\(M=\left(a^2+b^2-c^2\right)^2-4a^2b^2\)

\(M=\left(a^2+b^2-c^2\right)^2-\left(2ab\right)^2\)

\(M=\left(a^2+b^2-c^2-2ab\right)\left(a^2+b^2-c^2+2ab\right)\)

\(M=\left(\left(a^2-2ab+b^2\right)-c^2\right)\left(\left(a^2+2ab+b^2\right)-c^2\right)\)

\(M=\left(\left(a-b\right)^2-c^2\right)\left(\left(a+b\right)^2-c^2\right)\)

\(M=\left(a-b-c\right)\left(a-b+c\right)\left(a+b-c\right)\left(a+b+c\right)\)

\(M=\left(a^2+b^2-c^2\right)^2-4a^2b^2\)

\(=\left(a^2+b^2-c^2\right)^2-\left(2ab\right)^2\)

\(=\left(a^2+b^2-c^2-2ab\right)\left(a^2+b^2-c^2+2ab\right)\)

\(=\left(\left(a^2-2ab+b^2\right)-c^2\right)\left(\left(a^2+2ab+b^2\right)-c^2\right)\)

\(=\left(\left(a-b\right)^2-c^2\right)\left(\left(a+b\right)^2-c^2\right)\)

\(=\left(a-b-c\right)\left(a-b+c\right)\left(a+b-c\right)\left(a+b+c\right)\)

\(4a^2b^2-\left(a^2+b^2-1\right)^2\)

\(=\left[2ab-\left(a^2+b^2-1\right)\right].\left[2ab+\left(a^2+b^2-1\right)\right]\)

\(=\left(2ab-a^2-b^2+1\right)\left(2ab+a^2+b^2+-1\right)\)

\(=\left[1-\left(a-b\right)^2\right]\left[\left(a+b\right)^2-1\right]\)

\(=\left(1-a+b\right)\left(1+a-b\right)\left(a+b+1\right)\left(a+b-1\right)\)

\(4a^2b^2-\left(a^2+b^2-1\right)^2=\left(2ab+a^2+b^2-1\right)\left(2ab-a^2-b^2+1\right)\)

\(=\left[\left(a+b\right)^2-1\right]\left[1-\left(a-b\right)^2\right]\)

\(=\left(a+b-1\right)\left(a+b+1\right)\left(1+a-b\right)\left(1-a+b\right)\)

4a²b² + 36a²b³ + 6ab⁴

= 2ab²(2a + 18ab + 3b²)

4a²b³ - 6a³b²

= 2a²b²(2b - 3a)

4a²b²-(a²+b²-c²)²

= (4ab-a²-b²+c²)(4ab+a²+b²-c²)

= -[(a-b)²-c²][(a+b)²-c²]

=-(a-b+c)(a-b-c)(a+b-c)(a+b+c)

=(b-a-c)(b+c-a)(a+b-c)(a+b+c)

\(4a^2b^2-\left(a^2+b^2-c^2\right)^2\)

\(=\left(2ab\right)^2-\left(a^2+b^2-c^2\right)^2\)

\(=\left(2ab-a^2-b^2+c^2\right)\left(2ab+a^2+b^2-c^2\right)\)

4a²b² + 36a²b³ + 6ab⁴

= 2ab²(2a + 18ab + 3b²)

3n(m - 3) + 5m(m - 3)

= (3n + 5m)(m - 3)

2a(x - y) - (y - x)

= (x - y)(2a + 1)

4a²b³ - 6a³b²

= 2a²b²(2b - 3a)

e) Ta có: \(a^3-a^2-a+1\)

\(=a^2\left(a-1\right)-\left(a-1\right)\)

\(=\left(a-1\right)\left(a^2-1\right)\)

\(=\left(a-1\right)^2\cdot\left(a+1\right)\)

f) Ta có: \(x^3-2xy-x^2y+2y^2\)

\(=x^2\left(x-y\right)-2y\left(x-y\right)\)

\(=\left(x-y\right)\left(x^2-2y\right)\)

a) \(\left(a^2+b^2\right)^2-4a^2b^2=\left(a^2+b^2+2ab\right)\left(a^2+b^2-2ab\right)=\left(a+b\right)^2.\left(a-b\right)^2\)

b) \(3x^2-3xy-5x+5y=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)

c) \(-x^3+3x^2-3x+1=\left(1-x\right)^3\)

d) Đề sai ko ???

e) \(a^3-a^2-a+1=a^2\left(a-1\right)-\left(a-1\right)=\left(a-1\right)\left(a^2-1\right)=\left(a-1\right)^2\left(a+1\right)\)

f) \(x^3-2xy-x^2y+2y^2=x^2\left(x-y\right)-2y\left(x-y\right)=\left(x-y\right)\left(x^2-2y\right)\)

a, \(=\left(a^2+b^2-2ab\right)\left(a^2+b^2+2ab\right)=\left(\left(a-b\right)\left(a+b\right)\right)^2=\left(a^2-b^2\right)^2\)

\(b,=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)

\(c,=-\left(x^2-3x^2+3x-1\right)=-\left(x-1\right)^3\)

\(d,=2\left(x^2+2xy+y^2-4z^2\right)=2\left(\left(x+y\right)^2-4z^2\right)=2\left(x+y-2z\right)\left(x+y+2z\right)\)

\(e,=a^2\left(a-1\right)-\left(a-1\right)=\left(a-1\right)\left(a^2-1\right)\)

\(f,=x^2\left(x-y\right)-2y\left(x-y\right)=\left(x^2-2y\right)\left(x-y\right)\)

\(a.25^2-4a^2+12ab-9b^2\\ =25^2-\left(4a^2+12ab-9b^2\right)\\ =25^2-\left(2a-3b\right)^2\\ =\left(25-2a+3b\right)\left(25+2a-3b\right)\\ b.x^3+x^2y-xy^2-y^3\\ =x^2\left(x+y\right)-y^2\left(x+y\right)\\ =\left(x+y\right)\left(x^2-y^2\right)\\ =\left(x+y\right)\left(x+y\right)\left(x-y\right)\\ =\left(x+y\right)^2\left(x-y\right)\)

a: Ta có: \(25x^2-4a^2+12ab-9b^2\)

\(=25x^2-\left(2a-3b\right)^2\)

\(=\left(5x-2a+3b\right)\left(5x+2a-3b\right)\)

b: Ta có: \(x^3+x^2y-xy^2-y^3\)

\(=\left(x-y\right)\left(x^2+xy+y^2\right)+xy\left(x-y\right)\)

\(=\left(x-y\right)\left(x+y\right)^2\)