NH
3
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
25 tháng 7 2017
Ta có : x3 - 4x2 + 12x - 27
= (x3 - 27) - (4x2 - 12x)
= (x3 - 33) - 4x(x - 3)
= (x - 3)(x2 + 3x + 9) - 4x(x - 3)
= (x - 3) (x2 + 3x + 9 - 4x)
= (x - 3)(x2 - x + 9)
30 tháng 9 2018
\(x^4+2x^3+2x^2+2x+1\)
\(=\left(x^4+2x^3+x^2\right)+\left(x^2+2x+1\right)\)
\(=\left(x^2+x\right)^2+\left(x+1\right)^2\)
\(=x^2\left(x+1\right)^2+\left(x+1\right)^2\)
\(=\left(x+1\right)^2\left(x^2+1\right)\)
BM
25 tháng 7 2017
a, \(x^3-4x^2+12x-27\) \(=x^3-3x^2-x^2+3x+9x-27\)
= \(x^2\left(x-3\right)-x\left(x-3\right)+9\left(x-3\right)\) \(=\left(x-3\right)\left(x^2-x+9\right)\)
b, \(x^3+2x^2+2x+1\) \(=x^3+x^2+x^2+x+x+1\)
= \(x^2\left(x+1\right)+x\left(x+1\right)+x+1=\) \(\left(x+1\right)\left(x^2+x+1\right)\)
c, \(x^4-2x^3+2x-1=\) \(x^4-x^3-x^3+x^2-x^2+x+x-1\)
= \(x^3\left(x-1\right)-x^2\left(x-1\right)-x\left(x-1\right)+x-1\)
= \(\left(x-1\right)\left(x^3-x^2-x+1\right)\)
d, \(x^4+2x^3+2x^2+2x+1=\) \(x^4+x^3+x^3+x^2+x^2+x+x+1\)
= \(x^3\left(x+1\right)+x^2\left(x+1\right)+x\left(x+1\right)+x+1\)
= \(\left(x+1\right)\left(x^3+x^2+x+1\right)\)
25 tháng 7 2017
Ta có : x3 - 4x2 + 12x - 27
= (x3 - 27) - (4x2 - 12x)
= (x - 3)(x2 + 3x + 9) - 4x(x - 3)
= (x - 3)(x2 + 3x + 9 - 4x)
= (x - 3)(x2 - x + 9)
b) https://olm.vn/hoi-dap/question/1004349.html tôi tự coppy tôi
A
1 tháng 10 2020
b, \(x^3+2x^2+2x+1=\left(x^2+x+1\right)\left(x+1\right)\)
c, \(x^3-4x^2+12x-27=\left(x^2-x+9\right)\left(x-3\right)\)
d, \(x^4-2x^3+2x-1=\left(x-1\right)^3\left(x+1\right)\)
e, sai đề
1 tháng 10 2020
a, \(\left(ab-1\right)^2+\left(a+b\right)^2=\left(a^2+1\right)\left(b^2+1\right)\)
b, \(x^3+2x^2+2x+1=\left(x+1\right)\left(x^2+x+1\right)\)
c, \(x^3-4x^2+12x-27=\left(x-3\right)\left(x^2-x+9\right)\)
d, \(x^4-2x^3+2x-1=\left(x-1\right)^3\left(x+1\right)\)
e, cho mình sửa đề xíu
\(x^4+2x^3+2x^2+2x+1=\left(x+1\right)^2\left(x^2+1\right)\)
1 tháng 10 2020
a)(ab−1)2+(a+b)2
=a2b2−2ab+1+a2+2ab+b2
=a2b2+1+a2+b2=a2(b2+1)+(b2+1) = (a2+1)(b2+1)
c)x3−4x2+12x−27
=x3−27+(−4x2+12x)
=(x−3)(x2+3x+9)−4x(x−3)
=(x−3)(x2+3x+9−4x)
=(x−3)(x2−x+9)
b)x3+2x2+2x+1
=x3+2x2+x+x+1
=x(x2+2x+1)+(x+1)
=x(x+1)2+(x+1)
=(x+1)(x(x+1)+1)
=(x+1)(x2+x+1)
d)x4−2x3+2x−1
=x4−2x3+x2−x2+2x−1
=x2(x2−2x+1)−(x2−2x+1)
=(x2−2x+1)(x2−1)
=(x−1)2(x−1)(x+1)
=(x−1)3(x+1)
e)x4+2x3+2x2+2x+1
=x4+2x3+x2+x2+2x+1
=x2(x2+2x+1)+(x2+2x+1)
=(x2+2x+1)(x2+1)
=(x+1)2(x2+1)
8 tháng 8 2016
Đây là cách hiện đại :
\(x^4-2x^3+2x-1\)
\(=\left(x^4-1\right)-\left(2x^3-2x\right)\)
\(=\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(\left(x^2+1\right)-2x\right)\)
\(=\left(x+1\right)\left(x-1\right)\left(\left(x^2+1\right)-2x\right)\)
TH
7 tháng 8 2016
a,=\(x^4-x^3-x^3+x^2-x^2+x+x-1\)
cu hai so nhom 1 nhom roi dat thua so chung la xong
b,x^4+x^3+x^3+x^2+x^2+x+x+1
cu hai so lai nhom 1 nhom va dat thua so chung
TT
1
15 tháng 6 2018
a,\(=x^4-3x^3+3x^3-9x^2-4x^2+12x-12x+36\)
\(=x^3\left(x-3\right)+3x^2\left(x-3\right)-4x\left(x-3\right)-12\left(x-3\right)\)
\(=\left(x-3\right)\left(x^3+3x^2-4x-12\right)\)
\(=\left(x-3\right)[x^2\left(x+3\right)-4\left(x+3\right)]\)
\(=\left(x^2-9\right)\left(x^2-4\right)\)
22 tháng 6 2016
x4 + 2x3 + 2x2 + 2x + 1
= x4 - 2x2 =
= x2 x x2 - x2 - x2 + 1 = x2 (1- x2 ) + ( 1 - x2 )
= ( 1 - x2 ) x ( 1 - x2 )
= ( 1 - x2 ) 2
22 tháng 6 2016
- SKT_Twisted Fate Âm Phủ
- Sai rồi :
- \(x^4-2x^2=?\)
a)\(x^4-2x^3+2x-1=x^4-x^3-x^3+x+x-1\)
\(=x^3\left(x-1\right)-x\left(x^2-1\right)+\left(x-1\right)\)
\(=x^3\left(x-1\right)-x\left(x-1\right)\left(x+1\right)+\left(x-1\right)\)
\(=x^3\left(x-1\right)-\left(x^2+x\right)\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right)\left[x^3-\left(x^2+x\right)+1\right]\)
\(=\left(x-1\right)\left(x^3-x^2-x+1\right)\)
\(=\left(x-1\right)\left[x^2\left(x-1\right)-\left(x-1\right)\right]\)
\(=\left(x-1\right)\left(x-1\right)\left(x^2-1\right)\)
\(=\left(x-1\right)\left(x-1\right)\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)^3\left(x+1\right)\)
b)\(x^4+2x^3+2x^2+2x+1=x^4+x^3+x^3+x^2+x^2+x+x+1\)
\(=x^3\left(x+1\right)+x^2\left(x+1\right)+x\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^3+x^2+x+1\right)\)
\(=\left(x+1\right)\left[x^2\left(x+1\right)+\left(x+1\right)\right]\)
\(=\left(x+1\right)\left(x+1\right)\left(x^2+1\right)\)
\(=\left(x+1\right)^2\left(x^2+1\right)\)
a) =
\(x^4-x^3-x^3+x^2-x^2+x+x-1=x^3\left(x-1\right)-x^2\left(x-1\right)-x\left(x-1\right)+\left(x-1\right)=\left(x-1\right)\left(x^3-x^2-x+1\right)\)