\(4\left(x+5\right)\left(x+6\right)\left(x+10\right)\left(x+12\right)-3x^2\)

\(=4\left[\left(x+5\right)\left(x+12\right)\right]\left[\left(x+6\right)\left(x+10\right)\right]-3x^2\)

\(=4\left(x^2+17x+60\right)\left(x^2+16x+60\right)-3x^2\)

\(=\left(2x^2+34x+120\right)\left(2x^2+32x+60\right)-3x^2\)

\(=\left(2x^2+33x+120\right)^2-x^2-3x^2\)

\(=\left(2x^2+33x+120-2x\right)\left(2x^2+33x+120+2x\right)\)

\(=\left(2x+15\right)\left(x+8\right)\left(2x^2+35x+120\right)\)

\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

Ta có: \(2x^3+3x^2+2x+3=0\)

\(\Leftrightarrow x^2\left(2x+3\right)+\left(2x+3\right)=0\)

\(\Leftrightarrow2x+3=0\)

hay \(x=-\dfrac{3}{2}\)

\(4\left(x+5\right)\left(x+12\right)\left(x+6\right)\left(x+10\right)-3x^2\)

\(=2\left(x^2+60+17x\right).2\left(x^2+60+16x\right)-3x^2\)

\(=\left(2x^2+120+33x+x\right)\left(2x^2+120+33x-x\right)-3x^2\)

\(=\left(2x^2+120+33x\right)^2-x^2-3x^2\)

\(=\left(2x^2+120+33x\right)^2-4x^2\)

\(=\left(2x^2+120+33x+2x\right)\left(2x^2+120+33x-2x\right)\)

\(=\left(2x^2+35x+120\right)\left(2x^2+31x+120\right)\)

\(=\left(2x^2+35x+120\right)\left(x+8\right)\left(2x+15\right)\)

4x²-3x-1=(3x²-3x)+(x²-1)=3x(x-1)+(x-1)(x+1)=(x-1)(3x+x+1)=(x-1)(4x+1)

a) \(x^3+6x^2+3x-10\)

\(=x^3-x^2+7x^2-7x+10x-10\)

\(=x^2\left(x-1\right)+7x\left(x-1\right)+10\left(x-1\right)\)

\(=\left(x-1\right)\left(x^2+7x+10\right)\)

\(=\left(x-1\right)\left(x^2+2x+5x+10\right)\)

\(=\left(x-1\right)\left(x+2\right)\left(x+5\right)\)

b)  \(x^3+3x^2-33x-35\)

\(=x^3-5x^2+8x^2-40x+7x-35\)

\(=x^2\left(x-5\right)+8x\left(x-5\right)+7\left(x-5\right)\)

\(=\left(x-5\right)\left(x^2+8x+7\right)\)

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

\(=\left(x-5\right)\left(x+1\right)\left(x+7\right)\)