\(2x^3-2xy^2-8x^2+8xy\)

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

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

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

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

2x^3 - 2xy^2 - 8x^2 + 8xy 

= 2x^2 ( x - y ) - 8x ( x - y ) 

= ( x - y ) ( 2x^2 - 8x ) 

= ( x - y ) 2x ( x - 4 ) 

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

\(2x^3-2xy^2-8x^2+8xy\)

\(=2x\left(x^2-y^2\right)-8x\left(x-y\right)\)

\(=2x\left(x+y\right)\left(x-y\right)-8x\left(x-y\right)\)

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

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

20% dd

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Ta có:

D=2x2+3y2+4xy−8x−2y+18C=2x2+3y2+4xy−8x−2y+18

D=2(x2+2xy+y2)+y2−8x−2y+18C=2(x2+2xy+y2)+y2−8x−2y+18

D=2[(x+y)2−4(x+y)+4]+(y2+6y+9)+1C=2[(x+y)2−4(x+y)+4]+(y2+6y+9)+1

D=2(x+y−2)2+(y+3)2+1≥1C=2(x+y−2)2+(y+3)2+1≥1

Dấu "=" xảy ra ⇔x+y=2⇔x+y=2và y=−3y=−3

Hay x = 5 , y = -3

Đc chx bạn

\(A=\left(-6x^7y^6\right)\left(8x^3y^3\right)=\left(-6.8\right).\left(x^7.x^3\right).\left(y^6.y^3\right)=-48x^{10}y^9\).

\(B=-7xy^2-2xy+6xy^2+5xy+6=\left(-7xy^2+6xy^2\right)+\left(-2xy+5xy\right)+6=-xy^2+3xy+6\)

Lời giải:

$2x^3y+2xy^3+4x^2y^2-8xy$

$=2xy(x^2+y^2+2xy-4)$

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

$=2xy[(x+y)^2-2^2]=2xy(x+y-2)(x+y+2)$

P.s: lần sau bạn lưu ý ghi đầy đủ yêu cầu đề.

\(a,\dfrac{2}{3}xy^2.\dfrac{2}{3}xy=\dfrac{4}{9}x^2y^3\)

\(b,-\dfrac{1}{2}x^2y.2xy^2=-x^3y^3\)

\(c,8xy^3.2x^3y^2=16x^4y^5\)

\(d,-\dfrac{1}{4}x^2y^3.2x^3y^2=-\dfrac{1}{2}x^5y^5\)

\(e,4x^2y^4.\dfrac{1}{2}x^2y^3=2x^4y^7\)

\(f,-8xy.\dfrac{1}{4}x^2y=-2x^3y^2\)

\(Ayumu\)