\(1,=-\left(y^2+12y+36\right)=-y^2-12y-36\)

\(2,=-\left(16-8y+y^2\right)=-16+8y-y^2\)

\(3,=-\left(\dfrac{4}{9}+\dfrac{4}{3}x+x^2\right)=-\dfrac{4}{9}-\dfrac{4}{3}x-x^2\)

\(4,=-\left(x^2-3x+\dfrac{9}{4}\right)=-x^2+3x-\dfrac{9}{4}\)

\(5,-\left(2+3y\right)^2=-\left(4+12y+9y^2\right)=-4-12y-9y^2\)

.... mấy ý còn lại bn tự lm nhé, tương tự thhooi

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

2) \(-\left(4-y\right)^2=-y^2+8y-16\)

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

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

5) \(-\left(3y+2\right)^2=-9y^2-12y-4\)

6) \(-\left(2y-3\right)^2=-4y^2+12y-9\)

7) \(-\left(5x+2y\right)^2=-25x^2-20xy-4y^2\)

8) \(-\left(2x-\dfrac{3}{2}\right)^2=-4x^2+6x-\dfrac{9}{4}\)

a: \(9x^3y^2+3x^2y^2\)

\(=3x^2y^2\cdot3x+3x^2y^2\cdot1\)

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

b: \(x^2-2x+1-y^2\)

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

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

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

làm ơn giúp e vs

\(1,=\left(x-2\right)\left(5-y\right)\\ 2,=2\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(2x-2y-z\right)\\ 3,=5xy\left(x-2y\right)\\ 4,=3\left(x^2-2xy+y^2-4z^2\right)=3\left[\left(x-y\right)^2-4z^2\right]\\ =3\left(x-y-2z\right)\left(x-y+2z\right)\\ 5,=\left(x+2y\right)^2-16=\left(x+2y-4\right)\left(x+2y+4\right)\\ 6,=-\left(6x^2-3x-4x+2\right)=-\left(2x-1\right)\left(3x-2\right)\\ 7,=\left(2x+y\right)\left(2x+y+x\right)=\left(2x+y\right)\left(3x+y\right)\\ 8,=\left(x-y\right)\left(x+5\right)\\ 9,=\left(x+1\right)^2-y^2=\left(x-y+1\right)\left(x+y+1\right)\\ 10,=\left(x^2-9\right)x=x\left(x-3\right)\left(x+3\right)\\ 11,=\left(x-2\right)\left(y+1\right)\\ 12,=\left(x-3\right)\left(x^2-4\right)=\left(x-3\right)\left(x-2\right)\left(x+2\right)\\ 13,=3\left(x+y\right)-\left(x+y\right)^2=\left(x+y\right)\left(3-x-y\right)\)

1) Thay x = -8 , ta có ;

(-8) - 2 x (-8) - 3 x (-8) - 4 x (-8 ) - 5 x (-8) = 104

2) Thay x = -8 , ta có ;

(-8 + 1)-2(-8 + 1)-3( -8 + 1 )- 4( -8 + 1 ) - 5( -8 + 1 ) =91

Đúng thì tick ko đúng thì thôi 

1) \(\left(x+1\right)^2=x^2+2x+1\)

2) \(\left(2x+1\right)^2=4x^2+4x+1\)

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

4) \(\left(2x+3\right)^2=4x^2+12x+9\)

5) \(\left(3x+2y\right)^2=9x^2+12xy+4y^2\)

6) \(\left(2x^2+1\right)^2=4x^4+4x^2+1\)

7) \(\left(x^3+1\right)^2=x^6+2x^3+1\)

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

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

10) \(\left(\dfrac{1}{2}x+\dfrac{1}{3}y\right)^2=\dfrac{1}{4}x^2+\dfrac{1}{3}xy+\dfrac{1}{9}y^2\)

2)3x²-6xy+3y²=3(x²-2xy+y²)=3(x-y)²

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

5)(x+y)³-(x-y)³=[(x+y)-(x-y)][(x+y)²+(x+y)(x-y)+(x-y)²]=(x+y-x+y)(x²+2xy+y²+x²-y²+x²-2xy+y²)=2y(3x²+y²)

6)3x²-5x+2=3x²-2x-3x+2=(3x²-3x)-(2x-2)=3x(x-1)-2(x-1)=(x-1)(3x-2)

1) x-2x-3x-4x-5x=-13x thay x=-8 => -13x=(-8).(-13)=104

2)(x+1)-2(x+1)-3(x+1)-4(x+1)-5(x+1)=-13(x+1) thay x=-8 => -13(x+1)=(-13).(-7)-91

a,  x²+2xy+y²+2x+2y-15

<=> (x+y )²+2(x+y)+1-16

Đặt x+y =a

<=> a²+2a+1-4²

<=> (a+1)²-4²

<=> (a+5)(a-3) =>( x+y+5)(x+y-3)

b, x²-4xy+4y²-2x-4y-35

<=> (x-2y)²-2(x-2y)+1-36

Đặt (x-2y)  =b 

=> b²-2b+1-6²

<=> (b-1)²-6²

<=> (b-7)(b+5)=> (x-2y-7)(x-2y+5)

c, 

a,A= x^2+2xy+y^2+2x+2y-15

= (x+y)^2+(x+y)-15

Đặt x+y=a, ta có:

A=a^2+2a-15

  =a^2+2a+1-16

  =(a+1)^2-4^2

  =(a+1+4)(a+1-4)

  =(a+5)(a-3)

Thay a=x+y, ta có: A=(x+y+5)(x+y-3).

b,B= x^2 - 4xy+4y^2-2x-4y-35

   Hình như là sai đề đó bạn. Phải là x^2 - 4xy+4y^2-2x+4y-35 hoặc x^2 - 4xy+4y^2+2x-4y-35 hoặc x^2 + 4xy+4y^2-2x-4y-35 mới đúng đó bạn. Bạn xem lại đi nha.

c,C=6x^4 - 5x^3+8x^2-5x+6

C= x^2(6x^2-5x+8-5/x+6/x^2)

  =x^2(6(x^2+2+1/x^2)-5(x+1/x)-4)

  =x^2(6(x+1/x)^2-5(x+1/x)-4)

Đặt x+1/x=a, ta có:

C=x^2(6a^2-5a-4)

  =x^2(6a^2+3a-8a-4)

  =x^2(2a+1)(3a-4)

Thay a=x+1/x vào là được bạn nhé.