Chứng minh:x2+2xy+y2+3x−3y+9\(\ge\)0
Giải phương trình: 4x2 + 2xy + y2 - 3x + 3y - 6 = 0
3x-x2-y2+3y-2xy
3x - x2 - y2 + 3y - 2xy
= (3x + 3y) - (x2 + 2xy + y2)
= 3(x + y) - (x + y)2
= (3 - x - y)(x + y)
phân tích đa thức thành nhân tử
x2 - 2xy + y2 + 3x - 3y – 4
\(x^2-2xy+y^2+3x-3y-4\)
\(=\left(x-y\right)^2-1+3x-3y-3\)
\(=\left[\left(x-y\right)^2-1^2\right]+\left(3x-3y-3\right)\)
\(=\left[\left(x-y\right)-1\right]\left[\left(x-y\right)+1\right]+3\left(x-y-1\right)\)
\(=\left(x-y-1\right)\left(x-y+1\right)+3\left(x-y-1\right)\)
\(=\left(x-y-1\right)\left[\left(x-y+1\right)+3\right]\)
\(=\left(x-y-1\right)\left(x-y+4\right)\)
Phân tích đa thức thành nhân tử : x2 – 2xy + y2 + 3x – 3y – 10
x2-2xy+y2+3x-3y-10
= (x-y)2+3(x-y)-10
= [(x-y)2+5(x-y)]-[2(x-y)+10]
= (x-y)(x-y+5)-2(x-y+5)
= (x-y+5)(x-y-2)
Ta có: \(x^2-2xy+y^2+3x-3y-10\)
\(=\left(x-y\right)^2+3\left(x-y\right)-10\)
\(=\left(x-y+5\right)\left(x-y-2\right)\)
c) 3x + 3y – x2 – 2xy – y2 d) x3 – x + 3x2y + 3xy2 – y + y3
c) \(3x+3y-x^2-2xy-y^2=3\left(x+y\right)-\left(x+y\right)^2=\left(x+y\right)\left(3-x-y\right)\)d) \(=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)
\(=\left(x+y\right)\left(x+y+1\right)\left(x+y-1\right)\)
\(c,=3\left(x+y\right)-\left(x+y\right)^2=\left(3-x-y\right)\left(x+y\right)\\ d,=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\\ =\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)
c) \(=3\left(x+y\right)-\left(x+y\right)^2=\left(x+y\right)\left(3-x-y\right)\)
d) \(\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)
phân tích các đa thức sau thành nhân tử
a x2 - y2 -3x + 3y
b 2x + 2y -x2 + y2
c x2 -16 + y2 + 2xy
cứuuu
a) \(x^2-y^2-3x+3y\)
\(=\left(x-y\right)\left(x+y\right)-3\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-3\right)\)
b) \(2x+2y-x^2+y^2\)
\(=2\left(x+y\right)-\left(x^2-y^2\right)\)
\(=2\left(x+y\right)-\left(x-y\right)\left(x+y\right)\)
\(=\left(x+y\right)\left(2-x+y\right)\)
c) \(x^2-16+y^2+2xy\)
\(=x^2+y^2+2xy-16\)
\(=\left(x+y\right)^2-16\)
\(=\left(x+y+4\right)\left(x+y-4\right)\)
a) \(x^2-y^2-3x+3y\)
\(=\left(ax+y\right)\left(ax-y\right)-3.\left(x-y\right)\)
b) \(2x+2y-x^2+y^2\)
\(=2\left(x+y\right)-\left(x+y\right)\left(x-y\right)\)
c) \(x^2-16+y^2+2xy\)
\(=\left(x+y\right)\left(x-y\right)+2xy-16\)
cho x,y khac nhau biet x2-y=y2-x. tinh gia tri cua bieu thuc A=x2+2xy+y2-3x-3y
Theo bài ra, ta có: \(x^2-y=y^2-x\Leftrightarrow x^2-y^2=-x+y\)
\(\Leftrightarrow\left(x-y\right)\left(x+y\right)=-\left(x-y\right)\)
\(\Leftrightarrow\left(x+y\right)=-1\)
Ta lại có: \(A=x^2+2xy+y^2-3x-3y=\left(x+y\right)^2-3\left(x+y\right)\)
Thay x+y=-1 vào biểu thức A, ta được: \(A=\left(-1\right)^2-3.\left(-1\right)=1+3=4\)
Vậy A=4
Phân tích các đa thức sau thành nhân tử:
a) 3x - 3y + x 2 - y 2 ; b) x 2 -4 x 2 y 2 + y 2 + 2xy
c) x 6 - x 4 + 2 x 3 + 2 x 2 ; d) x 3 - 3x 2 +3x - 1 - y 3 .
a) (x - y)(x + y + 3). b) (x + y - 2xy)(2 + y + 2xy).
c) x 2 (x + l)( x 3 - x 2 + 2). d) (x – 1 - y)[ ( x - 1 ) 2 + ( x - 1 ) y + y 2 ].
Phân tích đa thức thành nhân tử
a)A=x2+7x+7y-y2
b)B=x2+2xy+y2-3x-3y
\(a,A=x^2+7x+7y-y^2\\ =x^2-y^2+7x+7y\\ =\left(x-y\right)\left(x+y\right)+7\left(x+y\right)\\ =\left(x+y\right)\left(x-y+7\right)\)
\(b,B=x^2+2xy+y^2-3x-3y\\ =\left(x+y\right)^2-3\left(x+y\right)\\ =\left(x+y\right)\left(x+y-3\right)\)
Phân tích các đa thức sau thành nhân tử a.3x²-6x+9x². B.3x²+5y-3xs-5x C.3y²-3z²+3x²+6xy. D.x²-25-2xy+y2
\(a,3x^2-6x+9x^2=12x^2-6x=6x\left(2x-1\right)\\ b,3x^2+5y-3xy-5x=\left(3x^2-3xy\right)-\left(5x-5y\right)=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\\ c,3y^2-3z^2+3x^2+6xyz=3\left(y^2-z^2+x^2+2xyz\right)\\ d,x^2-25-2xy+y^2=\left(x-y\right)^2-5^2=\left(x-y-5\right)\left(x-y+5\right)\)