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a) \(x^5-2x^4+3x^3-4x^2+2\)
\(=x^5-x^4-x^4+x^3+2x^3-2x^2-2x^2+2\)
\(=x^4\left(x-1\right)-x^3\left(x-1\right)+2x^2\left(x-1\right)-2\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)\left(x^4-x^3+2x^2-2x-2\right)\)
b) \(x^4+1997x^2+1996x+1997\)
\(=\left(x^4+x^2+1\right)+1996\left(x^2+x+1\right)\)
\(=\left(x^2-x+1\right)\left(x^2+x+1\right)+1996\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+1997\right)\)
c) \(x^8+x^4+1\)
\(=x^8+2x^4+1-x^4\)
\(=\left(x^4+1\right)-x^4\)
\(=\left(x^4-x^2+1\right)\left(x^4+x^2+1\right)\)
\(=\left(x^4-x^2+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)\)
c) \(x^5+x+1\)
\(=x^5-x^2+x^2+x+1\)
\(=x^2\left(x^3-1\right)+\left(x^2+x+1\right)\)
\(=x^2\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^3-x^2+1\right)\)
x4-3x3-x+3 = (x4-3x3)-(x-3) = x3(x-3)-(x-3) = (x-3)(x3-1) = (x-3)(x-1)(x2+x+1)
3x+3y-x2-2xy-y2 = (3x+3y)-(x2+2xy+y2) = 3(x+y)-(x+y)2 = (x+y)( 3-x-y)
x2-x-12 = x(x-1)-12
b, \(\left(x^2+x\right)^2+4x^2+4x-12=x^4+2x^3+x^2+4x^2+4x-12\)
\(=x^4+2x^3+5x^2+4x-12\)
\(=\left(x^4-x^3\right)+\left(3x^3-3x^2\right)+\left(8x^2-8x\right)+\left(12x-12\right)\)
\(=x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)\)
\(=\left(x^3+3x^2+8x+12\right)\left(x-1\right)\)
\(=\left[\left(x^3+2x^2\right)+\left(x^2+2x\right)+\left(6x+12\right)\right]\left(x-1\right)\)
\(=\left[x^2\left(x+2\right)+x\left(x+2\right)+6\left(x+2\right)\right]\left(x-1\right)\)
\(=\left(x^2+x+6\right)\left(x+2\right)\left(x-1\right)\)
c, \(x^3+3x^2-4=\left(x^3+2x^2\right)+\left(x^2+2x\right)-\left(2x+4\right)\)
\(=x^2\left(x+2\right)+x\left(x+2\right)-2\left(x+2\right)\)
= \(\left(x^2+x-2\right)\left(x+2\right)\)
a)\(x^5+x^4+1=x^5-\left(-x^3+x^3\right)+x^4+\left(x^2-x^2\right)+\left(x-x\right)+1\)
\(=x^5-x^3+x^2+x^4-x^2+x+x^3-x+1\)
\(=x^2\left(x^3-x+1\right)+x\left(x^3-x+1\right)+\left(x^3-x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^3-x+1\right)\)
b,c có ng lm rồi
d)\(2x^4-3x^3-7x^2+6x+8\)
Ta thấy x=-1 là nghiệm của đa thức
=>đa thức có 1 hạng tử là x+1
\(\Rightarrow\left(x+1\right)\left(2x^3-5x^2-2x+8\right)\)
\(\Rightarrow\left(x+1\right)\left[2x^3-x^2-4x-4x^2+2x+8\right]\)
\(\Rightarrow\left(x+1\right)\left[x\left(2x^2-x-4\right)-2\left(2x^2-x-4\right)\right]\)
\(\Rightarrow\left(x+1\right)\left(x-2\right)\left(2x^2-x-4\right)\)
phần còn lại bạn tự lo nhé
a) \(45+x^3-5x^2-9x\)
\(=\left(x^3-5x^2\right)-\left(9x-45\right)\)
\(=x^2\left(x-5\right)-9\left(x-5\right)\)
\(=\left(x-5\right)\left(x^2-9\right)=\left(x-5\right)\left(x-3\right)\left(x+3\right)\)
\(a,45+x^3-5x^2-9x\)
\(=\left(x^3-5x^2\right)+\left(45-9x\right)\)
\(=x^2\left(x-5\right)+9\left(5-x\right)\)
\(=x^2\left(x-5\right)-9\left(x-5\right)\)
\(=\left(x-5\right)\left(x^2-9\right)\)
\(=\left(x-5\right)\left(x^2-3^2\right)\)
\(=\left(x-5\right)\left(x+3\right)\left(x-3\right)\)
\(c,2x^2+3x-5\)
\(=2x^2-2x+5x-5\)
\(=2x\left(x-1\right)+5\left(x-1\right)\)
\(=\left(x-1\right)\left(2x+5\right)\)
\(e,\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)
\(=\left[\left(x+2\right)\left(x+8\right)\right]\left[\left(x+4\right)\left(x+6\right)\right]+16\)
\(=\left(x^2+8x+2x+16\right)\left(x^2+6x+4x+24\right)+16\)
\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\left(1\right)\)
\(\text{Đặt }x^2+10x+\frac{16+24}{2}=t\)
\(\text{hay }x^2+10x+20=t\left(2\right)\)
(1)\(\Leftrightarrow\left(t-4\right)\left(t+4\right)+16\)
\(=t^2-4^2+16\)
\(=t^2-16+16\)
\(=t^2\left(3\right)\)
Thay (3) vào (2),ta được:\(\left(x^2+10x+20\right)^2\)
a) \(x^2-6x+8\)
\(=x^2-2\cdot x\cdot3+3^2-1\)
\(=\left(x-3\right)^2-1^2\)
\(=\left(x-3-1\right)\left(x-3+1\right)\)
\(=\left(x-4\right)\left(x-2\right)\)
Còn lại tương tự
a) \(x^2-6x+8=x^2-2x-4x+8\)
\(=\left(x^2-2x\right)-\left(4x-8\right)\)
=x(x-2)-4(x-2) = (x-2)(x-4)