phan tich da thuc thanh nhan tu
a)x^4+x^2y^2+y^4
b)x^3+3x-4
c)x^2+9x+8
d)x^2+x-42
e)y^2-13y+12
f)x^2-x-30
g)2x^2+xy-y^2
h)y^2-y-12
i)x^2+x-2
j)x^3+3x^2-2
k)x^3-6x^2+16
l)x^3+3x+4
m)x^4+6x^3-12x^2-8x
minh can gap lam, chiu nay la minh hoc roi
K
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D
4 tháng 8 2019
a. x4 + x2y2 + y4 = (x4 + 2x2y2 + y4) - x2y2
= (x2 + y2)2 – (xy)2
= [(x2 + y2) + xy] [(x2 + y2) – xy]
= (x2 + xy + y2)(x2 –xy + y2)
AN
28 tháng 9 2017
phan tich cac da thuc sau thanh nhan tu theo mau:
a)\(2x^3-x\)
\(=x\left(2x^2-1\right)\)
\(=x\left(\left(\sqrt{2}x\right)^2-1^2\right)\)\
\(=x\left(\sqrt{2}x-1\right)\left(\sqrt{2}x+1\right)\)
b)\(5x^2\left(x-1\right)-15x\left(x-1\right)\)
\(=\left(5x^2-15x\right)\left(x-1\right)\)
\(=5x\left(x-3\right)\left(x-1\right)\)
d)\(3x\left(x-2y\right)+6y\left(2y-x\right)\)
\(=3x\left(x-2y\right)-6y\left(x-2y\right)\)
\(=\left(3x-6y\right)\left(x-2y\right)\)
\(=3\left(x-2y\right)\left(x-2y\right)\)
\(=3\left(x-2y\right)^2\)
PD
1
MT
29 tháng 7 2015
x^3 - x + 3x^2y + 3xy^2 + y^3 - y
=x3+y3+3x2y+3xy2-x-y
=(x+y)(x2-xy+y2)+3xy(x+y)-(x+y)
=(x+y)(x2-xy+y2+3xy-1)
=(x+y)(x2+2xy+y2-1)
=(x+y)[(x+y)2-1]
=(x+y)(x+y-1)(x+y+1)
x^2 + 5x - 6
=x2-x+6x-6
=x.(x-1)+6.(x-1)
=(x-1)(x+6)
15 tháng 10 2018
\(1.x^3+2x+x^2=x\left(x^2+x+2\right)\)
\(2.2x^3+4x^2+2x=2x\left(x^2+2x+1\right)=2x\left(x+1\right)^2\)
\(3.-3x^3-5x^2+8x=-3x^3+3x^2-8x^2+8x\)
\(=-3x^2\left(x-1\right)-8x\left(x-1\right)=\left(3x^2+8x\right)\left(1-x\right)\)
\(=x\left(3x+8\right)\left(1-x\right)\)
\(4.x^2+4x-5=x^2-x+5x-5=\left(x-1\right)\left(x+5\right)\)
\(5.6x^2-3x-3=6x^2-6x+3x-3=3\left(x-1\right)\left(2x+1\right)\)
\(6.3x^2-2x-5=3x^2+3x-5x-5=\left(x+1\right)\left(3x-5\right)\)
\(8.x^2-2x-4y^2-4y=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)\(=\left(x+2y\right)\left(x-y-2\right)\)
\(9.x^3+2x^2y+xy^2-9x=x\left(x^2+2xy+y^2-9\right)\)
\(=x\left(x+y-3\right)\left(x+y+3\right)\)
\(10.x^2-y^2+6x+9=\left(x+3-y\right)\left(x+3+y\right)\)
dài quá bạn
\(\text{a) }x^4+x^2y^2+y^4=x^4+2x^2y^2-x^2y^2+y^4=\left(x^4+2x^2y^2+y^4\right)-\left(x^2y^2\right)=\left(x^2+y^2\right)^2-\left(xy\right)^2\)
\(=\left(x^2+y^2+xy\right)\left(x^2+y^2-xy\right)\)
\(\text{b) }x^3+3x-4=x^3+3x-1-3=\left(x^3-1\right)+\left(3x-3\right)=\left(x-1\right)\left(x^2+x+1\right)+3\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x+1+3\right)=\left(x-1\right)\left(x^2+x+4\right)\)
\(\text{c) }x^2+9x+8=x^2+8x+x+8=\left(x^2+8x\right)+\left(x+8\right)=x\left(x+8\right)+\left(x+8\right)\)
\(=\left(x+8\right)\left(x+1\right)\)
\(\text{d) }x^2+x-42=x^2+7x-6x-42=\left(x^2+7x\right)-\left(6x+42\right)=x\left(x+7\right)-6\left(x+7\right)\)
\(=\left(x+7\right)\left(x-6\right)\)
\(\text{e) }y^2-13y+12=y^2-y-12y+12=\left(y^2-y\right)-\left(12y-12\right)=y\left(y-1\right)-12\left(y-1\right)\)
\(=\left(y-1\right)\left(y-12\right)\)
Mấy câu sau mk sẽ giải tiếp, bạn ráng chờ nha