Phân tích đa thức thành nhân tử
x2 + 4y2 -5x -10y + 4xy
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\(x^2+4xy+4y^2-25\)
\(=\left(x^2+4xy+4y^2\right)-25\)
\(=\left(x+2y\right)^2-5^2\)
\(=\left(x+2y+5\right)\left(x+2y-5\right)\)
a: \(x^2-4xy+4y^2-2x+4y-35\)
\(=\left(x^2-4xy+4y^2\right)-\left(2x-4y\right)-35\)
\(=\left(x-2y\right)^2-2\left(x-2y\right)-35\)
\(=\left(x-2y\right)^2-7\left(x-2y\right)+5\left(x-2y\right)-35\)
\(=\left(x-2y\right)\left(x-2y-7\right)+5\left(x-2y-7\right)\)
\(=\left(x-2y-7\right)\left(x-2y+5\right)\)
c: \(\left(xy+ab\right)^2+\left(ay-bx\right)^2\)
\(=x^2y^2+a^2b^2+2xyab+a^2y^2-2aybx+b^2x^2\)
\(=x^2y^2+a^2y^2+a^2b^2+b^2x^2\)
\(=y^2\left(x^2+a^2\right)+b^2\left(a^2+x^2\right)\)
\(=\left(x^2+a^2\right)\left(y^2+b^2\right)\)
\(x^2-4y^2-2x+1=\left(x-1\right)^2-4y^2=\left(x-1-2y\right)\left(x-1+2y\right)\)
a, \(x^4+2x^2+1-x^2\)
= \(\left(x^2+1\right)^2-x^2\)
= \(\left(x^2+x+1\right)\left(x^2-x+1\right)\)
b, \(x^4+x^2+1\)
= \(x^4+2x^2+1-x^2\)
= .. ( như phần a )
c, \(y^4+64\)
= \(\left(y^2+8\right)\left(y^2-8\right)\)
d, \(4xy+3z-12y-xz\)
\(=4y\left(x-3\right)-z\left(x-3\right)\)
\(=\left(x-3\right)\left(4y-z\right)\)
e, \(x^2-4xy+4y^2-z^2+6z-9\)
\(=\left(x-2y\right)^2-\left(z-3\right)^2\)
g, \(x^2-4xy+5x+4y^2-10y\)
\(=\left(x^2-4xy+4y^2\right)+\left(5x-10y\right)\)
\(=\left(x-2y\right)^2+5\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x-2y+5\right)\)
h, \(x^2-7x+6\)
\(=x^2-6x-x+6\)
\(=x\left(x-6\right)-\left(x-6\right)\)
\(=\left(x-6\right)\left(x-1\right)\)
i, \(x^3+5x^2+6x+2\)
\(=x^3+x^2+4x^2+4x+2x+2\)
\(=x^2\left(x+1\right)+4x\left(x+1\right)+2\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+4x+2\right)\)
\(=x^2+x-6x+6\\ =x\left(x+1\right)-6\left(x+1\right)\\ =\left(x+1\right)\left(x+6\right)\)
\(=\left(x+2y\right)^2-4z^2=\left(x+2y+2z\right)\left(x+2y-2z\right)\)
Câu 1:
$x^2+4y^2+4xy-16=[x^2+(2y)^2+2.x.2y]-16$
$=(x+2y)^2-4^2=(x+2y-4)(x+2y+4)$
Câu 2:
$x^3+x^2+y^3+xy=(x^3+y^3)+(x^2+xy)$
$=(x+y)(x^2-xy+y^2)+x(x+y)=(x+y)(x^2-xy+y^2+x)$
Câu 1:
\(x^2+4y^2+4xy-16\)
\(=\left(x+2y\right)^2-16\)
\(=\left(x+2y+4\right)\left(x+2y-4\right)\)
Câu 2:
\(x^3+x^2+y^3+xy\)
\(=\left(x^3+y^3\right)\left(x^2+xy\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)+x\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2+x\right)\)