BT2: Phân tích các đa thức sau thành nhân tử bằng phương pháp tách hạng tử. a, x^2 + 4xy - 21y^2 b, 5x^2 + 6xy + y^2 c, x^2 + 2xy - 15y^2 d, x^2 - 7xy + 10y^2
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27 tháng 7 2016
X2+4xy-21y2=(x2+4xy+4y2)-25y2=(x+2)2-(5y)2=(x+2-5y)(x+2+5y)
5x2+6xy+y2=9x2+6xy+y2-4x2=(3x+y)2-4x2=(3x+y+2x)(3x+y-2x)
(x-y)2+4(x-y)-12=(x-y+2)2-16=(x-y+2+4)(x-y+2-4)
x2-7xy+10y2=x2-7xy+\(\frac{49y^2}{4}-\frac{9y^2}{4}\)= \(\left(x-\frac{7}{2}\right)^2-\left(\frac{3y}{2}\right)^2\)=\(\left(x-\frac{7}{2}-\frac{3y}{2}\right)\left(x-\frac{7}{2}+\frac{3y}{2}\right)\)
x2+2xy-15y2=(x+y)2-16y2=(x+y-4y)(x+y+4y
KN
29 tháng 7 2018
a) \(x^2+4xy-21y^2=x^2-3xy+7xy-21y^2=x\left(x-3y\right)+7y\left(x-3y\right)\)\(=\left(x-3y\right)\left(x+7y\right)\)
b)\(5x^2+6xy+y^2\)
=\(5x^2+5xy+xy+y^2\)
=\(5x^{ }\left(x+y\right)+y\left(x+y\right)\)
=\(\left(5x+y\right)\left(x+y\right)\)
KN
29 tháng 7 2018
c) \(x^2-7xy+10y^2\)
\(=\left(x^2-2xy\right)\left(5xy-10y^2\right)\)
\(=x\left(x-2y\right)-5y\left(x-2y\right)\)
\(=\left(x-5y\right)\left(x-2y\right)\)
d)\(x^2+2xy-15y^2\)
\(=x^2+2xy+y^2-16y^2\)
\(\left(x+y\right)^2-\left(4y\right)^2\)
\(=\left(x-3y\right)\left(x+5y\right)\)
MT
19 tháng 8 2015
a) 5x^2 + 6xy + y^2
=5x2+5xy+xy+y2
=5x.(x+y)+y.(x+y)
=(x+y)(5x+y)
b) x^2 + 2xy - 15y^2.
=x2-3xy+5xy-15y2
=x.(x-3y)+5y.(x-3y)
=(x-3y)(x+5y)
c) (x-y)^2 + 4(x-y) - 12
=(x-y)2+4(x-y)+4-16
=(x-y+2)2-16
=(x-y+2-4)(x-y+2+4)
=(x-y-2)(x-y+6)
d) x^3 - 2x - 4.
=x3+2x2+2x-2x2-4x-4
=x.(x2+2x+2)-2.(x2+2x+2)
=(x2+2x+2)(x-2)
4 tháng 8 2020
Bài 3:
a) Ta có: \(x^2+4xy-21y^2\)
\(=x^2+7xy-3xy-21y^2\)
\(=x\left(x+7y\right)-3y\left(x+7y\right)\)
\(=\left(x+7y\right)\left(x-3y\right)\)
b) Ta có: \(5x^2+6xy+y^2\)
\(=5x^2+5xy+xy+y^2\)
\(=5x\left(x+y\right)+y\left(x+y\right)\)
\(=\left(x+y\right)\left(5x+y\right)\)
c) Ta có: \(x^2+2xy-15y^2\)
\(=x^2+5xy-3xy-15y^2\)
\(=x\left(x+5y\right)-3y\left(x+5y\right)\)
\(=\left(x+5y\right)\left(x-3y\right)\)
d) Ta có: \(\left(x-y\right)^2+4\left(x-y\right)-12\)
\(=\left(x-y\right)^2+6\left(x-y\right)-2\left(x-y\right)-12\)
\(=\left(x-y\right)\left(x-y+6\right)-2\left(x-y+6\right)\)
\(=\left(x-y+6\right)\left(x-y-2\right)\)
e) Ta có: \(x^2-7xy+10y^2\)
\(=x^2-2xy-5xy+10y^2\)
\(=x\left(x-2y\right)-5y\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x-5y\right)\)
f) Ta có: \(x^2yz+5xyz-14yz\)
\(=yz\left(x^2+5x-14\right)\)
\(=yz\left(x^2+7x-2x-14\right)\)
\(=yz\left[x\left(x+7\right)-2\left(x+7\right)\right]\)
\(=yz\left(x+7\right)\left(x-2\right)\)
OF
18 tháng 10 2020
a)3x2-5x-2=3x2-6x+x-2
=3x(x-2)+(x-2)
=(3x+1)(x-2)
b)x2-7xy-10y2=x2-2xy-5xy-10y2
=x(x-2y)-5y(x-2y)
=(x-5y)(x-2y)
c)c) x2 +4xy-21y2=x2+7xy-3xy-21y2
=x(x+7y)-3y(x+7y)
=(x-3y)(x+7y)
d) 5x2 +6xy+y2=5x2+5xy+xy+y2
=5x(x+y)+y(x+y)
=(5x+y)(x+y)
e) x2 +2xy-15y2=x2+5xy-3xy-15y2
=x(x+5y)-3y(x+5y)
=(x-3y)(x+5y)
CHÚC BẠN HỌC TỐT
23 tháng 7 2018
\(x^4+x^2-2=x^4-x^2+2x^2-2 \)
\(=x^2\left(x^2-1\right)+2\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2+2\right)\)
\(=\left(x+1\right)\left(x-1\right)\left(x^2+2\right)\)
H9
HT.Phong (9A5)
CTVHS
22 tháng 10 2023
a) \(x^3+4x^2-21x\)
\(=x\left(x^2+4x-21\right)\)
\(=x\left(x^2-3x+7x-21\right)\)
\(=x\left[x\left(x-3\right)+7\left(x-3\right)\right]\)
\(=x\left(x-3\right)\left(x+7\right)\)
b) \(5x^3+6x^2+x\)
\(=x\left(5x^2+6x+1\right)\)
\(=x\left(5x^2+5x+x+1\right)\)
\(=x\left[5x\left(x+1\right)+\left(x+1\right)\right]\)
\(=x\left(x+1\right)\left(5x+1\right)\)
c) \(x^3-7x+6\)
\(=x^3+2x^2-3x-2x^2-4x+6\)
\(=x\left(x^2+2x-3\right)-2\left(x^2+2x-3\right)\)
\(=\left(x-2\right)\left(x^2+2x-3\right)\)
\(=\left(x-2\right)\left(x-1\right)\left(x+3\right)\)
d) \(3x^3+2x-5\)
\(=3x^3+3x^2+5x-3x^2-3x-5\)
\(=x\left(3x^2+3x+5\right)-\left(3x^2+3x+5\right)\)
\(=\left(x-1\right)\left(3x^2+3x+5\right)\)
2 tháng 10 2016
a) \(xy+y^2-x-y=y\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(y-1\right)\)
b) \(25-x^2+4xy-4y^2=25-\left(x-2y\right)^2=\left(5-x+2y\right)\left(5+x-2y\right)\)
c) \(x^2-4x+3=x^2-x-3x+3=x\left(x-1\right)-3\left(x-1\right)=\left(x-1\right)\left(x-3\right)\)
d) \(y^2\left(x-1\right)-7y^3+7xy^3\)
\(=y^2\left(x-1-7y+7xy\right)\)
\(=y^2\left[\left(x-1\right)-7y\left(1-x\right)\right]=y^2\left(x-1\right)\left(1+7y\right)\)
2 tháng 10 2016
a)
\(xy+y^2-x-y\\ =\left(xy-x\right)+\left(y^2-y\right)\\ =x\left(y-1\right)+y\left(y-1\right)\\ =\left(y-1\right)\left(x+y\right)\)
a: x^2+4xy-21y^2
\(=x^2+7xy-3xy-21y^2\)
\(=x\left(x+7y\right)-3y\left(x+7y\right)\)
\(=\left(x+7y\right)\left(x-3y\right)\)
b: \(5x^2+6xy+y^2\)
\(=5x^2+5xy+xy+y^2\)
=5x(x+y)+y(x+y)
=(x+y)(5x+y)
c: \(x^2+2xy-15y^2\)
\(=x^2+5xy-3xy-15y^2\)
=x(x+5y)-3y(x+5y)
=(x+5y)(x-3y)
d: \(x^2-7xy+10y^2\)
\(=x^2-2xy-5xy+10y^2\)
=x(x-2y)-5y(x-2y)
=(x-2y)(x-5y)
a) \(x^2+4xy-21y^2\)
\(=x^2+7xy-3xy-21y^2\)
\(=x\left(x+7y\right)-3y\left(x+7y\right)\)
\(=\left(x+7y\right)\left(x-3y\right)\)
b) \(5x^2+6xy+y^2\)
\(=5x^2+5xy+xy+y^2\)
\(=5x\left(x+y\right)+y\left(x+y\right)\)
\(=\left(5x+y\right)\left(x+y\right)\)
c) \(x^2+2xy-15y^2\)
\(=x^2+5xy-3xy-15y^2\)
\(=x\left(x+5y\right)-3y\left(x+5y\right)\)
\(=\left(x+5y\right)\left(x-3y\right)\)
d) \(x^2-7xy+10y^2\)
\(=x^2-2xy-5xy+10y^2\)
\(=x\left(x-2y\right)-5y\left(x-2y\right)\)
\(=\left(x-5y\right)\left(x-2y\right)\)