đề sai rùi phải là : \(36\left(x-y\right)^2-25\left(2x-1\right)^2\)

\(=>\left[6\left(x-y\right)\right]^2-\left[5\left(2x-1\right)\right]^2=\left[6\left(x-y\right)-5\left(2x-1\right)\right]\left[6\left(x-y\right)+5\left(2x-1\right)\right]\)

\(=>\left(6x-6y-10x+5\right)\left(6x-6y+10x-5\right)=\left(5-4x-6y\right)\left(16x-6y-5\right)\)

Áp dụng HDT : x^2 -y^2 =(x-y) (x+y)

Ủng hộ = 1 cái t i c k nha cảm ơn

36(x-y)²-25(2x-y)²

= 36(x-y)² - 100(x-y)²

=(36-100)(x-y)²

= -64(x-y)²

\(9\left(x-3y\right)^2-25\left(2x+y\right)^2\)

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

\(=\left(3x-9y\right)^2-\left(10x+5y\right)^2\)

\(=\left[3x-9y+10x+5y\right]\left[3x-9y-\left(10x+5y\right)\right]\)

\(=\left(13x-4y\right)\left(-7x-14y\right)\)

\(=-7\left(x+2y\right)\left(13x-4y\right)\)

9(x - 3y)² - 25(2x + y)²

= 3².(x - 3y)² - 5².(2x + y)²

= (3x - 9y)² - (10x + 5y)²

= (3x - 9y - 10x - 5y)(3x - 9y + 10x + 5y)

= (-7x - 14y)(13x - 4y)

= -7(x + 2y)(13x - 4y)

a,\(A=x^3-2x^2=x^2\left(x-2\right)\)

\(b,\) \(y^2+2y-x^2+1=-x^2-xy-x+xy+y^2+y+x+y+1\)

\(=-x\left(x+y+1\right)+y\left(x+y+1\right)+\left(x+y+1\right)\)

\(=\left(-x+y+1\right)\left(x+y+1\right)\)

\(c,\) \(\left(x+1\right)^2-25=\left(x+1+5\right)\left(x+1-5\right)=\left(x+6\right)\left(x-4\right)\)

a,x³−2x²=x²(x−2)

b,y²+2y-x²+1=(y+1)²-x²=(y+x+1)(y-+x+1)

c, (x+1)²−25=(x+1+5)(x+1−5)=(x+6)(x−4)

A, \(x^3-4x^2-9x+36=x^2\left(x-4\right)-9\left(x-4\right)\)

   \(=\left(x-4\right)\left(x^2-9\right)=\left(x-4\right)\left(x-3\right)\left(x+3\right)\)

\(b,x^2-y^2-2x-2y=\left(x-y\right)\left(x+y\right)-2\left(x+y\right)\)

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

Bài 4:

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

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

\(x^2-12x+36=\left(x-6\right)^2\)

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

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

     y\(^2\) - ( x\(^2\) - 2x + 1 )

⇔ y\(^2\) - ( x - 1 )\(^2\)

⇔ ( y - x - 1 ) ( y + x - 1 )

a) \(xy+y^2-x-y\)

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

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

\(=\left(y-1\right)\left(x+y\right)\)

a) xy +y² - x-y

y(x+y) -(x+y)

(x+y)(y-1)

c) x² - 4x +3

 x² -3x - x - 3

x(x-3) -(x-3)

(x-3)(x-1)

câu 2

ĐỂ phép chia hết thì m+12 = 0 => m = -12

có thể đúng cũng có thể sai ,có j sai hoặc ko đúng ib mk nhé

\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)

1.A

2.C

3.B

4.C

a

c

b

c

(x-1)y^2-4(x-1)y