Bài 2. Phân tích các đa thức sau thành nhân tử: x^5+27x^2
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\(x^4+x^2-27x-9\)
\(=x^4-27x+\left(x-3\right)\left(x+3\right)\)
\(=x\left(x^3-27\right)+\left(x-3\right)\left(x+3\right)\)
\(=x\left(x-3\right)\left(x^2+3x+9\right)+\left(x-3\right)\left(x+3\right)\)
\(=\left(x-3\right)\left(x^3+3x^2+10x+3\right)\)
\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
a.
\(1-4x^2=\left(1-2x\right)\left(1+2x\right)\)
b.
\(8-27x^3=\left(2\right)^3-\left(3x\right)^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)
c.
\(27+27x+9x^2+x^3=x^3+3.x^2.3+3.3^2.x+3^3\)
\(=\left(x+3\right)^3\)
d.
\(2x^3+4x^2+2x=2x\left(x^2+2x+1\right)=2x\left(x+1\right)^2\)
e.
\(x^2-y^2-5x+5y=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-5\right)\)
f.
\(x^2-6x+9-y^2=\left(x-3\right)^2-y^2=\left(x-3-y\right)\left(x-3+y\right)\)
Sửa đề: \(A=10x^4-27x^3y-110x^2y^2-27xy^3+10y^4\)
\(=10x^4+20x^2y^2+10y^4-27xy\left(x^2+y^2\right)-130x^2y^2\)
\(=10\left(x^4+2x^2y^2+y^4\right)-27xy\left(x^2+y^2\right)-130x^2y^2\)
\(=10\cdot\left(x^2+y^2\right)^2-27xy\left(x^2+y^2\right)-130x^2y^2\)
\(=10\left(x^2+y^2\right)^2-52xy\left(x^2+y^2\right)+25xy\left(x^2+y^2\right)-130x^2y^2\)
\(=2\left(x^2+y^2\right)\left\lbrack5\left(x^2+y^2\right)-26xy\right\rbrack+5xy\left\lbrack5\left(x^2+y^2\right)-26xy\right\rbrack\)
\(=\left(5x^2-26xy+5y^2\right)\left(2x^2+5xy+2y^2\right)\)
\(=\left(5x^2-25xy-xy+5y^2\right)\left(2x^2+4xy+xy+2y^2\right)\)
\(=\left\lbrack5x\left(x-5y\right)-y\left(x-5y\right)\right\rbrack\left\lbrack2x\left(x+2y\right)+y\left(x+2y\right)\right\rbrack\)
=(x-5y)(5x-y)(2x+y)(x+2y)
a) \(27x^3+27x^2+9x+1=\left(3x+1\right)^3\)
b) \(-x^3-3x^2-3x-1=-\left(x^3+3x^2+3x+1\right)=-\left(x+1\right)^3\)
c) \(-8+12x-6x^2+x^3=\left(x-2\right)^3\)


x5 + 27x2
= x2(x3 + 27)
= x2(x + 3)(x2 - 3x + 9)