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MM
2
29 tháng 9 2018
\(d, x^3+4x^2-29x+24\)
\(=x^3+8x^2-4x^2-32x+3x+24\)
\(=x^2\left(x+8\right)-4x\left(x+8\right)+3\left(x+8\right)\)
\(=\left(x+8\right)\left(x^2-4x+3\right)\)
\(=\left(x+8\right)\left(x^2-x-3x+3\right)\)
\(=\left(x+8\right)\left[x\left(x-1\right)-3\left(x-1\right)\right]\)
\(=\left(x+8\right)\left(x-1\right)\left(x-3\right)\)
\(e, x^3+6x^2+11x+6\)
\(=x^3+3x^2+3x^2+9x+2x+3\)
\(=x^2\left(x+3\right)+3x\left(x+3\right)+2\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2+3x+2\right)\)
\(=\left(x+3\right)\left(x^2+x+2x+2\right)\)
\(=\left(x+3\right)\left[x\left(x+1\right)+2\left(x+1\right)\right]\)
\(=\left(x+3\right)\left(x+1\right)\left(x+2\right)\)
KB
29 tháng 9 2018
Mik ko giỏi p/t bậc 3 cho lắm nên có sai thì mong bạn sửa giúp :
d ) \(x^3+4x^2-29x+24\)
\(=x^2\left(x-1\right)+5x^2-10x+5-19x+19\)
\(=x^2\left(x-1\right)+5\left(x-1\right)^2-19\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+5x-5-19\right)\)
\(=\left(x-1\right)\left(x^2+5x-14\right)\)
\(=\left(x-1\right)\left(x^2+7x-2x-14\right)\)
\(=\left(x-1\right)\left(x+7\right)\left(x-2\right)\)
e ) \(x^3+6x^2+11x+6\)
\(=x^2\left(x+1\right)+5x^2+10x+5+x+1\)
\(=x^2\left(x+1\right)+5\left(x+1\right)^2+x+1\)
\(=\left(x+1\right)\left(x^2+5x+5+1\right)\)
\(=\left(x+1\right)\left(x^2+5x+6\right)\)
PH
2
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) \(x^4+6x^3+7x^2-6x+1\)=\(x^4+3x^3-x^2+3x^3+9x^2-3x-x^2-3x+1\)=\(x^2\left(x^2+3x-1\right)+3x\left(x^2+3x-1\right)-\left(x^2+3x-1\right)\)=\(\left(x^2+3x-1\right)\left(x^2+3x-1\right)\)=\(\left(x^2+3x-1\right)^2\)
CC
0
LN
3
5 tháng 7 2018
\(x^3+6x^2+11x+6=x^3+x^2+5x^2+5x+6x+6\)
\(=x^2\left(x+1\right)+5x\left(x+1\right)+6\left(x+1\right)=\left(x+1\right)\left(x^2+5x+6\right)\)
\(=\left(x+1\right)\left(x^2+2x+3x+6\right)=\left(x+1\right)\left[x\left(x+2\right)+3\left(x+2\right)\right]\)
\(=\left(x+1\right)\left(x+2\right)\left(x+3\right)\)
FF
0
VT
1
ZR
12 tháng 10 2015
x^3 + 6x^2 + 11x + 6
= x^3 + x^2 + 5x^2 + 5x + 6x + 6
= x^2(x + 1) + 5x(x + 1) + 6(x + 1)
= (x + 1)(x^2 + 5x + 6)
= (x + 1)(x^2 + 2x + 3x + 6)
= (x + 1)[x(x + 2) + 3(x + 2)
= (x + 1)(x + 2)(x + 3)
a, \(x^3+4x^2-29x+24\)
\(=x^3-x^2+5x^2-5x-24x+24\)
\(=x^2\left(x-1\right)+5x\left(x-1\right)-24\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+5x-24\right)\)
\(=\left(x-1\right)\left[x\left(x-3\right)+8\left(x-3\right)\right]\)
\(=\left(x-1\right)\left(x-3\right)\left(x+8\right)\)
\(x^3+6x^2+11x+6\)
\(=x^3+x^2+5x^2+5x+6x+6\)
\(=x^2\left(x+1\right)+5x\left(x+1\right)+6\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+5x+6\right)\)
\(=\left(x+1\right)\left[x\left(x+2\right)+3\left(x+2\right)\right]\)
\(=\left(x+1\right)\left(x+2\right)\left(x+3\right)\)
Chúc bạn học tốt.