Phân tích đa thức thành nhân tử .
- 3x2 + 2x – 1
- x3 + 6x2 + 11x + 6
- x4 + 2x2 – 3
- ab + ac +b2 + 2bc + c2
- a3 – b3 + c3 + 3abc
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
PS
3
Những câu hỏi liên quan
PT
1
27 tháng 7 2019
\(1,2x^2-3x-2\)
\(=2x^2-4x+x-2\)
\(=2x\left(x-2\right)+\left(x-2\right)\)
\(=\left(2x+1\right)\left(x-2\right)\)
\(2,4x^2-7x-2\)
\(=4x^2-8x+x-2\)
\(=4x\left(x-2\right)+x-2\)
\(\left(4x+1\right)\left(x-2\right)\)
KT
9 tháng 1 2018
1) \(3x^2+2x-1\)
\(=3x^2+3x-x-1\)
\(=3x\left(x+1\right)-\left(x+1\right)\)
\(=\left(x+1\right)\left(3x-1\right)\)
2) \(x^3+6x^2+11x+6\)
\(=x^3+x^2+5x^2+5x+6x+6\)
\(=x^2\left(x+1\right)+5x\left(x+1\right)+6\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+5x+6\right)\)
\(=\left(x+1\right)\left(x+2x+3x+6\right)\)
\(=\left(x+1\right)\left(x+2\right)\left(x+3\right)\)
3) \(x^4+2x^2-3\)
\(=\left(x^2+1\right)^2-4\)
\(=\left(x^2+1-2\right)\left(x^2+1+2\right)\)
\(=\left(x^2-1\right)\left(x^2+3\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+3\right)\)
4) \(ab+ac+b^2+2bc+c^2\)
\(=a\left(b+c\right)+\left(b+c\right)^2\)
\(=\left(b+c\right)\left(a+b+c\right)\)
9 tháng 1 2018
1, \(3x^2+2x-1\)
\(=3x^2+3x-x-1\)
\(=3x\left(x+1\right)-\left(x+1\right)\)
\(=\left(x+1\right)\left(3x-1\right)\)
2, \(x^3+6x^2+11x+6\)
\(=\left(x^3+3x^2\right)+\left(3x^2+9x\right)+\left(2x+6\right)\)
\(=x^2\left(x+3\right)+3x\left(x+3\right)+2\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2+3x+2\right)\)
\(=\left(x+3\right)\left(x+1\right)\left(x+2\right)\)
L
20 tháng 9 2020
.\(a\left(b^2+c^2\right)+b\left(c^2+a^2\right)+c\left(a^2+b^2\right)-2abc-a^3-b^3-c^3\)
=\(a\left(b^2-2bc+c^2-a^2\right)+b\left(a^2+2ac+c^2-b^2\right)+c\left(a^2-2ab+b^2-c^2\right)\)
=\(a\left[\left(b-c\right)^2-a^2\right]+b\left[\left(a+c\right)^2-b^2\right]+=c\left[\left(a-b^2\right)-c^2\right]\)
=\(a\left(c-b+a\right)\left(a+b-c\right)+b\left(a+c-b\right)\left(a+b+c\right)+c\left(a-b+c\right)\left(a-b-c\right)\)
=\(\left(a+c-b\right)\left[a\left(c-b+a\right)+b\left(a+b+c\right)+c\left(a-b-c\right)\right]\)
=\(\left(a+c-b\right)\left(b+a-c\right)\left(c+b-a\right)\)
CP
17 tháng 7 2021
a) \(x^4+2x^3-4x-4=\left(x^4+2x^3+x^2\right)-\left(x^2+4x+4\right)\)
\(=\left(x^2+x\right)^2-\left(x+2\right)^2=\left(x^2+x-x-2\right)\left(x^2+x+x+2\right)\)
\(=\left(x^2-2\right)\left(x^2+2x+2\right)\)
17 tháng 7 2021
a) Ta có: \(x^4+2x^3-4x-4\)
\(=\left(x^4+2x^3+x^2\right)-\left(x^2+4x+4\right)\)
\(=\left(x^2+x\right)^2-\left(x+2\right)^2\)
\(=\left(x^2+x-x-2\right)\left(x^2+x+x+2\right)\)
\(=\left(x^2-2\right)\cdot\left(x^2+2x+2\right)\)
\(a,3x^2+2x-1\)
\(\Leftrightarrow3x^2+3x-x-1\)
\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)\)
\(\Leftrightarrow\left(x+1\right)\left(3x-1\right)\)
\(b,x^3+6x^2+11x+6\)
\(\Leftrightarrow x^3+3x^2+3x^2+9x+2x+6\)
\(\Leftrightarrow x^2\left(x+3\right)+3x\left(x+3\right)+2\left(x+6\right)\)
\(\Leftrightarrow\left(x+3\right)\left(x^2+3x+2\right)\)
\(\Leftrightarrow\left(x+3\right)\left(x+1\right)\left(x+2\right)\)
\(c,x^4+2x^2-3\)
\(\Leftrightarrow x^4-x^3+x^3-x^2+3x^2-3\)
\(\Leftrightarrow x^3\left(x-1\right)+x^2\left(x-1\right)+3\left(x-1\right)\left(x+1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+x^2+3x+3\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+3\right)\)
\(d,ab+ac+b^2+2bc+c^2\)
\(\Leftrightarrow a\left(b+c\right)+\left(b+c\right)^2\)
\(\Leftrightarrow\left(b+c\right)\left(a+b+c\right)\)
3x^2+2x-1=3x^2+3x-x-1=3x(x+1)-(x+1)=(x+1)(3x-1)
x^4+2x^2-3=x^4+3x^2-x^2 -3=x^2(x^2+3)-(x^2+3)=(x^2+3)(x^2-1)