Phân tích đa thức thành nhân tử
a. xy + y2 – x – y
b. 25 – x2 + 4xy – 4y2
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a: \(=4xy\left(1-5x^2y\right)\)
b: \(=\left(x-y\right)\left(x+y\right)+3\left(x-y\right)=\left(x-y\right)\left(x+y+3\right)\)
c: \(=x\left(x-a\right)+y\left(x-a\right)=\left(x-a\right)\left(x+y\right)\)
d: \(=\left(x+2y\right)^2-36=\left(x+2y+6\right)\left(x+2y-6\right)\)
\(a.x^3-2x^2-2x-4\\ =\left(x^3-2x^2\right)-\left(2x-4\right)\\ =x^2\left(x-2\right)-2\left(x-2\right)\\ =\left(x^2-2\right)\left(x-2\right)\)
\(b.xy+1-x-y\\ =\left(xy-x\right)+\left(-y+1\right)\\ =x\left(y-1\right)-\left(y-1\right)\\ =\left(x-1\right)\left(y-1\right)\)
\(c.x^2-4xy+4y^2-4y\\ =\left(x-2y\right)^2-4y\\ =\left(x-2y\right)^2-\left(2y\right)^2\\ =\left(x-2y+2y\right)\left(x-2y-2y\right)\\ =x\left(x-4y\right)\)
\(d.16-x^2+2xy-y^2\\ =4^2-\left(x-y\right)^2\\ =\left(4-x+y\right)\left(4-x-y\right)\)
b: =xy-x-y+1
=x(y-1)-(y-1)
=(x-1)(y-1)
c: =(x-2y)^2-4y
\(=\left(x-2y-2\sqrt{y}\right)\left(x-2y+2\sqrt{y}\right)\)
d: =16-(x^2-2xy+y^2)
=16-(x-y)^2
=(4-x+y)(4+x-y)
1, \(a^6+b^3=\left(a^2+b\right)\left(a^4-a^2b+b^2\right)\)
2, \(x^2-10x+25=\left(x-5\right)^2\)
3, \(8x^3-\dfrac{1}{8}=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)
4, \(x^2+4xy+4y^2=\left(x+2y\right)^2\)
1) \(a^6+b^3=\left(a^2\right)^3+b^3=\left(a^2+b\right)\left(a^4-a^2b+b^2\right)\)
2) \(x^2-10x+25=\left(x-5\right)^2\)
3) \(8x^3-\dfrac{1}{8}=\left(2x\right)^3-\left(\dfrac{1}{3}\right)^3=\left(2x-\dfrac{1}{3}\right)\left(4x^2+\dfrac{2x}{3}+\dfrac{1}{4}\right)\)
4) \(x^2+4xy+4y^2=\left(x+2y\right)^2\)
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:
\(x^3+x^2+y^3+xy\)
\(=\left(x^3+y^3\right)\left(x^2+xy\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)+x\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2+x\right)\)
= \(-\left(x^2+4xy+4y^2\right)\)
= \(-\left(x+2y\right)^2\)
Phân tích đa thức thành nhân tử
a,\(xy+y^2-x-y\)
=\(\left(xy+y^2\right)-\left(x+y\right)\)
=\(y\left(x+y\right)-\left(x+y\right)\)
=\(\left(x+y\right)\left(y-1\right)\)
b, \(25-x^2+4xy-4y^2\)
=\(-\left(x^2-4xy+4y^2-25\right)\)
=\(-\left[\left(x-2y\right)^2-5^2\right]\)
=\(-\left(x-2y-5\right)\left(x-2y+5\right)\)
a) xy + y2 - x - y
= x(x+y) - (x+y)
= (x+y)(x - 1)
b) 25 - x2 + 4xy - 4y2
= 52 -( x2 - 4xy + 4y2)
Tự làm nha![hehe hehe](https://hoc24.vn/media/cke24/plugins/smiley/images/hehe.png)