bài 1: Phân tích đa thức thành nhân tử
a, (xy-1)2+ (x+y)2
b, a2+2a2+2a+1
c, (1+2a).(1-2a)-a.(a+2).(a-2)
d, a2+b2-a2b2+ab-a-b
e, xy.(x+y)-yz.(y+z)+xz(x-z)
f, xyz-(xy+yz+zx)+(x+y+z)-1
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NH
1
7 tháng 9 2021
Câu 1:
\(a^2+b^2-a^2b^2+ab-a-b\)
\(=a^2\left(1-b^2\right)+b\left(b-1\right)+a\left(b-1\right)\)
\(=-a^2\left(b-1\right)\left(b+1\right)+\left(b-1\right)\left(a+b\right)\)
\(=\left(b-1\right)\left(-a^2b-a^2+a+b\right)\)
\(=\left(b-1\right)\cdot\left[-b\left(a^2-1\right)-a\left(a-1\right)\right]\)
\(=\left(b-1\right)\left(a-1\right)\left[-b\left(a+1\right)-a\right]\)
27 tháng 10 2023
b: \(xy\left(x+y\right)-yz\left(y+z\right)+xz\left(x-z\right)\)
\(=x^2y+xy^2-y^2z-yz^2+x^2z-xz^2\)
\(=x^2y-yz^2+xy^2-y^2z+x^2z-xz^2\)
\(=y\left(x-z\right)\left(x+z\right)+y^2\left(x-z\right)+xz\left(x-z\right)\)
\(=\left(x-z\right)\left(xy+yz+y^2+xz\right)\)
\(=\left(x-z\right)\left(x+y\right)\left(x+z\right)\)
13 tháng 10 2016
a) = (xyz+xy) +(z+1) +(yz+zx)+(x+y)
= xy(z+1) +(z+1)+z(x+y)+(x+y)
= (z+1)(xy+1)+(x+y)(Z+1)
=(z+1)(xy+1+x+y)
MT
8 tháng 10 2019
Hai câu đầu tham khảo
Câu hỏi của Bangtan Sonyeondan - Toán lớp 8 - Học toán với OnlineMath
MT
8 tháng 10 2019
c) \(E=\left(x+a\right)\left(x+2a\right)\left(a+3a\right)\left(x+4a\right)+a^4\)
\(=\left(x+a\right)\left(x+4a\right)\left(x+2a\right)\left(a+3a\right)+a^4\)
\(=\left(x^2+5ax+4a^2\right)\left(a^2+5ax+6a^2\right)+a^4\)(1)
Đặt \(x^2+5ax+4a^2=t\)
\(\Rightarrow\left(1\right)=t\left(t+2a^2\right)+a^4\)
\(=t^2+2a^2t+a^4=\left(t+a^2\right)^2\)(2)
Mà \(x^2+5ax+4a^2=t\)
Nên \(\left(2\right)=\left(x^2+5ax+5a^2\right)^2\)
10 tháng 7 2017
a) xy(x + y) + yz(z + y) + zx(z + x) + 3xyz
= [xy(x + y) + xyz] + [yz(z + y) + xyz] + [zx(z + x) + xyz]
= xy(x + y + z) + yz(x + y + z) + zx(x + y + z)
= (xy + yz + zx)(x + y + z)
b) Vô câu hỏi tương tự
26 tháng 7 2017
a) xy(x + y) + yz(z + y) + zx(z + x) + 3xyz
= [xy(x + y) + xyz] + [yz(z + y) + xyz] + [zx(z + x) + xyz]
= xy(x + y + z) + yz(x + y + z) + zx(x + y + z)
= (xy + yz + zx)(x + y + z)
b) tương tự
LN
2
MT
22 tháng 7 2015
A ) xy(z+y)+yz(y+z)+zx(z+x)
=y.[x(z+y)+z(y+z)]+zx(z+x)
=y.(xz+xy+zy+z2)+zx(z+x)
=y.(xz+z2+xy+zy)+zx(z+x)
=y.[z.(z+x)+y.(z+x)]+zx(z+x)
=y.(z+x)(z+y)+zx(z+x)
=(z+x)[y(z+y)+zx]
=(z+x)(yz+y2+zx)
B )xy(x+y)-yz(y+z)-zx(z-x)
=y.[x(x+y)-z(y+z)]-zx(z-x)
=y.(x2+xy-zy-z2)-zx(z-x)
=y.(x2-z2+xy-zy)-zx(z-x)
=y.[(x+z)(x-z)+y.(x-z)]-zx(z-x)
=y.(x-z)(x+z+y)+zx(x-z)
=(x-z)[y(x+z+y)+zx]
=(x-z)(yx+yz+y2+zx)
=(x-z)(yx+zx+yz+y2)
=(x-z)[x.(y+z)+y.(y+z)]
=(x-z)(y+z)(x+y)
LT
30 tháng 6 2021
b. \(\text{ xy(x+y)-yz(y+z)-xz(z-x) =xy(x+y+z-z)+yz(y+z)+xz(x-z) =xy(x-z)+xy(y+z)+yz(y+z)+xz(x-z) =(x+y)(y+z)(x-z) }\)
2 tháng 9 2018
\(yz\left(y+z\right)+zx\left(z-x\right)-xy\left(x+y\right)\)
\(=yz\left(y+z\right)+zx\left(z-x\right)-xy\left[\left(y+z\right)-\left(z-x\right)\right]\)
\(=yz\left(y+z\right)+zx\left(z-x\right)-xy\left(y+z\right)+xy\left(z-x\right)\)
\(=y\left(y+z\right)\left(z-x\right)+x\left(z-x\right)\left(z-y\right)\)
\(=\left(z-x\right)\left(yz-xy+xz-xy\right)\)
\(a,=\left(xy-1-x-y\right)\left(xy-1+x+y\right)\\ b,Sửa:a^3+2a^2+2a+1\\ =a^3+a^2+a^2+a+a+1=\left(a+1\right)\left(a^2+a+1\right)\\ c,=1-4a^2-a\left(a^2-4\right)=1-4a^2-a^3+4a\\ =\left(1-a\right)\left(1+a+a^2\right)+4a\left(1-a\right)\\ =\left(1-a\right)\left(1+5a+a^2\right)\\ d,=\left(a^2-a^2b^2\right)+\left(b^2-b\right)+\left(ab-a\right)\\ =a^2\left(1-b\right)\left(1+b\right)+b\left(b-1\right)+a\left(b-1\right)\\ =\left(b-1\right)\left(-a^2-ab+b+a\right)\\ =\left(b-1\right)\left(b-1\right)\left(a+b\right)\left(1-a\right)\)
\(e,=x^2y+xy^2-yz\left(y+z\right)+x^2z-xz^2\\ =\left(x^2y+x^2z\right)+\left(xy^2-xz^2\right)-yz\left(y+z\right)\\ =x^2\left(y+z\right)+x\left(y-z\right)\left(y+z\right)-yz\left(y+z\right)\\ =\left(y+z\right)\left(x^2+xy-xz-yz\right)\\ =\left(y+z\right)\left(x+y\right)\left(x-z\right)\)
\(f,=xyz-xy-yz-xz+x+y+z-1\\ =xy\left(z-1\right)-y\left(z-1\right)-x\left(z-1\right)+\left(x-1\right)\\ =\left(z-1\right)\left(xy-y-x+1\right)=\left(z-1\right)\left(x-1\right)\left(y-1\right)\)