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1. (a2+b2+ab)2-a2b2-b2c2-c2a2
=a4+b4+a2b2+2(a2b2+ab3+a3b)-a2b2-b2c2-c2a2
=a4+b4+2a2b2+2ab3+2a3b-b2c2-c2a2
=(a2+b2)2+2ab(a2+b2)-c2(a2+b2)
=(a2+b2)[(a+b)2-c2]
=(a2+b2)(a+b+c)(a+b-c)
2. a4+b4+c4-2a2b2-2b2c2-2a2c2=(a2-b2-c2)2
3. a(b3-c3)+b(c3-a3)+c(a3-b3)
=ab3-ac3+bc3-ba3+ca3-cb3
=a3(c-b)+b3(a-c)+c3(b-a)
=a3(c-b)-b3(c-a)+c3(b-a)
=a3(c-b)-b3(c-b+b-a)+c3(b-a)
=a3(c-b)-b3(c-b)-b3(b-a)+c3(b-a)
=(c-b)(a-b)(a2+ab+b2)-(b-a)(b-c)(b2+bc+c2)
=(a-b)(c-b)(a2+ab+2b2+bc+c2)
4. a6-a4+2a3+2a2=a4(a+1)(a-1)+2a2(a+1)=(a+1)(a5-a4+2a2)=a2(a+1)(a3-a2+2)
5. (a+b)3-(a-b)3=(a+b-a+b)[(a+b)2+(a+b)(a-b)+(a-b)2]
=2b(3a2+b2)
6. x3-3x2+3x-1-y3=(x-1)3-y3=(x-1-y)[(x-1)2+(x-1)y+y2]
=(x-y-1)(x2+y2+xy-2x-y+1)
7. xm+4+xm+3-x-1=xm+3(x+1)-(x+1)=(x+1)(xm+3-1)
(Đúng nhớ like nhá !)
Minh Hải,Lê Thiên Anh,Nguyễn Huy Tú,Ace Legona,...giúp mk vs mai mk đi hk rùi
b) a3 + b3 + c3 - 3abc
= ( a + b)3 - 3ab - 3ba + c - 3abc
= (a3 + 3a2b + 3ab2 + b3) + c3 - (3a2b + 3ab2 + 3ab)
= (a + b)3 + c2 - 3ab(a + b + c)
= (a + b + c) [ (a + b)2 - ( a + b )c + c^2 ] - 3ab(a + b + c)
= ( a + b + c ) ( a2 + b2 + 2ab - ac - bc + c2 -3ab )
= ( a + b + c ) ( a2 + b2 + c2 - ab - ac - bc
câu a (a+b+c)2 +(a+b-c)2 - 4c2= (a+b+c)2+(a+b-c+2c).(a+b-c-2c) =(a+b+c)2 +(a+b+c).(a+b-3c)=(a+b+c). (a+b+c+a+b-3c)=(a+b+c).2.(a+b-c)
câu b 4a2b2-(a2+b2-c2) = (2ab-a2-b2+c2).(2ab+a2+b2-c2)
= (c2-(a-b)2).((a+b)2-c2)
= (c-a+b).(c+a-b).(a+b-c).(a+b+c)
câu c a4+b4+c4-2a2b2+2b2c2-2a2c2-4b2c2=(a2-b2-c2)2-4b2c2=(a2-b2-c2-2bc).(a2-b2-c2+2bc)=(a2-(b+c)2).(a2-(b-c)2)=(a-b-c).(a+b+c).(a-b+c).(a+b-c)
câu d dùng pp xét giá trị riêng thay b =c (bạn tự giải ) thì đa thức này nếu coi là đa thức biến b thì đa thức A chia hết cho b-c
a,b,c bình đẳng => A chia hết cho c-a , a-b
=>A= k(a-b)(b-c)(c-a)
thay thử một bộ a,b,c bất kì => k=? (mình đang vội )
thay k tính đc vàoA= k(a-b)(b-c)(c-a)
\(B=x^3+3x^2-4\)
\(B=x^3-x^2+4x^2-4\)
\(B=x^2\left(x-1\right)+4\left(x^2-1\right)\)
\(B=x^2\left(x-1\right)+4\left(x+1\right)\left(x-1\right)\)
\(B=\left(x-1\right)\left(x^2+4x+4\right)\)
\(B=\left(x-1\right)\left(x+2\right)^2\)
a) \(A=a^3-b^3-c^3-3abc\)
\(=\left(a-b\right)^3+3ab\left(a-b\right)-c^3-3abc\)
\(=\left(a-b-c\right)\left[\left(a-b\right)^2+c\left(a-b\right)+c^2\right]+3ab\left(a-b-c\right)\)
\(=\left(a-b-c\right)\left(a^2-2ab+b^2+ac-bc+c^2+3ab\right)\)
\(=\left(a-b-c\right)\left(a^2+b^2+c^2+ab+ac-bc\right)\)
b) \(B=a^2b^2\left(a-b\right)-c^2b^2\left(c-b\right)+a^2c^2\left(c-a\right)\)
\(=a^2b^2\left(a-b\right)+c^2b^2\left(b-c\right)+a^2c^2\left(c-a\right)\)
\(=a^2b^2\left(a-b\right)+c^2b^2\left(b-c\right)-a^2c^2\left[\left(a-b\right)+\left(b-c\right)\right]\)
\(=a^2b^2\left(a-b\right)+c^2b^2\left(b-c\right)-a^2c^2\left(a-b\right)-a^2c^2\left(b-c\right)\)
\(=a^2\left(a-b\right)\left(b^2-c^2\right)+c^2\left(b-c\right)\left(b^2-a^2\right)\)
\(=a^2\left(a-b\right)\left(b-c\right)\left(b+c\right)+c^2\left(b-c\right)\left(b-a\right)\left(b+a\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(a^2b+a^2c-bc^2-ac^2\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\left(ab+bc+ca\right)\)