KC
1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
31 tháng 7 2018
\(4x^3-13x^2+9x-18 \)
\(=4x^2\left(x-3\right)-x\left(x-3\right)+6\left(x-3\right)\)
\(=\left(x-3\right)\left(4x^2-x+6\right)\)
TH
3
28 tháng 7 2020
a) 12x3 + 4x2 + 9x + 3 = 4x2(3x + 1) + 3(3x + 1) = (4x2 + 3)(3x + 1)
b) x3 + 2x2 - x - 2 = x2(x + 2) - (x + 2) = (x2 - 1(x + 2) = (x - 1)(x + 1)(x + 2)
c) a3 + (a - b)3 = (a + a - b)[a2 - a(a - b) + (a - b)2] = (2a - b)(a2 - a2 + ab + a2 - 2ab + b2)
= (2a - b)(a2 - ab + b2)
XO
28 tháng 7 2020
a) 12x3 + 4x2 + 9x + 3
= 4x2(3x + 1) + 3(3x + 1)
= (4x2 + 3)(3x + 1)
b) x3 + 2x2 - x - 2
= x2(x + 2) - (x + 2)
= (x2 - 1)(x + 2)
c) a3 + (a - b)3
= a3 - a2(a - b) + a(a - b)2 + (a - b)a2 - (a - b)2a + (a - b)3
= a[(a2 - a(a - b) + (a - b)2] + (a - b)[a2 - a(a - b) + (a - b)2]
= (a + a - b)[(a2 - a(a - b) + (a - b)2]
NH
2
26 tháng 7 2018
\(x^4+x^2+1\)
\(=\left[\left(x^2\right)^2+2x^2.1+1^2\right]-x^2\)
\(=\left(x^2+1\right)^2-x^2\)
\(=\left(x^2+x+1\right)\left(x^2-x+1\right)\)
\(\left(x^2-8\right)^2+36\)
\(=x^4-16x^2+64+36\)
\(=\left[\left(x^2\right)^2-2.10x^2+10^2\right]-\left(2x\right)^2\)
\(=\left(x^2-10\right)^2-\left(2x\right)^2\)
\(=\left(x^2-10-2x\right)\left(x^2-10+2x\right)\)
\(4x^4+81\)
\(=\left[\left(2x^2\right)^2+2.2x^2.9+9^2\right]-\left(6x\right)^2\)
\(=\left(2x^2+9\right)-\left(6x\right)^2\)
\(=\left(2x^2+9-6x\right).\left(2x^2+9+6x\right)\)
Tham khảo nhé~
SL
2
27 tháng 6 2020
Bài làm:
\(8x^3+4x^2-9x+30=8x^3+16x^2-12x^2-24x+15x+30\)
\(=8x^2\left(x+2\right)-12x\left(x+2\right)+15\left(x+2\right)\)
\(=\left(x+2\right)\left(8x^2-12x+15\right)\)
27 tháng 6 2020
\(8x^3+4x^2-9x+30\)
\(=8x^3+16x^2-12x^2-24x+15x+30\)
\(=8x^2\cdot\left(x+2\right)-12x\cdot\left(x+2\right)+15x\cdot\left(x+2\right)\)
\(=\left(x+2\right)\cdot\left(8x^2-12x+15\right)\)
HN
1
18 tháng 8 2017
\(x^3+9x^2+6x-16\)
\(=x^3+x^2-2x+8x^2+8x-16\)
\(=x\left(x^2+x-2\right)+8\left(x^2+x-2\right)\)
\(=\left(x^2+x-2\right)\left(x+8\right)\)
\(=\left(x^2-x+2x-2\right)\left(x+8\right)\)
\(=\left[x\left(x-1\right)+2\left(x-1\right)\right]\left(x+8\right)\)
\(=\left(x-1\right)\left(x+2\right)\left(x+8\right)\)
21 tháng 7 2018
\(x^2-y^2+6x+9=\left(x+3\right)^2-y^2=\left(x+3+y\right)\left(x+3-y\right)\)
21 tháng 7 2018
\(x^3+3x^2-9x-27=\left(x-3\right)\left(x^2+3x+9\right)+3x\left(x-3\right)=\left(x-3\right)\left(x^2+6x+9\right)=\left(x-3\right)\left(x+3\right)^2\)
23 tháng 8 2020
a) -x2 + 2x - 1
= -( x2 - 2x + 1 )
= -( x - 1 )2
b) 12y - 36 - y2
= -( y2 - 12y + 36 )
= -( y - 6 )2
c) -x3 + 9x2 - 27x + 27
= -( x3 - 9x2 + 27x - 27 )
= -( x - 3 )3
d) x3 - 6x2 + 9x
= x( x2 - 6x + 9 )
= x( x - 3 )2
e) a3b - ab3
= ab( a2 - b2 )
= ab( a - b )( a + b )
f) a2 + 2a + 1 - b2
= a2 + ab + a - ab - b2 - b + a + b + 1
= a( a + b + 1 ) - b( a + b + 1 ) + 1( a + b + 1 )
= ( a - b + 1 )( a + b + 1 )
23 tháng 8 2020
a)\(-x^2+2x-1\)
\(=-\left(x^2-2x+1\right)\)
\(=-\left(x-1\right)^2\)
b) \(12y-36-y^2\)
\(=-\left(y^2-12y+36\right)\)
\(=-\left(y^2-2\cdot1\cdot6+6^2\right)\)
\(=-\left(y-6\right)^2\)
c) \(-x^3+9x^2-27x+27\)
\(=-x^3+3x^2+6x^2-18x-9x+27\)
\(=-x^2\left(x-3\right)+6x\left(x-3\right)-9\left(x-3\right)\)
\(=\left(x-3\right)\left(-x^2+6x-9\right)\)
\(=\left(x-3\right)\cdot-\left(x^2-6x+9\right)\)
\(=\left(x-3\right)\cdot-\left(x^2-2\cdot x\cdot3+3^2\right)\)
\(=-\left(x-3\right)\left(x-3\right)^2\)
\(=\left(x-3\right)^3\)
d) \(x^3-6x^2+9\)
\(=x\left(x^2-6x+9\right)\)
\(=x\left(x-3\right)^2\)
e) \(a^3b-ab^3\)
\(=ab\left(a^2-b^2\right)\)
\(=ab\left(a-b\right)\left(a+b\right)\)
f) \(a^2+2a+1-b^2\)
\(=a^2+2\cdot a\cdot1+1^2-b^2\)
\(=\left(a+1\right)^2-b^2\)
\(=\left(a+1-b\right)\left(a+1+b\right)\)
21 tháng 7 2018
\(x^2-4y^2+4y-1=x^2-\left(2y-1\right)^2=\left(x+2y-1\right)\left(x-2y+1\right)\)
21 tháng 7 2018
\(x^4+3x^3-9x-9\)
\(=x^4-9+3x^3-9x\)
\(=\left(x^2-3\right)\left(x^2+3\right)+3x\left(x^2-3\right)\)
\(=\left(x^2-3\right)\left(x^2+3+3x\right)\)
\(=x^3+3x^2-7x^2-21x+12x+36\)
\(=x^2\left(x+3\right)-7x\left(x+3\right)+12\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2-7x+12\right)\)
\(=\left(x+3\right)\left(x^2-4x-3x+12\right)\)
\(=\left(x+3\right)[x\left(x-4\right)-3\left(x-4\right)]\)
\(=\left(x+3\right)\left(x-4\right)\left(x-3\right)\)