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

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

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

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

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

2x^3 - 2xy^2 - 8x^2 + 8xy 

= 2x^2 ( x - y ) - 8x ( x - y ) 

= ( x - y ) ( 2x^2 - 8x ) 

= ( x - y ) 2x ( x - 4 ) 

20% dd

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

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

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

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

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

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

Bài 2:

a: Sửa đề: \(A=-4x^2-5y^2+8xy+10y+12\)

\(=-4x^2+8xy-4y^2-y^2+10y-25+37\)

\(=-\left(2x-2y\right)^2-\left(y-5\right)^2+37\le37\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}2x-2y=0\\ y-5=0\end{cases}\Rightarrow\begin{cases}y=5\\ x=y=5\end{cases}\)

b: \(B=-x^2-y^2+xy+2x+2y\)

\(=-\frac14\left(4x^2+4y^2-4xy-8x-8y\right)\)

\(=-\frac14\left(4x^2-4xy+y^2-8x+4y+3y^2-12y\right)\)

\(=-\frac14\left\lbrack\left(2x-y\right)^2-4\left(2x-y\right)+4+3y^2-12y+12-16\right\rbrack\)

\(=-\frac14\left\lbrack\left(2x-y-2\right)^2+3\left(y-2\right)^2-16\right\rbrack=-\frac14\left(2x-y-2\right)^2-\frac34\left(y-2\right)^2+4\le4\forall x,y\)

Dấu '=' xảy ra khi y-2=0 và 2x-y-2=0

=>y=2 và 2x=y+2=2+2=4

=>x=2 và y=2

Bài 1:

d: \(D=2x^2+3y^2+4xy-8x-2y\)

\(=2x^2+4xy+2y^2-8x-8y+y^2+6y\)

\(=2\left(x+y\right)^2-8\left(x+y\right)+8+y^2+6y+9-17\)

\(=2\left(x+y-2\right)^2+\left(y+3\right)^2-17\ge-17\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}y+3=0\\ x+y-2=0\end{cases}\Rightarrow\begin{cases}y=-3\\ x=-y+2=-\left(-3\right)+2=3+2=5\end{cases}\)

f: \(F=2x^2+8xy+11y^2-4x-2y+6\)

\(=2x^2+8xy+8y^2-4x-8y+3y^2+6y+6\)

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

\(=2\left(x+2y-1\right)^2+3\left(y+1\right)^2+1\ge1\forall x,y\)

Dấu '=' xảy ra khi y+1=0 và x+2y-1=0

=>y=-1 và x=-2y+1=-2*(-1)+1=2+1=3

h: \(H=x^2+y^2-xy-x+y+1\)

\(=\frac14\left(4x^2+4y^2-4xy-4x+4y+4\right)\)

\(=\frac14\left(4x^2-4xy+y^2-4x+2y+3y^2+2y+4\right)\)

\(=\frac14\left\lbrack\left(2x-y\right)^2-2\left(2x-y\right)+1+3y^2+2y+\frac13+\frac83\right\rbrack\)

\(=\frac14\cdot\left\lbrack\left(2x-y-1\right)^2+3\left(y+\frac13\right)^2+\frac83\right\rbrack\ge\frac23\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}y+\frac13=0\\ 2x-y-1=0\end{cases}\Rightarrow\begin{cases}y=-\frac13\\ 2x=y+1=-\frac13+1=\frac23\end{cases}\Rightarrow\begin{cases}y=-\frac13\\ x=\frac13\end{cases}\)

Bài 2:

a: Sửa đề: \(A=-4x^2-5y^2+8xy+10y+12\)

\(=-4x^2+8xy-4y^2-y^2+10y-25+37\)

\(=-\left(2x-2y\right)^2-\left(y-5\right)^2+37\le37\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}2x-2y=0\\ y-5=0\end{cases}\Rightarrow\begin{cases}y=5\\ x=y=5\end{cases}\)

b: \(B=-x^2-y^2+xy+2x+2y\)

\(=-\frac14\left(4x^2+4y^2-4xy-8x-8y\right)\)

\(=-\frac14\left(4x^2-4xy+y^2-8x+4y+3y^2-12y\right)\)

\(=-\frac14\left\lbrack\left(2x-y\right)^2-4\left(2x-y\right)+4+3y^2-12y+12-16\right\rbrack\)

\(=-\frac14\left\lbrack\left(2x-y-2\right)^2+3\left(y-2\right)^2-16\right\rbrack=-\frac14\left(2x-y-2\right)^2-\frac34\left(y-2\right)^2+4\le4\forall x,y\)

Dấu '=' xảy ra khi y-2=0 và 2x-y-2=0

=>y=2 và 2x=y+2=2+2=4

=>x=2 và y=2

Bài 1:

d: \(D=2x^2+3y^2+4xy-8x-2y\)

\(=2x^2+4xy+2y^2-8x-8y+y^2+6y\)

\(=2\left(x+y\right)^2-8\left(x+y\right)+8+y^2+6y+9-17\)

\(=2\left(x+y-2\right)^2+\left(y+3\right)^2-17\ge-17\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}y+3=0\\ x+y-2=0\end{cases}\Rightarrow\begin{cases}y=-3\\ x=-y+2=-\left(-3\right)+2=3+2=5\end{cases}\)

f: \(F=2x^2+8xy+11y^2-4x-2y+6\)

\(=2x^2+8xy+8y^2-4x-8y+3y^2+6y+6\)

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

\(=2\left(x+2y-1\right)^2+3\left(y+1\right)^2+1\ge1\forall x,y\)

Dấu '=' xảy ra khi y+1=0 và x+2y-1=0

=>y=-1 và x=-2y+1=-2*(-1)+1=2+1=3

h: \(H=x^2+y^2-xy-x+y+1\)

\(=\frac14\left(4x^2+4y^2-4xy-4x+4y+4\right)\)

\(=\frac14\left(4x^2-4xy+y^2-4x+2y+3y^2+2y+4\right)\)

\(=\frac14\left\lbrack\left(2x-y\right)^2-2\left(2x-y\right)+1+3y^2+2y+\frac13+\frac83\right\rbrack\)

\(=\frac14\cdot\left\lbrack\left(2x-y-1\right)^2+3\left(y+\frac13\right)^2+\frac83\right\rbrack\ge\frac23\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}y+\frac13=0\\ 2x-y-1=0\end{cases}\Rightarrow\begin{cases}y=-\frac13\\ 2x=y+1=-\frac13+1=\frac23\end{cases}\Rightarrow\begin{cases}y=-\frac13\\ x=\frac13\end{cases}\)

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

\(A=2x^4+4x^3-7x^3-14x^2+8x^2+16x\)

\(=2x^2\left(x^2+2x\right)-7x\left(x^2+2x\right)+8\left(x^2+2x\right)\)

\(=\left(2x^2-7x+8\right)\left(x^2+2x\right)\)

\(=x\left(x+2\right)\left(2x^2-7x+8\right)\)

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

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

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

\(C=x\left(8y^2+8xy+2x^2-z^2\right)\)

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

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

\(=x\left(\sqrt{2}x+2\sqrt{2}y-z\right)\left(\sqrt{2}x+2\sqrt{2}y+z\right)\)

\(D=4a^4+10a^3+6a^2-6a^2-15a-9\)

\(=2a^2\left(2a^2+5a+3\right)-3\left(2a^2+5a+3\right)\)

\(=\left(2a^2-3\right)\left(2a^2+5a+3\right)\)

\(E=4a^3-ab^2+2ab-4a^2\)

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

\(=a\left(2a+b\right)\left(2a-b\right)-2a\left(2a-b\right)\)

\(=\left(2a-b\right)\left(2a^2+ab-2a\right)\)

\(F=5a^2-10a-4a+8\)

\(=5a\left(a-2\right)-4\left(a-2\right)\)

\(=\left(5a-4\right)\left(a-2\right)\)

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

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

a>(8x^2y+10xy6^2-6xy):2xy=4xy+5y-3

b>(3x^2-4x).(2x-6)=6x^3-26x^2+24x

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.