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

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

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

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

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

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

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

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

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

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

a: =x^2+2xy+y^2-4x^2y^2

=(x+y)^2-(2xy)^2

=(x+y+2xy)(x+y-2xy)

b: =49-(a^2-2ab+b^2)

=49-(a-b)^2

=(7-a+b)(7+a-b)

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

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

d: 

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

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

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

=2(a+b)^2-2c^2

=2[(a+b)^2-c^2]

=2(a+b-c)(a+b+c)

(b-a)*(c-a)*(c-b)*(c+b+a)

a(b+c)^2(b-c)+b(c+a)^2(C-a)+c(a+b)^2(a-b)

=a(b+c)^2(b-a)+bc(c+a)^2-ab(c+a)^2+ac(a+b^2-cb(a+b)^2

=a(b+c)^2(b-c)+cd(c-d)(2a+b+c)-a(bc^2+ba^2-ca^2-cb^2)

=(b-c)(ab^2+ac^2+2abc-2abc-cb^2-c^2b)a(b-c)(a^2-cb)

=(b-c)(ab^2+ac^2-ab^2-c^2b-a^3+abc)

=(b-c)(a-c)[b^2-(ac+a^2)+cb]

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

Ta có: \(\left(a-b\right)\left(b-c\right)\left(a-c\right)+\left(a+b\right)\left(b+c\right)\left(a-c\right)+\left(a+b\right)\left(a+c\right)\left(c-b\right)\)

    \(=\left(a-c\right).\left[\left(a-b\right)\left(b-c\right)+\left(a+b\right)\left(b+c\right)\right]+\left(a+b\right)\left(a+c\right)\left(c-b\right)\)

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

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

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

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

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

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

    \(=\left(a+c\right)\left[a.\left(b+c\right)-b.\left(b+c\right)\right]\)

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

\(x^2-10x+16=x^2-8x-2x+16=x\left(x-8\right)-2\left(x-8\right)=\left(x-8\right)\left(x-2\right)\)

\(x^2-2x-15=x^2-5x+3x-15=x\left(x-5\right)+3\left(x-5\right)=\left(x-5\right)\left(x+3\right)\)

\(2x^2+7x+3=2x^2+x+6x+3=x\left(2x+1\right)+3\left(2x+1\right)=\left(x+3\right)\left(2x+1\right)\)

a) \(x^2-10x+16=x^2-8x-2x+16=\left(x^2-8x\right)-\left(2x-16\right)=x\left(x-8\right)-2\left(x-8\right)=\left(x-8\right)\left(x-2\right)\)b) \(x^2-2x-15=x^2+3x-5x-15=\left(x^2+3x\right)-\left(5x+15\right)=x\left(x+3\right)-5\left(x+3\right)=\left(x+3\right)\left(x-5\right)\)c) \(2x^2+7x+3=2x^2+x+6x+3=\left(2x^2+x\right)+\left(6x+3\right)=x\left(2x+1\right)+3\left(2x+1\right)=\left(2x+1\right)\left(x+3\right)\)

(a+b+c)^3 thì viết được thành [(a+b)+c)]^3 rồi AD hằng đẳng thức để tính. Còn với (a^3+b^3+c^3) ta viết được (a+b)^3 -3a^2b -3ab^2 + c^3=(a+b)^3 -3ab(a+b)+c^3 ...thay vào rồi đổi biến

 k bt nhoak