Phân tích đa thức thành nhân tử:
a2 - a - 6
2x2 + x - 6
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a^2-9-8ab+16b^2\)
\(=a^2-8ab+16b^2-9\)
\(=\left(a-4b\right)^2-9\)
\(=\left(a-4b-3\right)\left(a-4b+3\right)\)
a2 - 9 - 8ab + 16b2
⇔ (a2 - 8ab + 16b2) - 9
⇔ (a - 4b)2 - 32
⇔ (a - 4b - 3)(a - 4b + 3)
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
\(x^2+x-6\)
\(=x^2-2x+3x-6\)
\(=x\left(x-2\right)+3.\left(x-2\right)\)
\(=\left(x+3\right)\left(x-2\right)\)
\(x^6-y^6\)
\(=\left(x^3-y^3\right)\left(x^3+y^3\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(x-6\sqrt{x-3}+6\text{=}x-3-6\sqrt{x-3}+9\)
\(\text{=}\left(\sqrt{x-3}\right)^2-2.3.\sqrt{x-3}+\left(3\right)^2\)
\(\text{=}\left(\sqrt{x-3}-3\right)^2\)
A = \(x-6\)\(\sqrt{x-3}\) + 6 (đkxd \(x>3\))
A = (\(x\) - 3) - 2.3.\(\sqrt{x-3}\) + 9
A = (\(\sqrt{x-3}\))2 - 2.3.\(\sqrt{x-3}\) + 32
A = (\(\sqrt{x-3}\)- 3)2
Lời giải:
$x-5\sqrt{x}+6=x-2\sqrt{x}-3\sqrt{x}+6$
$=\sqrt{x}(\sqrt{x}-2)-3(\sqrt{x}-2)$
$=(\sqrt{x}-2)(\sqrt{x}-3)$
\(a^2-a-6=a^2-3a+2a-6=\left(a-3\right)\left(a+2\right)\\ 2x^2+x-6=2x^2+4x-3x-6=\left(x+2\right)\left(2x-3\right)\)