PHÂN TÍCH ĐA THỨC SAU THÀNH NHÂN TỬ
a) ab(a+b)-bc(b+c)+ac(a-c)
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LV
1
LV
1
3 tháng 9 2023
a: =(x+y)^3+z^3-3xy(x+y)-3xyz
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)\)
b: \(=\left(x+y+y-z\right)^3-3\left(x+y\right)\left(y-z\right)\left(x+y+y-z\right)+\left(z-x\right)^3\)
\(=\left(x-z\right)^3+\left(z-x\right)^3-3\left(x+y\right)\left(y-z\right)\left(x-z\right)\)
\(=-3\left(x+y\right)\left(y-z\right)\left(x-z\right)\)
c: \(=\left(x^2+x\right)^2+3\left(x^2+x\right)+2-12\)
\(=\left(x^2+x\right)^2+3\left(x^2+x\right)-10\)
=(x^2+x+5)(x^2+x-2)
=(x^2+x+5)(x+2)(x-1)
d: =b^2c+bc^2+ac^2-a^2c-a^2b-ab^2
=b^2c-b^2a+bc^2-a^2b+ac^2-a^2c
=b^2(c-a)+b(c^2-a^2)+ac(c-a)
=(c-a)(b^2+ac)+b(c-a)(c+a)
=(c-a)(b^2+ac+bc+ba)
=(c-a)[b^2+bc+ac+ab]
=(c-a)[b(b+c)+a(b+c)]
=(c-a)(b+c)(b+a)
28 tháng 10 2021
\(a,=x\left(2x-y\right)+\left(2x-y\right)=\left(x+1\right)\left(2x-y\right)\\ b,=\left(a+b\right)\left(c-2\right)\\ c,=x\left(x+4y\right)+2\left(x+4y\right)=\left(x+2\right)\left(x+4y\right)\\ d,=x\left(x+2y\right)+3\left(x+2y\right)=\left(x+3\right)\left(x+2y\right)\)
25 tháng 12 2021
\(=a^2b-ab^2+b^2c-bc^2+ac^2-a^2c\)
\(=a^2\left(b-c\right)+bc\left(b-c\right)-a\left(b-c\right)\left(b+c\right)\)
\(=\left(b-c\right)\left(a^2-bc-ab-ac\right)\)
\(=\left(b-c\right)\left[a\left(a-b\right)-c\left(a-b\right)\right]\)
C
1
MN
2 tháng 3 2020
\(ab\left(b-a\right)-bc\left(b-c\right)-ac\left(c-a\right)\)
\(=ab\left(b-a\right)-b^2c+bc^2-ac^2+a^2c\)
\(=ab\left(b-a\right)+c^2\left(b-a\right)-c\left(b^2-a^2\right)\)
\(=\left(b-a\right)\left(ab+c^2-bc-ca\right)\)
\(=\left(b-a\right)\left[b\left(a-c\right)-c\left(a-c\right)\right]\)
\(=\left(b-a\right)\left(a-c\right)\left(b-c\right)\)
LT
1
B
1
2 tháng 3 2020
ab(b - a) - bc(b - c) - ac(c - a)
= ab2 - a2b - b2c + bc2 + ac(a - c)
= b2(a - c) - b(a2 - c2) + ac(a - c)
= b2(a - c) - b(a - c)(a + c) + ac(a - c)
= (b2 - ab - bc + ac)(a - c)
= [b(b - a) - c(b - a)](a - c)
= (b - c)(b -a)(a - c)
C
1
CC
2 tháng 3 2020
\(ab\left(b-a\right)-bc\left(b-c\right)-ac\left(c-a\right)\)
\(=ab\left(b-a\right)-\left(b^2c-bc^2\right)-\left(ac^2-a^2c\right)\)
\(=ab\left(b-a\right)-b^2c+bc^2-ac^2+a^2c\)
\(=ab\left(b-a\right)-\left(b^2c-a^2c\right)+\left(bc^2-ac^2\right)\)
\(=ab\left(b-a\right)-c\left(b^2-a^2\right)+c^2\left(b-a\right)\)
\(=ab\left(b-a\right)-c\left(b-a\right)\left(b+a\right)+c^2\left(b-a\right)\)
\(=\left(b-a\right)\left[ab-c\left(b+a\right)+c^2\right]=\left(b-a\right)\left[ab-\left(bc+ac\right)+c^2\right]\)
\(=\left(b-a\right)\left(ab-bc-ac+c^2\right)=\left(b-a\right)\left[\left(ab-bc\right)-\left(ac-c^2\right)\right]\)
\(=\left(b-a\right)\left[b\left(a-c\right)-c\left(a-c\right)\right]=\left(b-a\right)\left(b-c\right)\left(a-c\right)\)
