a) 5xy ( x - y ) - 2x + 2y

= 5xy ( x - y ) - 2 ( x - y )

= ( x - y ) ( 5xy - 2 )

b) 6x-2y-x(y-3x)

= 2 ( y - 3x ) - x ( y - 3x )

= ( y - 3x ( ( 2 - x )

c)  x² + 4x - xy-4y

= x ( x + 4 ) - y ( x + 4 )

( x + 4 ) ( x - y )

d) 3xy + 2z - 6y - xz 

= ( 3xy - 6y ) + ( 2z - xz )

= 3y ( x - 2 ) + z ( x - 2 )

= ( x - 2 ) ( 3y + z )

a,5xy(x-y)-2x+2y=5xy(x-y)-2(x-y)=(x-y)(5xy-2)

b,6x-2y-x(y-3x)=-2(y-3x)-x(y-3x)=(y-3x)(-2-x)

c,x^2+4x-xy-4y=x(x+4)-y(x+4)=(x+4)(x-y)

d,3xy+2z-6y-xz=(3xy-6y)+(2z-xz)=3y(x-2)+z(2-x)=3y(x-2)-z(x-2)=(x-2)(3y-z)

11)

a,4-9x^2=0

(2-3x)(2+3x)=0

2-3x=0=>x=2/3 hoặc 2+3x=0=>x=-2/3

b,x^2 +x+1/4=0

(x+1/2)^2 =0

x+1/2=0

x=-1/2

c,2x(x-3)+(x-3)=0

(x-3)(2x+1)=0

x-3=0=>x=3 hoặc 2x+1=0=>x=-1/2

d,3x(x-4)-x+4=0

3x(x-4)-(x-4)=0

(x-4)(3x-1)=0

x-4=0=>x=4 hoặc 3x-1=0=>x=1/3

e,x^3-1/9x=0

x(x^2-1/9)=0

x(x+1/3)(x-1/3)=0

x=0 hoặc x+1/3=0=>x=-1/3 hoặc x-1/3=0=>x=1/3

f,(3x-y)^2-(x-y)^2 =0

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

2x(4x-2y)=0

4x(2x-y)=0

x=0hoặc 2x-y=0=>x=y/2

thu thủy học lớp 9 chưa có long lone

Sửa đề: Tìm cặp \(x,y\in Z\) thỏa mãn \(x^2+3xy+2y^2+3x+6y-4=0\).

\(x^2+3xy+2y^2+3x+6y-4=0\)

\(\Leftrightarrow x^2+2xy+xy+2y^2+3x+6y=4\)

\(\Leftrightarrow\left(x^2+2xy\right)+\left(xy+2y^2\right)+\left(3x+6y\right)=4\)

\(\Leftrightarrow x\left(x+2y\right)+y\left(x+2y\right)+3\left(x+2y\right)=4\)

\(\Leftrightarrow\left(x+2y\right)\left(x+y+3\right)=4\)

Vì \(x,y\in Z\Rightarrow\left(x+2y\right)\left(x+y+3\right)\in Z\)

Trường hợp 1: \(\left\{{}\begin{matrix}x+2y=1\\x+y+3=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\) (thỏa mãn)

Trường hợp 2: \(\left\{{}\begin{matrix}x+2y=4\\x+y+3=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-8\\y=6\end{matrix}\right.\) (thỏa mãn)

Trường hợp 3: \(\left\{{}\begin{matrix}x+2y=2\\x+y+3=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=3\end{matrix}\right.\) (thỏa mãn)

Trường hợp 4: \(\left\{{}\begin{matrix}x+2y=-2\\x+y+3=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-8\\y=3\end{matrix}\right.\) (thỏa mãn)

Vậy: \(\left(x,y\right)=\left[\left(1;0\right),\left(-8;6\right),\left(-4;3\right),\left(-8;3\right)\right]\)

a) (x - 1)(y + 2) = 7

=> (x-1)\(\in\)Ư(7)={1;7 ; -1; -7}

Nếu x-1= 1 thì x= 1+1 => x=2

Nếu x-1= 7 thì x= 7+1=> x=8

Nếu x-1= -1 thì x= -1+1 => x= 0

Nếu x-1= -7 thì x= -7+1 => x= -6

Sau đó bn thay x r tính xong đó tự tìm y+2 nha

Bài 10 :

Câu a :

\(5xy\left(x-y\right)-2x+2y\)

\(=5xy\left(x-y\right)-2\left(x-y\right)\)

\(=\left(x-y\right)\left(5xy-2\right)\)

Câu b :

\(6x-2y-x\left(y-3x\right)\)

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

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

Câu c :

\(x^2+4x-xy-4y\)

\(=x\left(x+4\right)-y\left(x+4\right)\)

\(=\left(x+4\right)\left(x-y\right)\)

Câu d :

\(3xy+2z-6y-xz\)

\(=\left(3xy-6y\right)-\left(xz-2z\right)\)

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

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

Bài 11 :

Câu a :

\(4-9x^2=0\)

\(\Leftrightarrow\left(2-3x\right)\left(2+3x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2-3x=0\\2+3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)

Vậy ........................

Câu b :

\(x^2+x+\dfrac{1}{4}=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=0\)

\(\Leftrightarrow x+\dfrac{1}{2}=0\)

\(\Leftrightarrow x=-\dfrac{1}{2}\)

Vậy........................

Câu c :

\(2x\left(x-3\right)+\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{2}\end{matrix}\right.\)

Vậy..................

Câu d :

\(3x\left(x-4\right)-x+4=0\)

\(\Leftrightarrow3x\left(x-4\right)-\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy................................

Câu e :

\(x^3-\dfrac{1}{9}x=0\)

\(\Leftrightarrow x\left(x^2-\dfrac{1}{9}\right)=0\)

\(\Leftrightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-\dfrac{1}{3}=0\\x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)

Vậy........................

Câu f :

\(\left(3x-y\right)^2-\left(x-y\right)^2=0\)

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

\(\Leftrightarrow2x\left(4x-2y\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\4x-2y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

Vậy..........................

Đặt \(\hept{\begin{cases}\sqrt{4x^2+3xy-7y^2}=a\\\sqrt{3x^2-2xy-y^2}=b\end{cases}}\)

\(\Rightarrow a^2-b^2=x^2+5xy-6y^2\)

Từ đó ta có pt (1)

\(\Leftrightarrow a-b+4\left(a^2-b^2\right)=0\)

\(\Leftrightarrow\left(a-b\right)\left(1+4a+4b\right)=0\)

\(\Leftrightarrow\)a = b

\(\Leftrightarrow x^2+5xy-6y^2=0\)

\(\Leftrightarrow x^2-2xy+y^2+7xy-7y^2=0\)

\(\Leftrightarrow\left(x-y\right)\left(x+6y\right)=0\)

Tới đây thì bài toán đơn giản rồi bạn làm tiếp nhé

a) 9x² - 36

=(3x)²-6²

=(3x-6)(3x+6)

=4(x-3)(x+3)

b) 2x³y-4x²y²+2xy³

=2xy(x²-2xy+y²)

=2xy(x-y)²

c) ab - b²-a+b

=ab-a-b²+b

=(ab-a)-(b²-b)

=a(b-1)-b(b-1)

=(b-1)(a-b)

P/s đùng để ý đến câu trả lời của mình

\(2x^2+2y^2+3x-6y=5xy-7\)

\(\Leftrightarrow2x^2+2y^2+3x-6y-5xy=-7\)

\(\Leftrightarrow2x^2-4xy+2y^2-xy+3x-6y=-7\)

\(\Leftrightarrow2x\left(x-2y\right)-y\left(x-2y\right)+3\left(x-2y\right)=-7\)

\(\Leftrightarrow\left(2x-y+3\right)\left(x-2y\right)=-7\)

vì x,y nguyên nên \(\hept{\begin{cases}2x-y+3\\x-2y\end{cases}\in Z}\)

Ta có : -7 = ( -7 ) . 1 = (-1 ) . 7

Tới đây bạn tự làm nhé