![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(x^3+9x^2+27x+27=\left(x+3\right)^3\)
b) \(3\sqrt{3x^3}+18x^2+12\sqrt{3x}+8=\left(\sqrt{3x}+2\right)^3\)
c) \(\dfrac{1}{4}-x^2=\left(\dfrac{1}{2}-x\right)\left(\dfrac{1}{2}+x\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\left(x+2\right)^3+y^3\)
\(=\left(x+2+y\right)\left(x^2+4x+4-xy-2y+y^2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(\left\{{}\begin{matrix}a=x+y\\b=y+z\\c=x+z\end{matrix}\right.\Leftrightarrow x+y+z=\dfrac{a+b+c}{2}\)
\(8\left(x+y+z\right)^3-\left(x+y\right)^3-\left(y+z\right)^3-\left(z+x\right)^3\\ =8\left(\dfrac{a+b+c}{2}\right)^3-a^3-b^3-c^3\\ =\left(a+b+c\right)^3-a^3-b^3-c^3\\ =\left(a+b\right)^3+c^3+3\left(a+b\right)c\left(a+b+c\right)-\left(a+b\right)^3+3ab\left(a+b\right)-c^3\\ =3\left(a+b\right)\left(ac+bc+c^2+ab\right)\\ =3\left(a+b\right)\left(b+c\right)\left(a+c\right)\\ =3\left(x+y+y+z\right)\left(y+z+z+x\right)\left(z+x+x+y\right)\\ =3\left(x+2y+z\right)\left(x+y+2z\right)\left(2x+y+z\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
1: \(x-9=\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\)
2: \(x-16=\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)\)
3: \(9x-1=\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)\)
4: \(x\sqrt{x}+1=\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)\)
\(1,x-9=\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\\ 2,x-16=\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)\\ 3,9x-1=\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)\\ 4,x\sqrt{x}+1=\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
=(x^3+6x^2+12x+8)+y^3
=(x^3+3x^2+3x2^2+2^3)+y^3
=(x+2)^3+y^3
=(x+2+y)((x+2)^2-(x+2)y+y^2)
=(x+2+y)(x^2+4x+4-xy-2y+y^2)
=(x+2+y)(x^2+y^2-xy+4x-2y+4)
\(8+3\sqrt{3}x^3\)
\(=2^3+\left(1\sqrt{3}x\right)^3\)
\(=\left(2+1\sqrt{3}x\right)\left(4-2\sqrt{3}x+3x^2\right)\)
(Nhớ k cho mình với nhá!)