phân tích đa thức thành nhân tử.
1) 3x^2 + 2x – 1
2) x^3 + 6x^2 + 11x + 6
3) x^4 + 2x^2 – 3
4) ab + ac +b^2 + 2bc + c^2
5) a^3 – b^3 + c^3 + 3abc
hộ mk cái
phân tích đa thức thành nhân tử.
3x2 + 2x – 1x3 + 6x2 + 11x + 6x4 + 2x2 – 3ab + ac +b2 + 2bc + c2a3 – b3 + c3 + 3abcTa có
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, \(x^3+2x^2+4x^2+8x+3x+6\)
\(=x^2\left(x+2\right)+4x\left(x+2\right)+3\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2+4x+3\right)\)
\(=\left(x+2\right)\left(x^2+x+3x+3\right)\)
\(=\left(x+2\right)\text{[}x\left(x+1\right)+3\left(x+1\right)\text{]}\)
\(=\left(x+2\right)\left(x+1\right)\left(x+3\right)\)
3,\(x^4+2x^2-3=x^4-x^2+3x^2-3\)
\(=x^2\left(x^2-1\right)+3\left(x^2-1\right)\)
\(\left(x^2-1\right)\left(x^2+3\right)=\left(x-1\right)\left(x+1\right)\left(x^2+3\right)\)
4,\(ab+ac+b^2+2bc+c^2\)
\(=a\left(b+c\right)+\left(b+c\right)^2\)
\(=\left(b+c\right)\left(a+b+c\right)\)
phân tích đa thức thành nhân tử.
3x2 + 2x – 1x3 + 6x2 + 11x + 6x4 + 2x2 – 3ab + ac +b2 + 2bc + c2a3 – b3 + c3 + 3abc 3x^2+2x-1
=3x^2+3x-x-1
=3x(x+1)-(x+1)
=(x+1)(3x-1)
x^3+6x^2+11x+6
=x^3+5x^2+6x+x^2+5x+6
=x(x^2+5x+6)+(x^2+5x+6)
=(x+1)(x^2+5x+6)
=(x+1)(x^2+3x+2x+6)
=(x+1)(x+2)(x+3)
x^4+2x^2-3
=x^4-x^2+3x^2-3
=x^2(x^2-1)+3(x^2-1)
=(x^2-1)(x^2+3)
=(x+1)(x-1)(x^2+3)
ab+ac+b^2+2bc+c^2
=a(b+c)+(b+c)^2
=(b+c)(a+b+c)
a^3-b^3+c^3+3abc
=(a-b)^3+3ab(a-b)+c^3+3abc
=(a-b+c)^3-3(a-b)c(a-b+c)+3ab(a-b+c)
=(a-b+c)(a^2+b^2+c^2-2ab+2ac-2bc-3ac+3...
=(a-b+c)(a^2+b^2+c^2+ab+bc-ca)
=1/2.(a-b+c)(a^2+2ab+b^2+b^2+2bc+c^2+c...
=1/2.(a-b+c)[(a+b)^2+(b+c)^2+(c-a)^2]
Bài 1 : phân tích đa thức thành nhân tử.
3x2 + 2x – 1x3 + 6x2 + 11x + 6x4 + 2x2 – 3ab + ac +b2 + 2bc + c2a3 – b3 + c3 + 3abc3, \(=x^4-x^2+3x^2-3\)
\(=x^2\left(x^2-1\right)+3\left(x^2-1\right)\)
\(=\left(x^2+3\right)\left(x-1\right)\left(x+1\right)\)
5, nhận xét : \(\left(a-b\right)^3=a^3-3a^2b+3ab^2-b^3\Rightarrow a^3-b^3=\left(a-b\right)^3+3a^2b-3ab^2\)
thay vào đầu bài ta có: \(\left(a-b\right)^3+c^3+3a^2b-3ab^2+3abc\)
\(=\left(a-b+c\right)\left[\left(a-b\right)^2-\left(a-b\right)c+c^2\right]+3ab\left(a-b+c\right)\)
\(=\left(a-b+c\right)\left(a^2-2ab+b^2-ac+bc+c^2+3ab\right)\)
\(=\left(a-b+c\right)\left(a^2+b^2+c^2+ab-ac+bc\right)\)
1, \(=3x^2+3x-x-1\)
\(=3x\left(x+1\right)-\left(x+1\right)\)
\(=\left(x+1\right)\left(3x-1\right)\)
: phân tích đa thức thành nhân tử.
3x2 + 2x – 1x3 + 6x2 + 11x + 6x4 + 2x2 – 3ab + ac +b2 + 2bc + c2a3 – b3 + c3 + 3abcphân tích đa thức thành nhân tử.
3x2 + 2x – 1x3 + 6x2 + 11x + 6x4 + 2x2 – 3ab + ac +b2 + 2bc + c2a3 – b3 + c3 + 3abc
3x2 + 2x – 1
=3x2 -x +3x-1
=x.(3x-1)+(3x-1)
=(3x-1)(x+1)
Làm giúp 1 bài nữa nhé!
Phân tích đa thức thành nhân tử.
3x2 + 2x – 1x3 + 6x2 + 11x + 6x4 + 2x2 – 3ab + ac +b2 + 2bc + c2a3 – b3 + c3 + 3abc1)\(3x^2+2x-1=3x^2+3x-x-1=3x\left(x+1\right)-\left(x+1\right)=\left(3x-1\right)\left(x+1\right)\)
2)\(x^3+6x^2+11x+6=x^3+3x^2+3x^2+9x+2x+6\)
\(=x^2\left(x+3\right)+3x\left(x+3\right)+2\left(x+3\right)\)\(=\left(x^2+3x+2\right)\left(x+3\right)\)
\(=\left(x^2+2x+x+2\right)\left(x+3\right)\)\(=\left[x\left(x+2\right)+\left(x+2\right)\right]\left(x+3\right)\)
\(=\left(x+1\right)\left(x+2\right)\left(x+3\right)\)
3)\(x^4+2x^2-3=x^4+3x^2-x^2-3=x^2\left(x^2+3\right)-\left(x^2+3\right)=\left(x^2-1\right)\left(x^3+3\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+3\right)\)
4)\(ab+ac+b^2+2bc+c^2=a\left(b+c\right)+\left(b+c\right)^2=\left(b+c\right)\left(a+b+c\right)\)
5) câu này sau khi phân tích được (a-b+c)(a2+b2+c2+ab+bc-ac)
Bài 1 : phân tích đa thức thành nhân tử.
3x2 + 2x – 1x3 + 6x2 + 11x + 6x4 + 2x2 – 3ab + ac +b2 + 2bc + c2a3 – b3 + c3 + 3abc
Phân tích đa thức thành nhân tử .
3x2 + 2x – 1x3 + 6x2 + 11x + 6x4 + 2x2 – 3ab + ac +b2 + 2bc + c2a3 – b3 + c3 + 3abc\(a,3x^2+2x-1\)
\(\Leftrightarrow3x^2+3x-x-1\)
\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)\)
\(\Leftrightarrow\left(x+1\right)\left(3x-1\right)\)
\(b,x^3+6x^2+11x+6\)
\(\Leftrightarrow x^3+3x^2+3x^2+9x+2x+6\)
\(\Leftrightarrow x^2\left(x+3\right)+3x\left(x+3\right)+2\left(x+6\right)\)
\(\Leftrightarrow\left(x+3\right)\left(x^2+3x+2\right)\)
\(\Leftrightarrow\left(x+3\right)\left(x+1\right)\left(x+2\right)\)
\(c,x^4+2x^2-3\)
\(\Leftrightarrow x^4-x^3+x^3-x^2+3x^2-3\)
\(\Leftrightarrow x^3\left(x-1\right)+x^2\left(x-1\right)+3\left(x-1\right)\left(x+1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+x^2+3x+3\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+3\right)\)
\(d,ab+ac+b^2+2bc+c^2\)
\(\Leftrightarrow a\left(b+c\right)+\left(b+c\right)^2\)
\(\Leftrightarrow\left(b+c\right)\left(a+b+c\right)\)
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)
a) \(3x^2+2x-1=3x^2+3x-x\)\(-1\)
\(3x\left(x+1\right)-\left(x+1\right)\)
\(=\left(3x-1\right)\left(x+1\right)\)
b) x3 + 6x2 + 11x + 6 = x3+6x2+12x+8-x-2 = ( x3+6x2+12x+8)-(x+2) = (x+2)^3 - (x+2) = (x+2)[(x+2)^2 - 1]
=(x+2)(x^2 +4x + 4 - 1) = (x+2)(x^2 + 4x +3)
c) x4 + 2x2 – 3 = x4 + 2x2 +1 - 4 = (x^2 +1)^2 - 2^2
= (x^2 + 1 + 2)(x^2 + 1- 2)
=(x^2 + 3)( x^2 - 1)
ab + ac +b2 + 2bc + c2= a(b+c) + (b+c)^2
=(b+c)(a+b+c)
a3 – b3 + c3 + 3abc\(=a^3-3a^2b+3ab^2-b^3+3a^2b-3ab^2+c^3+3abc\)
\(=\left(a-b\right)^3+3ab\left(a-b\right)+c^3+3abc\)
\(=\left[\left(a-b\right)^3+c^3\right]+\left[3ab\left(a-b\right)+3abc\right]\)
\(=\left(a-b+c\right)\left[\left(a-b\right)^2-\left(a-b\right)c+c^2\right]\)\(+3ab\left(a-b+c\right)\)
\(=\left(a-b+c\right)\left(a^2-2ab+b^2-ac+bc+c^2\right)\)\(+3ab\left(a-b+c\right)\)
\(=\left(a-b+c\right)\left(a^2-2ab+b^2-ac+bc+c^2+3ab\right)\)
\(=\left(a-b+c\right)\left(a^2+b^2+c^2-ac+bc+ab\right)\)
câu cuối ko bik đúng hay sai đâu nha
Phân tích đa thức thành nhân tử (đừng để ý đến cái tên nha các bạn)
x3 + 6x2 + 11x + 6x4 + 2x2 – 3ab + ac +b2 + 2bc + c2a3 – b3 + c3 + 3aba]
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)
b thiếu đề bài nè x^4 - 2x^2 - 3 = 0
(x^2)(x^2)-2(x^2)-3=0
(x^2)(x^2)-3(x^2)+1(x^2)-3=0
(x^2)(x^2-3) + 1(x^2-3) = 0
(x^2-3) (x^2+1)=0
c bó tay
d (a3 + b3 + c3) - 3abc
= ( (a+b+c)3 - 3ab(a+b) -3bc(b+c) -3ac(a+c) - 6abc) - 3abc
= (a+b+c)3 - 3ab(a+b) - 3bc(b+c) - 3ac(a+c) - 9abc
= (a+b+c)3
- 3ab(a+b) - 3abc
- 3bc(b+c) - 3abc
- 3ac(a+c) - 3abc
= (a+b+c)3
- 3ab(a+b+c)
- 3bc(a+b+c)
- 3ac(a+b+c)
= (a+b+c)( (a+b+c)2 - 3ab -3bc 3ac)
=(a+b+c)( a2 + b2 + c2 + 2ab +2bc + 2ca -3ab - 3bc -3ac)
=(a+b+c) (a2 + b2 + c2 - ab - bc -ac)
ý d hình như đề sai
chị ui kết bạn với em đi em hết lượt kết bạn rùi
em học lớp 5
1. x3 + 6x2 + 11x + 6
= x3 + 3x2 + 3x2 + 9x + 2x + 6
= x2(x + 3) + 3x(x + 3) + 2(x + 3)
= (x + 3)(x2 + 3x + 2)