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A - B = ( 5x2 - 12xy + 2y2 ) + ( -4x2 + 12xy - y2 )
= 5x2 - 12xy + 2y2 - 4x2 + 12xy - y2
= 9x2 + y2
x2 > hoặc = 0 => 9x2 > hoặc = 0
y2 > hoặc = 0
<=> 9x2 + y2 > hoặc = 0
A + B > hoặc = 0
nên A và B không cùng có giá trị âm
tương tự 2) lấy A - B
3) lấy A + B
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\(P=\dfrac{1}{3}xy\left(x^2+y^2\right)-4x^2\left(xy^2-y\right)+2\left(x^2y-xy^2\right)\)
\(=\dfrac{1}{3}x^3y+\dfrac{1}{3}xy^3-4x^3y^2+4x^2y+2x^2y-2xy^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a) \(x^2+10x+26+y^2+2y=(x^2+10x+25)+(y^2+2y+1)\)
..................................................= \(\left(x+5\right)^2+\left(y+1\right)^2\)
b) \(z^2-6z+5-t^2-4t=(z^2-6t+9)-(t^2+4t+4)\)
............................................= \(\left(z-3\right)^2-\left(t+2\right)^2\)
c) \(x^2-2xy+2y^2+2y+1=(x^2-2xy+y^2)+(y^2+2y+1)\)
..................................................= \(\left(x-y\right)^2+\left(y+1\right)^2\)
d) \(4x^2-12x-y^2+2y+8=\left(4x^2-12x+9\right)-\left(y^2-2y+1\right)\)
.................................................= \(\left(2x-3\right)^2-\left(y-1\right)^2\)
Bài 2:
a) \(\left(x+y+4\right)\left(x+y-4\right)=\left(x+y\right)^2-16\)
b) \(\left(x-y+6\right)\left(x+y-6\right)=x^2-\left(y-6\right)^2\)
c) \(\left(y+2z-3\right)\left(y-2z+3\right)=y^2-\left(2z-3\right)^2\)
d) \(\left(x+2y+3z\right)\left(2y+3z-x\right)=\left(2y+3z\right)^2-x^2\)
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a. 4x3+4x2+xy2= x(4x2+4xy+y2)=x(2x+y)2
b. xy-3+x-3y= xy+x-3-3y= x(y+1)-3(y+1)=(x-3)(y+1)
c. x3-6x2y+9xy2= x(x2-6xy+9y2)= x(x-3y)2
d. 2xy-1+x-2y= 2xy+x-1-2y= x(2y+1)-(2y+1)=(x-1)(2y+1)
a, \(4x^2-17xy+13y^2\)
\(=4x^2-4xy-13xy+13y^2\)
\(=\left(4x^2-4xy\right)-\left(13xy-y^2\right)\)
\(=4x\left(x-y\right)-13y\left(x-y\right)\)
\(=\left(x-y\right).\left(4x-13y\right)\)
b, \(x^2-xy-2y^2\)
\(=x^2+xy-2xy-2y^2\)
\(=\left(x^2+xy\right)-\left(2xy+2y^2\right)\)
\(=x.\left(x+y\right)-2y.\left(x+y\right)\)
\(=\left(x+y\right).\left(x-2y\right)\)
Chúc bạn học tốt!!!