3 . (x+3) - x²+9

= 3.(x+3)-(x²-9)

=3.(x+3)-[(x-3).(x+3)]

=(x+3).[3-(x-3)]

=(x+3).(3-x+3)

=(x+3).(9-x)

nhân tử gì ở đây

\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)

x³ + 27 + (x + 3)(x - 9)

= (x + 3)(x² + 3x + 9) + (x + 3)(x - 9)

= (x + 3)(x² + 3x + 9 + x - 9)

= (x + 3)(x² + 4x)

= x(x + 3)(x + 4)

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

\(x^2-9-\left(x-3\right)\left(5-2x\right)=\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(5-2x\right)=\left(x-3\right)\left(x+3-5+2x\right)=\left(x-3\right)\left(3x-2\right)\)

\(x^2-9-\left(x-3\right)\left(5-2x\right)\\ =\left(x^2-9\right)-\left(x-3\right)\left(5-2x\right)\\ =\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(5-2x\right)\\ =\left(x-3\right)\left(x+3-5+2x\right)\\ =\left(x-3\right)\left(3x-2\right)\)

Bạn nên viết lại đa thức bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để được hỗ trợ tốt hơn.

Ta có \(f\left(x\right)=\left(x+1\right)\left(x-3\right)\left(x+5\right)\left(x+9\right)+256\)

\(f\left(x\right)=\left(x+1\right)\left(x+5\right)\left(x-3\right)\left(x+9\right)+256\)

\(f\left(x\right)=\left(x^2+6x+5\right)\left(x^2+6x-27\right)+256\)

\(f\left(x\right)=\left[\left(x^2+6x-11\right)^2-256\right]+256\)

\(f\left(x\right)=\left(x^2+6x-11\right)^2\)

\(9(x-y)^2-27(y-x)^3\\=9(x-y)^2+27(x-y)^3\\=(x-y)^2[9+27(x-y)]\\=(x-y)^2(9+27x-27y)\\=9(x-y)^2(1+3x-3y)\)

x⁹+x³+x²+x¹

= (x⁹-x⁶)+(x⁶-x³)+x²+x+1

= x⁶(x³-1)+x³(x³-1)+x²+x+1

= x⁶(x-1)(x²+x+1)+x³(x-1)(x²+x+1)+(x²+x+1)

= (x²+x+1)(x⁷-x⁶+x⁴-x³+1)