Phân tích đa thức thành nhân tử
1. x^7 + x^2 + 1
2. x^7 + x^5 +1
3. x^4 + 6x^3 + 7x^2 - 6x+1
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1. \(=3x\left(2x+5\right)\)
2. \(=\left(3x-1\right)\left(3x+1\right)\)
3. \(=\left(x+3\right)^2-y^2=\left(x-y+3\right)\left(x+y+3\right)\)
1, = 3x.(2x + 5)
2. = (3x)2 - 12 = (3x - 1).(3x +1 )
3, =(x2 + 6x + 9) - y2 = (x + 3)2 - y2 =(x + y -3 ). (x - y +3)
Ta có: \(1+6x-6x^2-x^3\)
\(=-x^3-6x^2+6x+1\)
\(=\left(-x^3+1\right)-6x\left(x-1\right)\)
\(=-\left(x-1\right)\left(x^2+x+1\right)-6x\cdot\left(x-1\right)\)
\(=\left(x-1\right)\left(-x^2-x-1-6x\right)\)
\(=-\left(x-1\right)\left(x^2+7x+1\right)\)
1) \(2xy^3-6x^2+10xy\)
\(=2x.y^3-2x.3x+2x.5y\)
\(=2x\left(y^3-3x+5y\right)\)
\(=2x[y\left(y^2-5\right)-3x]\)
1: \(x^2-3x+2=\left(x-1\right)\left(x-2\right)\)
2: \(x^2-x-6=\left(x-3\right)\left(x+2\right)\)
3: \(x^2+7x+12=\left(x+3\right)\left(x+4\right)\)
1) \(x^2-3x+2=\left(x^2-x\right)-\left(2x-2\right)=x\left(x-1\right)-2\left(x-1\right)=\left(x-1\right)\left(x-2\right)\)
2) \(x^2-x-6=\left(x^2-3x\right)+\left(2x-6\right)=x\left(x-3\right)+2\left(x-2\right)=\left(x-2\right)\left(x-3\right)\)
3) \(x^2+7x+12=\left(x^2+3x\right)+\left(4x+12\right)=x\left(x+3\right)+4\left(x+3\right)=\left(x+3\right)\left(x+4\right)\)
1: \(x^2-3x+2=\left(x-1\right)\left(x-2\right)\)
2: \(x^2-x-6=\left(x-3\right)\left(x+2\right)\)
3: \(x^2+7x+12=\left(x+3\right)\left(x+4\right)\)
2: =(2x+1)^2-y^2
=(2x+1+y)(2x+1-y)
3: =x^2(x^2+2x+1)
=x^2(x+1)^2
4: =x^2+6x-x-6
=(x+6)(x-1)
5: =-6x^2+3x+4x-2
=-3x(2x-1)+2(2x-1)
=(2x-1)(-3x+2)
6: =5x(x+y)-(x+y)
=(x+y)(5x-1)
7: =2x^2+5x-2x-5
=(2x+5)(x-1)
8: =(x^2-1)*(x^2-4)
=(x-1)(x+1)(x-2)(x+2)
9: =x^2(x-5)-9(x-5)
=(x-5)(x-3)(x+3)