6x+3xy\right)-y=15\)

\(\Leftrightarrow3x\left(2+y\right)-y=13+2\)

\(\Leftrightarrow3x\left(2+y\right)-y-2=13\)

\(\Leftrightarrow3x\left(2+y\right)-\left(y+2\right)=13\)

\(\Leftrightarrow\left(2+y\right)\left(3x-1\right)=13\)

\(\Rightarrow\left(2+y\right);\left(3x-1\right)\inƯ\left(13\right)=\left\{\pm1;\pm13\right\}\)

Xét từng trường hợp :

TH1 : \(\hept{\begin{cases}2+y=1\\3x-1=13\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-1\\x=\frac{14}{3}\end{cases}}\)

TH2:\(\hept{\begin{cases}2+y=13\\3x-1=1\end{cases}\Leftrightarrow\hept{\begin{cases}y=11\\x=\frac{2}{3}\end{cases}}}\)

TH3:\(\hept{\begin{cases}2+y=-1\\3x-1=-13\end{cases}\Leftrightarrow\hept{\begin{cases}y=-3\\x=-4\end{cases}}}\)

TH4:\(\hept{\begin{cases}2+y=-13\\3x-1=-1\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-15\\x=0\end{cases}}\)

Vậy................

/hok chắc/

~ học tốt~

3xy - 6x + y + 3 = 0

=> 3xy + y - 6x = -3

=> y(3x + 1) = 6x - 3

=> 6x - 3 chia hết cho 3x + 1

Mà 3x + 1 chia hết cho 3x + 1 => 6x + 2 chia hết cho 3x + 1

Do đó 6x + 2 - 6x + 3 chia hết cho 3x + 1

=> 5 chia hết cho 3x + 1

=> 3x + 1 thuộc {1; -1; 5; -5}

=> 3x thuộc {0; -2; 4; -6}

=> x thuộc {0; -2} (Vì x thuộc Z)

<=>(3x+1)y-6x+3=0

=>(3x+1)y-6x-0+3=0

=>3x+1=0

=>3x=-1

=>3(y-2)=0

=>3y=3.2( rut gon 3)

=>y=2

3xy-6x+y+3=0

=>3xy+y-6x=-3

=>y(3x+1)-6x=-3

=>6x-3 chia hết cho 3x+1

mà 3x+1 chia hết cho 3x+1 =>6x + 2 chia hết cho 3x+1

do đó 6x+2-6x+3 chia hết cho 3x+1

=>5 chia hết cho 3x+1

=>mx+1 thuộc {1;-1;5;-5}

=>3x thuộc {0;-2;4;-6}

=> x thuộc {0;-2} (vì x thuộc Z)

bài 5:

1: \(\dfrac{12x^3y^2}{18xy^5}=\dfrac{12x^3y^2:6xy^2}{18xy^5:6xy^2}=\dfrac{2x^2}{3y^3}\)

2: \(\dfrac{10xy-5x^2}{2x^2-8y^2}=\dfrac{5x\cdot2y-5x\cdot x}{2\left(x^2-4y^2\right)}\)

\(=\dfrac{5x\left(2y-x\right)}{-2\left(x+2y\right)\left(2y-x\right)}=\dfrac{-5x}{2\left(x+2y\right)}\)

3: \(\dfrac{x^2-xy-x+y}{x^2+xy-x-y}\)

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

\(=\dfrac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}=\dfrac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}=\dfrac{x-y}{x+y}\)

4: \(\dfrac{\left(x+1\right)\left(x^2-2x+1\right)}{\left(6x^2-6\right)\left(x^3-1\right)}\)

\(=\dfrac{\left(x+1\right)\left(x-1\right)^2}{6\left(x^2-1\right)\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\dfrac{\left(x+1\right)\left(x-1\right)}{6\left(x-1\right)\left(x+1\right)\cdot\left(x^2+x+1\right)}\)

\(=\dfrac{1}{6\left(x^2+x+1\right)}\)

5: \(\dfrac{2x^2-7x+3}{1-4x^2}\)

\(=-\dfrac{2x^2-7x+3}{4x^2-1}\)

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

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

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

Bài 3:

1: \(9x^3-xy^2\)

\(=x\cdot9x^2-x\cdot y^2\)

\(=x\left(9x^2-y^2\right)\)

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

2: \(x^2-3xy-6x+18y\)

\(=\left(x^2-3xy\right)-\left(6x-18y\right)\)

\(=x\left(x-3y\right)-6\left(x-3y\right)\)

\(=\left(x-3y\right)\left(x-6\right)\)

3: \(x^2-3xy-6x+18y\)

\(=\left(x^2-3xy\right)-\left(6x-18y\right)\)

\(=x\left(x-3y\right)-6\left(x-3y\right)\)

\(=\left(x-3y\right)\left(x-6\right)\)

4: \(6xy-x^2+36-9y^2\)

\(=36-\left(x^2-6xy+9y^2\right)\)

\(=36-\left(x-3y\right)^2\)

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

5: \(x^4-6x^2+5\)

\(=x^4-x^2-5x^2+5\)

\(=x^2\left(x^2-1\right)-5\left(x^2-1\right)\)

\(=\left(x^2-5\right)\left(x^2-1\right)\)

\(=\left(x^2-5\right)\left(x-1\right)\left(x+1\right)\)

6: \(9x^2-6x-y^2+2y\)

\(=\left(9x^2-y^2\right)-\left(6x-2y\right)\)

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

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

câu a :

a, suy ra x-7 và x+3 khác dấu

mà x-7 < x+3

suy ra x-7 <0     ;  x+3 > 0

suy ra x <7        ; x > -3

suy ra 7 > x > -3

vậy x = -2 ; -1 ; ... ; 6

nha rồi tui giải câu b cho  

mình chỉ làm ddcj phần a thôi 

Để (x-7)(x+3)=0 thì

x-7=0 hoặc x+3=0 

suy ra: x=7 hoặc x=3

nên đễ(x-7)(x-3)<0 thì x<3

ủng hộ nha

mình rồi đó trả lời phần b cho mình đc ko