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`a^2 + ab + 2a + 2b = a(a+2) + b(a+2) = (a+b)(a+2)`

\(=b\left(9a^2+6ab+b^2\right)-2b\left(3a+b\right)\)

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

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

\(9a^2b+6ab^2+b^3-6ab-2b^2\)

\(=b\left(9a^2+6ab+b^2-6a-2b\right)\)

\(=b\left[\left(3a+b\right)^2-2\left(3a+b\right)\right]\)

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

\(ab^3c^2-a^2b^2c^2+ab^2c^3-a^2bc^3\)

\(=abc^2\left(b^2-ab+abc-ac\right)\)

a: \(a^2+6ab+9b^2-1\)

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

\(=\left(a+3b+1\right)\left(a+3b-1\right)\)

b: \(4x^2-25+\left(2x+7\right)\left(5-2x\right)\)

\(=\left(2x-5\right)\left(2x+5\right)-\left(2x+7\right)\left(2x-5\right)\)

\(=\left(2x-5\right)\left(2x+5-2x-7\right)\)

\(=-2\left(2x-5\right)\)

c: \(5\left(x+3y\right)-15x\left(x+3y\right)\)

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

\(=-5\left(3x-1\right)\left(x+3y\right)\)

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

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

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

e: \(a^2-6a+9-b^2\)

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

\(=\left(a-3-b\right)\left(a-3+b\right)\)

f: \(x^3-y^3-3x^2+3x-1\)

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

\(=\left(x-1\right)^3-y^3\)

\(=\left(x-1-y\right)\left[\left(x-1\right)^2+y\left(x-1\right)+y^2\right]\)

a: Ta có: \(\left(a^2+b^2-5\right)^2-4\left(ab+2\right)^2\)

\(=\left(a^2+b^2-5-2ab-4\right)\left(a^2+b^2-5+2ab+4\right)\)

\(=\left[\left(a-b\right)^2-9\right]\cdot\left[\left(a+b\right)^2-1\right]\)

\(=\left(a-b-3\right)\left(a-b+3\right)\left(a+b-1\right)\left(a+b+1\right)\)

Bài 4: 

Ta có: \(\left(x^3-x^2\right)-4x^2+8x-4=0\)

\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)