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DM
3
DM
4
TN
19 tháng 6 2016
x3-6x2-x+30
=x3-5x2-x2+5x-6x+30
=(x-5)(x2-x-6)
=(x-5)(x-3)(x+2)
HT
19 tháng 6 2016
\(x^3-6x^2-x+30=x^3-3x^2-3x^2+9x-10x+30.\)
\(=x^2\left(x-3\right)-3x\left(x-3\right)-10\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-3x-10\right)\)
\(=\left(x-3\right)\left(x^2+2x-5x-10\right)\)
\(=\left(x-3\right)\left[x\left(x+2\right)-5\left(x+2\right)\right]\)
\(=\left(x-3\right)\left(x+2\right)\left(x-5\right)\)
Vậy \(x^3-6x^2-x+30=\left(x-3\right)\left(x+2\right)\left(x-5\right)\)
NH
2
21 tháng 7 2019
x3 - 6x2 - x + 30
= (x + 2).x2 - 6x2 - x + 30/x + 2
= x2 - 8x + 15
= (x + 2)(x - 3)(x - 5)
KN
21 tháng 7 2019
\(x^3-6x^2-x+30\)
\(=\left(x^3-8x^2+15x\right)+\left(2x^2-16x+30\right)\)
\(=x\left(x^2-8x+15\right)+2\left(x^2-9x+15\right)\)
\(=\left(x^2-8x+15\right)\left(x+2\right)\)
\(=\left(x^2-3x-5x+15\right)\left(x+2\right)\)
\(=\left[x\left(x-3\right)-5\left(x-3\right)\right]\left(x+2\right)\)
\(=\left(x-5\right)\left(x-3\right)\left(x+2\right)\)
KM
1
25 tháng 8 2017
\(\Leftrightarrow x^3-3x^2-3x^2+9x-10x+30\)
\(\Leftrightarrow x^2\left(x-3\right)-3x\left(x-3\right)-10\left(x-3\right)\)
\(\Leftrightarrow\left(x-3\right)\left(x^2-3x-10\right)\)
\(\Leftrightarrow\left(x-3\right)\left(x-5\right)\left(x+2\right)\)
HN
1
TM
18 tháng 8 2017
\(=\left(x^3-2x^2\right)+\left(8x^2-16x\right)+\left(15x-30\right)\)
\(=x^2\left(x-2\right)+8x\left(x-2\right)+15\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+8x+15\right)\)
\(=\left(x-2\right)\left(x^2+3x+5x+15\right)\)
\(=\left(x-2\right)\left[x\left(x+3\right)+5\left(x+3\right)\right]\)
\(=\left(x-2\right)\left(x+3\right)\left(x+5\right)\)
13 tháng 11 2016
a) nhận xét hệ số : 1 + 4 - 29 + 24 = 0
=> x3 + 4x2 - 29x + 24 = x2(x-1) + 5x(x-1) - 24(x-1)
= (x-1)(x2+5x-24) = (x-1)(x-3)(x+8)
b) ...
RZ
13 tháng 11 2016
a) \(x^3+4x^2-29x+24\)=\(\left(x+8\right)\left(x^2-4x+3\right)\)=\(\left(x+8\right)\left(x^2-x-3x+3\right)\)=\(\left(x+8\right)\left(x-1\right)\left(x-3\right)\)
b) \(x^4+6x^3+7x^2-6x+1\)=\(x^4+3x^3-x^2+3x^3+9x^2-3x-x^2-3x+1\)=\(x^2\left(x^2+3x-1\right)+3x\left(x^2+3x-1\right)-\left(x^2+3x-1\right)\)=\(\left(x^2+3x-1\right)\left(x^2+3x-1\right)\)=\(\left(x^2+3x-1\right)^2\)
NB
1
5 tháng 7 2016
a) \(x^2+4x+3=\left(x^2+4x+4\right)-1=\left(x+2\right)^2-1^2=\left(x+1\right)\left(x+3\right)\) (mình sửa lại)
b) \(x^2+8x-9=\left(x^2+8x+16\right)-25=\left(x+4\right)^2-5^2=\left(x-1\right)\left(x+9\right)\)
c) \(3x^2+6x-9=3\left[\left(x^2+2x+1\right)-4\right]=3\left[\left(x+1\right)^2-2^2\right]=3\left(x-1\right)\left(x+3\right)\)
d) \(2x^2+x-3=2x^2-4x+2+5x-5=2\left(x^2-2x+1\right)+5\left(x-1\right)=2\left(x-1\right)^2+5\left(x-1\right)=\left(x-1\right)\left(2x+3\right)\)
VT
0
NM
1
3 tháng 9 2015
ta co: \(F\left(x\right)=x^3-6x^2+11x-6\)
\(=x^3-x^2-5x^2+5x+6x-6\)
\(=x^2\left(x-1\right)-5x\left(x-1\right)+6x\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-5x+6\right)\)
\(=\left(x-1\right)\left(x^2-2x-3x+6\right)\)
\(=\left(x-1\right)\left(x-2\right)\left(x-3\right)\)
\(x^3-6x^2-x+30\\=x^3+2x^2-8x^2-16x+15x+30\\=x^2(x+2)-8x(x+2)+15(x+2)\\=(x+2)(x^2-8x+15)\\=(x+2)(x^2-3x-5x+15)\\=(x+2)[x(x-3)-5(x-3)]\\=(x+2)(x-3)(x-5)\\Toru\)
\(x^3-6x^2-x+30\)
\(=x^3-3x^2-3x^2+9x-10x+30\)
\(=x^2\left(x-3\right)-3x\left(x-3\right)-10\left(x-3\right)\)
\(=\left(x^2-3x-10\right)\left(x-3\right)\)
\(=\left(x^2-5x+2x-10\right)\left(x-3\right)\)
\(=\left(x\left(x-5\right)+2\left(x-5\right)\right)\left(x-3\right)\)
\(=\left(x+2\right)\left(x-5\right)\left(x-3\right)\)