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a,\(3x-3y+6xy-6y^2=3x\left(1+2y\right)-3y\left(1+2y\right)=\left(3x-3y\right)\left(1+2y\right)\)
b,
\(4x^2-8xy-4z^2+4y^2=\left(2x-2y\right)^2-\left(2z\right)^2=\left(2x-2y-2z^{ }\right)\left(2x-2y+2z\right)\)
c,\(125a^3-8=\left(5a^{ }\right)^3-2^3=\left(5a-2\right)\left(5a^2+10a+4\right)\)
d, \(x^2-y^2-x-y=\left(x-y\right)\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-y-1\right)\)
\(a.3x-3y+6xy-6y^2=3\left(x-y\right)+6y\left(x-y\right)=3\left(x-y\right)\left(2y+1\right)\) \(b.4x^2-8xy-4z^2+4y^2=\left(2x-2y\right)^2-4z^2=\left(2x-2y-2z\right)\left(2x-2y+2z\right)=4\left(x-y-z\right)\left(x-y+z\right)\) \(c.125a^3-8=\left(5a-2\right)\left(25a^2+10a+4\right)\)
\(d.x^2-y^2-x-y=\left(x-y\right)\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-y-1\right)\) \(e.a^2b+ab^2a^3+b^3=b\left(a^2+a^4b+b^2\right)\)
\(3y^3+6xy^2+3x^2y=3y\left(y^2+2xy+x^2\right)=3y\left(x+y\right)^2\)
\(x^3-3x^2-4x+12=x^2\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x^2-4\right)=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
\(x^3+3x^2-3x-1=\left(x-1\right)\left(x^2+x+1\right)+3x\left(x-1\right)=\left(x-1\right)\left(x^2+x+1+3x\right)\)
\(=\left(x-1\right)\left(x^2+4x+1\right)\)
Tham khảo nhé~
a, a(x-y) + bx - by = a(x - y ) +b.(x-y) = (x-y)(a-b)
b, ac+bc + a + b = c.(a+b) +(a+b) = (a+b)(c+1)
c, \(5a^2-5ax-7a+7x=5a.\left(a-x\right)-7.\left(a-x\right)=\left(a-x\right)\left(5a-7\right)\)
d, \(7z^2-7yz-4z+4y=7z.\left(z-y\right)-4.\left(z-y\right)=\left(z-y\right)\left(7z-4\right)\)
e, \(x^3+3x^2+3x+9=x^2.\left(x+3\right)+3\left(x+3\right)=\left(x+3\right)\left(x^2+3\right)\)
g, \(pq-p^2-5\left(p-q\right)=p.\left(q-p\right)+5\left(q-p\right)=\left(q-p\right)\left(p+5\right)\)
a,\(x^3-3x^2+3x-1-y^3=\left(x^3-1\right)-\left(3x^2-3x\right)-y^3\)
\(=\left(x-1\right)\left(x^2+x+1\right)-3x\left(x-1\right)-y^3\)
\(=\left(x-1\right)\left(x^2-2x+1\right)-y^3\)
\(=\left(x-1\right)^3-y^3=\left(x-1-y\right)\left[\left(x-1\right)^2+y\left(x-1\right)+y^2\right]\)
....
\(x^3+4x^2+4x+3\)
\(=x^3+3x^2+x^2+3x+x+3\)
\(=x^2\left(x+3\right)+x\left(x+3\right)+\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2+x+1\right)\)
\(x^2-y^2+4y-4\)
\(=x^2-\left(y^2-4y+4\right)\)
\(=x^2-\left(y-2\right)^2\)
\(=\left(x-y+2\right)\left(x+y-2\right)\)
\(x^4+x^3y-xy^3-y^4\)
\(=x^3\left(x+y\right)-y^3\left(x+y\right)\)
\(=\left(x+y\right)\left(x^3-y^3\right)\)
\(=\left(x+y\right)\left(x-y\right)\left(x^2+xy+y^2\right)\)
Chúc bạn học tốt.
Phân tích đa thức thành nhân tử:
a) Ta có: \(3x^2-8xy+5y^2\)
\(=3x^2-3xy-5xy+5y^2\)
\(=3x\left(x-y\right)-5y\left(x-y\right)\)
\(=\left(x-y\right)\left(3x-5y\right)\)
b) Ta có: \(8xy^3+x\left(x-y\right)^3\)
\(=x\left[8y^3-\left(x-y\right)^3\right]\)
\(=x\left[2y-\left(x-y\right)\right]\left[4y^2+2y\left(x-y\right)+\left(x-y\right)^2\right]\)
\(=x\left(2y-x+y\right)\left(4y^2+2xy-2y^2+x^2-2xy+y^2\right)\)
\(=x\left(3y-x\right)\left(3y^2+x^2\right)\)
c) Ta có: \(2x\left(x-3\right)-x+3\)
\(=2x\left(x-3\right)-\left(x-3\right)\)
\(=\left(x-3\right)\left(2x-1\right)\)
d) Ta có: \(x^4-4x^3+4x^2\)
\(=x^2\left(x^2-4x+4\right)\)
\(=x^2\cdot\left(x-2\right)^2\)
e) Ta có: \(4x^2+4xy-4z^2+y^2-4z-1\)
\(=\left(4x^2+4xy+y^2\right)-\left(4z^2+4z+1\right)\)
\(=\left(2x+y\right)^2-\left(2z+1\right)^2\)
\(=\left(2x+y-2z-1\right)\left(2x+y+2z+1\right)\)
f) Ta có: \(x^2-2xy+y^2-x+y-6\)
\(=\left(x-y\right)^2-\left(x-y\right)-6\)
\(=\left(x-y\right)^2-3\left(x-y\right)+2\left(x-y\right)-6\)
\(=\left(x-y\right)\left(x-y-3\right)+2\left(x-y-3\right)\)
\(=\left(x-y-3\right)\left(x-y+2\right)\)
g) Ta có: \(x^2\left(x+3\right)^2-\left(x+3\right)^2-\left(x^2-1\right)\)
\(=x^2\left(x^2+6x+9\right)-\left(x^2+6x+9\right)-x^2+1\)
\(=\left(x^2-6x+9\right)\left(x^2-1\right)-\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2-6x+9-1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2-6x+8\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x-4\right)\)
a) \(3x-3y+6xy+6y^2=3\left(x-y+2xy+2y^2\right)\)
b) \(4x^2-8xy-4z^2+4y^2=4\left(x^2-2xy+y^2-z^2\right)=4\left(\left(x-y\right)^2-z^2\right)=4\left(x-y-z\right)\left(x-y+z\right)\)
c) \(125a^3-8=\left(5a-2\right)\left(25a^2+10a+4\right)\)
d) \(x^2-y^2-x-y=\left(x-y\right)\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-y-1\right)\)
e) \(a^2b+ab^2+a^3+b^3=ab\left(a+b\right)+\left(a+b\right)\left(a^2-ab+b^2\right)=\left(a+b\right)\left(a^2+b^2\right)\)