\(=2\left(x^2-y^2\right)-6\left(x+y\right)=2\left(x-y\right)\left(x+y\right)-6\left(x+y\right)=\left(x+y\right)\left(2x-2y-6\right)\)                                                                                    Đảm bảo chuẩn ko cần chỉnh (•••

  check mk nhá
 

2X²-2Y²-6X-6Y

=2(X²-Y²) +6(X-Y)

=2(X-Y)(X+Y)+3.2(X-Y)

=2(X-Y)(X+Y+3X-3Y)

=2(X-Y)(4X-2Y)

=4(X-Y)(2X-Y)

\(2x^2-2y^2-6x-6y\)

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

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

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

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

\(=\left(x^2+5xy+4y^2\right)\left(x^2+5xy+6y^2\right)+y^4\)

\(=\left(x^2+5xy\right)^2+10y^2\left(x^2+5xy\right)+24y^4+y^4\)

\(=\left(x^2+5xy+5y^2\right)^2\)

B1 :

a, B = (x+1)^2+(y-2)^2 = (99+1)^2+(102-2)^2 =  100^2+100^2 = 20000

b, = (2x^2+16x+32)-2y^2

   = 2.(x+4)^2-2y^2

   = 2.[(x+4)^2-y^2] = 2.(x+4-y).(x+4+y)

c, <=> (x^2-3x)+(2x-6) = 0

<=> (x-3).(x+2) = 0

<=> x-3=0 hoặc x+2=0

<=> x=3 hoặc x=-2

B2 :

P = (3-x).(x+3)/x.(x-3) = -(x+3)/x = -x-3/x

k mk nha

Bai 1

a)B=(x+1)²+(y-2)²

     Voi x=99,y=102

=>B= 100²+100²

       =20000

b)\(2x^2-2y^2+16x+32\)

=\(2\left[\left(x^2+8x+16\right)-y^2\right]\)

=\(2\left[\left(x+4\right)^2-y^2\right]\)

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

c)\(x^2-3x+2x-6=0\)

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

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

=>x=-2;3

Bai 2

\(P=\frac{9-x^2}{x^2-3x}\)

    =\(-\frac{x^2-9}{x\left(x-3\right)}\)

   =\(-\frac{\left(x-3\right)\left(x+3\right)}{x\left(x-3\right)}\)

=\(\frac{-x-3}{x}\)

bafi 1: a) \(B=x^2+2x+1+y^2-4y+4\)

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

khi \(x=99\); \(y=102\)ta co:

\(B=\left(99+1\right)^2+\left(102-2\right)^2\)

\(B=100^2+100^2\)

\(B=10000+10000\)

\(B=20000\)

c) \(x^2-3x+2x-6=0\)

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

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

\(\Rightarrow\orbr{\begin{cases}x+2=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-2\\x=3\end{cases}}\)

vay \(\orbr{\begin{cases}x=-2\\x=3\end{cases}}\)

bai 2:

\(P=\frac{9-x^2}{x^2-3x}\)

\(P=\frac{\left(3-x\right)\left(x+3\right)}{x\left(x-3\right)}\)

\(P=\frac{-\left(x+3\right)}{x}\)\(\left(x\ne0\right)\)

\(P=\frac{-x-3}{x}\)

vay \(P=\frac{-x-3}{x}\)

c) \(-\frac{x^4}{4}+2x^2y^3-4y^6=-\left(\frac{x^4}{4}-2x^2y^3+4y^6\right)=-\left[\left(\frac{x^2}{2}\right)^2-2.\frac{x^2}{2}.2y^3+\left(2y^3\right)^2\right]=-\left(\frac{x^2}{2}-2y^3\right)\)

a)

\(a^2+b^2+2ab+2a+2b+1\)

\(=(a^2+2ab+b^2)+(2a+2b)+1\)

\(=(a+b)^2+2(a+b)+1^2=(a+b+1)^2\)

b)

\(3x(x-2y)+6y(2y-x)\)

\(=3x(x-2y)-6y(x-2y)=(3x-6y)(x-2y)=3(x-2y)(x-2y)\)

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

c)

\(16xy+4y^2-9+16x^2\)

\(=(16x^2+16xy+4y^2)-9\)

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

d)

\(x^4+64y^8=(x^2)^2+(8y^4)^2=(x^2)^2+(8y^4)^2+2.x^2.8y^4-2x^2.8y^4\)

\(=(x^2+8y^4)^2-16x^2y^4=(x^2+8y^4)^2-(4xy^2)^2\)

\(=(x^2+8y^4-4xy^2)(x^2+8y^4+4xy^2)\)

e)

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

\(=3x(x-2)-(x-2)=(3x-1)(x-2)\)

đề đúng ko

Đa thức này không phân tích được nhé bạn

a)\(5x^2-4\left(x^2-2x+1\right)-5=5\left(x^2-1\right)-4\left(x-1\right)^2=5\left(x-1\right)\left(x+1\right)-4\left(x-1\right)^2=\left(x-1\right)\left(5x+5-4x+4\right)=\left(x-1\right)\left(x+9\right)\)

b) \(9x^2+6x-4y^2+4y=\left(9x^2+6x+1\right)-\left(4y^2-4y+1\right)=\left(3x+1\right)^2-\left(2y-1\right)^2=\left(3x+1-2y+1\right)\left(3x+1+2y-1\right)=\left(3x-2y+2\right)\left(3x+2y\right)\)

a: \(5x^2-4\left(x^2-2x+1\right)-5\)

\(=5x^2-4x^2+8x-4-5\)

\(=x^2+8x-9\)

\(=\left(x+9\right)\left(x-1\right)\)

b: \(9x^2+6x-4y^2+4y\)

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

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