![](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)
\(3x^4+6x^3-7x^2+8x-10\)
\(=\left(3x^4-3x^3\right)+\left(9x^3-9x^2\right)+\left(2x^2-2x\right)+\left(10x-10\right)\)
\(=\left(x-1\right)\left(3x^3+9x^2+2x+10\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(-3x^2-7x+10\)
\(=3x-3x^2+10-10x\)
\(=3x.\left(1-x\right)+10.\left(1-x\right)=\left(3x+10\right).\left(1-x\right)\)
\(-3x^2+3y^2-4xz-4yz\)
\(=3\left(y^2-x^2\right)-4z\left(x+y\right)\)
\(=3\left(y-x\right)\left(x+y\right)-4z\left(x+y\right)\)
\(=\left(x+y\right)\left(3y-3x-4z\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(7x-3x^2-2\)
\(=6x+x-3x^2-2\)
\(=\left(6x-3x^2\right)+\left(x-2\right)\)
\(=-3x\left(x-2\right)+\left(x-2\right)\)
\(=\left(-3x+1\right)\left(x-2\right)\)
\(7x - 3x^2 - 2\)
\(= 6x + x - 3x^2 - 2\)
\(=(6x-3x^2)+(x-2)\)
\(= -3x(x-2)+(x-2)\)
\(=(-3x+1)(x-2)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(\dfrac{3x^2-12x+12}{x^4-8x}=\dfrac{3\left(x^2-4x+4\right)}{x\left(x^3-8\right)}=\dfrac{3\left(x-2\right)^2}{x\left(x^3-2^3\right)}=\dfrac{3\left(x-2\right)^2}{x\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{3\left(x-2\right)}{x\left(x^2+2x+4\right)}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3x^3-7x^2+17x-5\)
\(=3x^3-x^2-6x^2+2x+15x-5\)
\(=x^2\left(3x-1\right)-2x\left(3x-1\right)+5\left(3x-1\right)\)
\(=\left(3x-1\right)\left(x^2-2x+5\right)\)
3x^3-7x^2+17x-5
= (3x^3-x^2)-(6x^2-2x)+(15x-5)
= (3x-1).(x^2-6x+15)
Tk mk nha
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(2x^2-5xy+3y^2\)
\(=2x^2-2xy-3xy+3y^2=2x\left(x-y\right)-3y\left(x-y\right)\)
\(=\left(x-y\right)\left(2x-3y\right)\)
\(x^3-7x-6=x^3+1-7x-7\)
\(=\left(x+1\right)\left(x^2-x+1\right)-7\left(x+1\right)=\left(x+1\right)\left(x^2-x-6\right)\)
\(=\left(x+1\right)\left(x-3\right)\left(x+2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(4x^2-3x-1\)
\(=4x^2-4x+x-1\)
\(=4x\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right)\left(4x+1\right)\)
Ta có : \(x^2-7x+12\)
\(=x^2-3x-4x+12\)
\(=x\left(x-3\right)-\left(4x-12\right)\)
\(=x\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-4\right)\left(x-3\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2x^4+3x^3-7x^2-6x+8\)
\(=2x^4+5x^3-2x^2-8x-2x^3-5x^2+2x+8\)
\(=x\left(2x^3+5x^2-2x-8\right)-\left(2x^3+5x^2-2x-8\right)\)
\(=\left(x-1\right)\left(2x^3+5x^2-2x-8\right)\)
\(=\left(x-1\right)\left(2x^3+x^2-4x+4x^2+2x-8\right)\)
\(=\left(x-1\right)\left[x\left(2x^2+x-4\right)+2\left(2x^2+x-4\right)\right]\)
\(=\left(x-1\right)\left(x+2\right)\left(2x^2+x-4\right)\)
\(\Rightarrow3x^2-10x+3x-10\)
\(\Rightarrow x\left(3x-10\right)+\left(3x-10\right)\)
\(\Rightarrow\left(3x-10\right)\left(x+1\right)\)
T I C K nha cảm ơn rất nhìu