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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)
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: \(\left(a-b\right)^3+c^3+3a^2b-3ab^2+3abc\)
\(=\left(a-b+c\right)\left[\left(a-b\right)^2-\left(a-b\right)c+c^2\right]+3ab\left(a-b+c\right)\)
\(=\left(a-b+c\right)\left(a^2-2ab+b^2-ac+bc+c^2+3ab\right)\)
\(=\left(a-b+c\right)\left(a^2+b^2+c^2+ab-ac+bc\right)\)
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]
1.
3x2 + 2x - 1
= 3x2 - x + 3x - 1
= x(3x-1) + (3x-1)
= (3x-1)(x+1)
2.
x3 + 6x2 + 11x + 6
= x3 + 3x2 + 3x2 + 9x + 2x + 6
= x2(x+3) + 3x(x+3) + 2(x+3)
= (x+3)(x2 + 3x + 2)
= (x+3)(x2 + 2x + x + 2)
= (x+3)[x(x+2) + (x+2)]
= (x+3)(x+2)(x+1)
3.
x4 + 2x2 - 3
= x4 + 2x2 + 1 - 4
= (x2 + 1)2 - 22
= (x2 +3)(x2-1)
= (x2 + 3)(x-10(x+1)
4.
ab + ac + b2 + 2bc + c2
= a(b+c) + (b+c)2
= (b+c)(a+b+c)
5.
a3 - b3 + c3 + 3abc
= (a+c)3 - 3ac(a+c) - b3 + 3abc
= (a+c-b) [(a+c)2 + b(a+c) + b2] -3ac (a+c-b)
= (a+c-b) (a2 + 2ac + c2 + ab+ bc + b2) - 3ac(a+c-b)
= (a+c-b) (a2 + b2 + c2 + 2ac + ab + bc - 3ac)
= (a+c-b) (a2 + b2 + c2 - ac + ab + bc)
Bài 3a)
\(a+b+c=0\Leftrightarrow a+b=-c\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)
\(\Leftrightarrow a^3+b^3+c^3=-3ab\left(a+b\right)\)
mà \(a+b=-c\Rightarrow a^3+b^3+c^3=3abc\)
1 . 3x2-x+2x-1
=3x2-x+3x-1
=(3x2-x)+(3x-1)
=x(3x-1)+(3x-1)
=(3x-1)(x+1)
2. x3+6x2+11x+6
=x3+x2+5x2+5x+6x+6
=(x3+x2)+(5x2+5x)+(6x+6)
=x2(x+1)+5x(x+1)+6(x+1)
=(x+1)(x2+5x+6)
=(x+1)(x2-x+6x+6)
=(x+1)(x2+2x+3x+6)
=(x+1)[(x2+2x)+(3x+6)]
=(x+1)[x(x+2)+3(x+2)
=(x+1)(x+2)(x+3)
1. 3x^2+2x-1
=3x^2+3x-x-1
=3x(x+1)-(x+1)
=(x+1)(3x-1)
2. 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)
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)
4. ab+ac+b^2+2bc+c^2
=a(b+c)+(b+c)^2
=(b+c)(a+b+c)
5. 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]
P/s: Ko chắc đâu nhé :)
1. 3x^2 + 2x – 1
3x^2 + 3x – x – 1
3x(x + 1) – (x + 1)
(x + 1)(3x – 1)
2. x^3 + 6x^2 +11x + 6
x^3 + 3x^2 + 3x^2 + 9x + 2x + 6
x^2(x + 3) + 3x(x + 3) + 2(x + 3)
(x + 3)(x^2 + 3x + 2)
(x + 3)(x^2 + 2x + x + 2)
(x + 3)[x(x + 2) + (x+2)]
(x + 3)(x + 2)(x + 1)
x^4 + 2x^2 – 3
=x^4 -x + 2x^2 +x -3.
= x(x^3 – 1 ) +(2x^2 + x -3)
=x(x-1)(x^2+X+1) + (x-1)(x+3/2)
=(x-1) (x(x^2 +x +1) +3+ 3/2)…
đến đó thì mình tự nhân nha\
4. ab + ac + b^2 + 2bc + c^2
a(b + c) + (b + c)^2
(b + c)(a + b + c)
Le Nhat Phuong cái 5 thì mình ko chắc nhưng vì bn nhanh nhất và đúng nhiều nên được thưởng :)