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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 2 nè
a+b+c = 0
=>(a+b+c)^3 = 0
a^3 + b^3 + c^3 + 3(a+b)(b+c)(a+c) = 0
vì a+b = -c
a+c = -b
b+c = -a
thay vào => a^3 + b^3 + c^3 - 3abc = 0
=> a^3 + b^3 + c^3 = 3abc
3, \(=x^4-x^2+3x^2-3\)
\(=x^2\left(x^2-1\right)+3\left(x^2-1\right)\)
\(=\left(x^2+3\right)\left(x-1\right)\left(x+1\right)\)
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: \(\left(a-b\right)^3+c^3+3a^2b-3ab^2+3abc\)
\(=\left(a-b+c\right)\left[\left(a-b\right)^2-\left(a-b\right)c+c^2\right]+3ab\left(a-b+c\right)\)
\(=\left(a-b+c\right)\left(a^2-2ab+b^2-ac+bc+c^2+3ab\right)\)
\(=\left(a-b+c\right)\left(a^2+b^2+c^2+ab-ac+bc\right)\)
a) \(x^2+2x+1=x^2+x+x+1=x\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(x+1\right)=\left(x+1\right)^2\) *Câu này có thể áp dụng hằng đẳng thức \(a^2+2ab+b^2=\left(a+b\right)^2\) cho nhanh*
b) \(a^3-b^3+c^3+3abc=\left(a^3-3a^2b+3ab^2-b^2\right)+3a^2b-3ab^2+c^3+3abc\)
\(=\left(a-b\right)^3+c^3+\left(3a^2b-3ab^2+3abc\right)\)
\(=\left(a-b+c\right)\left[\left(a-b\right)^2-\left(a-b\right)c+c^2\right]+3ab\left(a-b+c\right)\)
\(=\left(a-b+c\right)\left(a^2-2ab+b^2-ac+bc+c^2+3ab\right)\)
\(=\left(a-b+c\right)\left(a^2+b^2+c^2-ac+bc+ab\right)\)
c) \(a^3-b^3-c^3-3abc=\left[a^3-3a^2b+3ab^2-b^3\right]+3a^2b-3ab^2-c^3-3abc\)
\(=\left[\left(a-b\right)^3-c^3\right]+3ab\left(a-b-c\right)=\left(a-b-c\right)\left[\left(a-b\right)^2+\left(a-b\right)c+c^2\right]+3ab\left(a-b-c\right)\)
\(=\left(a-b-c\right)\left[a^2-2ab+b^2+ac-bc+c^2+3ab\right]=\left(a-b-c\right)\left(a^2+b^2+c^2+ab+ac-bc\right)\)
a,(x+1)2
b,(a+c-b).{(a+c)^2+(a+c)b+b^2-3ac}
c,(a-c-b).{(a-c)^2+(a-c)b+b^2+3ac}
1)\(3x^2+2x-1=3x^2+3x-x-1=3x\left(x+1\right)-\left(x+1\right)=\left(3x-1\right)\left(x+1\right)\)
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)
\(a,3x^2+2x-1\)
\(\Leftrightarrow3x^2+3x-x-1\)
\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)\)
\(\Leftrightarrow\left(x+1\right)\left(3x-1\right)\)
\(b,x^3+6x^2+11x+6\)
\(\Leftrightarrow x^3+3x^2+3x^2+9x+2x+6\)
\(\Leftrightarrow x^2\left(x+3\right)+3x\left(x+3\right)+2\left(x+6\right)\)
\(\Leftrightarrow\left(x+3\right)\left(x^2+3x+2\right)\)
\(\Leftrightarrow\left(x+3\right)\left(x+1\right)\left(x+2\right)\)
\(c,x^4+2x^2-3\)
\(\Leftrightarrow x^4-x^3+x^3-x^2+3x^2-3\)
\(\Leftrightarrow x^3\left(x-1\right)+x^2\left(x-1\right)+3\left(x-1\right)\left(x+1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+x^2+3x+3\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+3\right)\)
\(d,ab+ac+b^2+2bc+c^2\)
\(\Leftrightarrow a\left(b+c\right)+\left(b+c\right)^2\)
\(\Leftrightarrow\left(b+c\right)\left(a+b+c\right)\)
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)
câu d thêm ở cuối bài -24x2 nữa nha mình viết thiếu
Áp dụng hằng đẳn thức này: (a+b)^ 3 = a^3 + 3a^2b+3ab^2+b^3 = a^3 + b^3 +3ab(a+b)
a/. Có: a3+b3 +c3-3abc = (a+b)3-3ab(+b)+c3-3abc
= (a+b)3+c3-3ab(a+b) - 3abc= (a+b+c)[(a+b)2-(a+b)c+c2] - 3ab(a+b+c)
=(a+b+c)(a^2+2ab+b^2-ac-bc+c^2 - 3ab)
=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)
b/. tương tự a. khi nhóm thì nhóm (a^3 - c^3) trước
c/. 6x^4 - 11x^2 + 3 = 6t^2 -11t + 3 (Với t = x^2 >=0)
=6t^2 - 2t - 9t +3 = (6t^2 -2t) -(9t - 3) = 2t(3t - 1) - 3(3t-1) = (3t-1)(2t-3)