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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
a: \(=a^2-b^4\)
b: \(=\left(a^2+2a\right)^2-9\)
c: \(=a^2-\left(2a+3\right)^2\)
d: \(=a^4-\left(2a-3\right)^2\)
e: \(=\left(-a^2-2a+3\right)^2\)
g: \(=4a^2-a^4\)
Theo đề ra ta có : a2+b2=2(8+ab)
⇔a2+b2-2ab=16
⇔(a-b)2=16
⇔a-b=4
Ta có P=a2(a+1)−b2(b−1)+ab−3ab(a−b+1)+64
⇔P=a3+a2-b3+b2+ab-3a2b+3ab2-3ab+64
⇔P=(a3-b3)+(a2-2ab+b2)-(3a2b-3ab2)+64
⇔P=(a-b)(a2+ab+b2)+(a-b)2-3ab(a-b)+64
⇔P=(a-b)(a2+ab+b2+1-3ab)+64
⇔P=4[(a-b)2+1]+64
⇔P=4(16+1)+64= 132
⇔P= 132
#)Giải :
\(a^2+b^2\le1+ab\)
\(\Leftrightarrow a^2-ab+b^2\le1\)
\(\Leftrightarrow\left(a+b\right)\left(a^2-ab+b^2\right)\le a+b\)
\(\Leftrightarrow a^3+b^3\le a+b\)
\(\Leftrightarrow\left(a^3+b^3\right)\left(a^3+b^3\right)\le\left(a+b\right)\left(a^5+b^5\right)\left(a^3+b^3=a^5+b^5\right)\)
\(\Leftrightarrow a^6+2a^3b^3+b^6\le a^6+ab^5+a^5b+b^6\)
\(\Leftrightarrow a^5b+ab^5\ge2a^3b^3\)
\(\Leftrightarrow a^5b+ab^5-2a^3b^3\ge0\)
\(\Leftrightarrow ab\left(a^4-2a^2b^2+b^4\right)\ge0\)
\(\Leftrightarrow ab\left(a^2-b^2\right)^2\ge0\)( luôn đúng \(\forall a;b>0\))
Vậy \(a^2+b^2\le1+ab\left(đpcm\right)\)
P/s : Bài này mk tham khảo trên mạng ( tại thấy rảnh nên chép hộ ^^ )
1: \(a^3-2a^2+a=a\left(a^2-2a+1\right)=a\left(a-1\right)^2\)
2: \(2b^2+4b+2-2c^2\)
\(=2\left(b^2+2b+1-c^2\right)\)
\(=2\left(b+1-c\right)\left(b+1+c\right)\)
3: \(x^2+5x+6=\left(x+2\right)\left(x+3\right)\)
4: \(x^2+x-6=\left(x+3\right)\left(x-2\right)\)
5: \(x^2+6xy+9y^2=\left(x+3y\right)^2\)
6: \(y^2\left(x^2+y\right)-x^2z-yz\)
\(=y^2\left(x^2+y\right)-z\left(x^2+y\right)\)
\(=\left(x^2+y\right)\left(y^2-z\right)\)
a/ x3 + x2 z + y2 z - xyz + y3
= (x + y)(x2 - xy + y2) + z(x2 - xy + y2)
= (x2 - xy + y2)(x + y + z)
a)
Có:
\(\left(a^3-a\right)^2-12\left(a^3-a\right)+36=0\)
\(\Rightarrow\left(a^3-a-6\right)^2=0\)
\(\Rightarrow a^3-a-6=0\)
\(\Rightarrow a^3-a=6\)
sao suy ra dc o dong 2 z ?