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a: \(=\dfrac{\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)-3abc}{a^2+b^2+c^2-ab-bc-ac}\)
\(=\dfrac{\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)}{a^2+b^2+c^2-ab-bc-ac}\)
\(=\dfrac{\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)}{a^2+b^2+c^2-ab-bc-ac}\)
=a+b+c
b:
Sửa đề: \(=\dfrac{x^3-y^3+z^3+3xyz}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)
\(=\dfrac{\left(x-y\right)^3+z^3+3xy\left(x-y\right)+3xyz}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)
\(=\dfrac{\left(x-y+z\right)\left(x^2-2xy+y^2-xz+yz+z^2\right)+3xy\left(x-y+z\right)}{2\left(x^2+y^2+z^2+xy+yz-xz\right)}\)
\(=\dfrac{\left(x-y+z\right)\left(x^2+y^2+z^2+xy-xz+yz\right)}{2\left(x^2+y^2+z^2+xy+yz-xz\right)}\)
\(=\dfrac{x-y+z}{2}\)
a) \(\dfrac{a^3+b^3+c^3-3abc}{a^2+b^2+c^2-ab-bc-ca}\)
\(=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)}{a^2+b^2+c^2-ab-bc-ca}\)
\(=a+b+c\)
\(\frac{a^3+b^3-c^3+3abc}{\left(a-b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2}=\frac{\left(a+b\right)^3-c^3-3ab\left(a+b\right)+3abc}{2a^2+2b^2+2c^2-2ab+2bc+2ac}\)
\(=\frac{\left(a+b-c\right)\left[\left(a+b\right)^2+c\left(a+b\right)+c^2\right]-3ab\left(a+b-c\right)}{2a^2+2b^2+2c^2-2ab+2bc+2ac}\)
\(=\frac{\left(a+b-c\right)\left(a^2+2ab+b^2+ac+bc+c^2-3ab\right)}{2\left(a^2+b^2+c^2-ab+bc+ac\right)}\)
\(=\frac{a+b-c}{2}\)
a^3 +c^3 = (a+c). (a^2 -a.c+c^2)
= (a+c)^3 -3 ac.(a+c)
=> a^3+c^3-3abc+b^3 =(a+c)^3-3ac (a+c)-3abc +b^3
=(a+c)^3+b^3 -3ac (b+(a+c))
=(a+c+b). ((a+c)^2-(a+c).b+b^2) -3ac (a+c+b)
=(a+c+b)^3-3(a+c)b. (a+c+b)-3ac (a+c+b)
=(a+c+b)((a+c+b)^2 -3ab-3bc-3ac) (1)
(a-b)^2 + (b-c)^2 +(a-c)^2
= 2a^2 +2b^2+2c^2 -2ab-2bc-2ac
=2 (a^2+b^2+c^2-ac-ab-bc)
=2((a+b)^2-3ab +c^2 -ac-bc)
=2 ((a+b+c)^2-2(ac+bc)-3ab-ac-bc)
=2 (( a+c+b)^2 -3ab-3bc -3ac) (2)
Từ (1),(2) =>(a^3+b^3+c^3-3abc)/((a-b)^2
+(b-c)^2+(c-a)^2)
=(a+b+c)/2
\(A=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)}{2\left(a^2+b^2+c^2-ab-bc-ca\right)}=\dfrac{a+b+c}{2}=2\)
Lời giải:
\(\frac{a^2(b-c)+b^2(c-a)+c^2(a-b)}{ab^2-ac^2-b^3+bc^2}=\frac{a^2(b-c)-b^2[(b-c)+(a-b)]+c^2(a-b)}{a(b^2-c^2)-b(b^2-c^2)}\)
\(=\frac{(a^2-b^2)(b-c)-(b^2-c^2)(a-b)}{(a-b)(b^2-c^2)}=\frac{(a-b)(b-c)(a+b-b+c)}{(a-b)(b-c)(b+c)}=\frac{(a-b)(b-c)(a-c)}{(a-b)(b-c)(b+c)}\)
\(=\frac{a-c}{b+c}\)
Ta có: \(\dfrac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{ab^2-ac^2-b^3+bc^2}\)
\(=\dfrac{a^2\left(b-c\right)-b^2\left[\left(b-c\right)+\left(a-b\right)\right]+c^2\left(a-b\right)}{a\left(b^2-c^2\right)-b\left(b^2-c^2\right)}\)
\(=\dfrac{a^2\left(b-c\right)-b^2\left(b-c\right)-b^2\left(a-b\right)+c^2\left(a-b\right)}{\left(b^2-c^2\right)\left(a-b\right)}\)
\(=\dfrac{\left(b-c\right)\left(a^2-b^2\right)-\left(a-b\right)\left(b^2-c^2\right)}{\left(b-c\right)\left(b+c\right)\left(a-b\right)}\)
\(=\dfrac{\left(b-c\right)\left(a-b\right)\left(a+b\right)-\left(a-b\right)\left(b-c\right)\left(b+c\right)}{\left(b-c\right)\left(b+c\right)\left(a-b\right)}\)
\(=\dfrac{\left(a-b\right)\left(b-c\right)\left(a+b-b-c\right)}{\left(b-c\right)\left(a-b\right)\left(b+c\right)}\)
\(=\dfrac{a-c}{b+c}\)
\(B=\dfrac{a^3+c^3+3ac\left(a+c\right)-b^3-3ac\left(a+c\right)+3abc}{\left(a+b\right)^2+\left(b+c\right)^2+\left(c-a\right)^2}\)
\(=\dfrac{\left(a+c\right)^3-b^3-3ac\left(a+c-b\right)}{\left(a+b\right)^2+\left(b+c\right)^2+\left(c-a\right)^2}\)
\(=\dfrac{\left(a+c-b\right)\left[\left(a+c\right)^2+b\left(a+c\right)+b^2\right]-3ac\left(a+c-b\right)}{\left(a+b\right)^2+\left(b+c\right)^2+\left(c-a\right)^2}\)
\(=\dfrac{\left(a+c-b\right)\left(a^2+b^2+c^2+ab+bc-ac\right)}{\left(a+b\right)^2+\left(b+c\right)^2+\left(c-a\right)^2}\)
\(=\dfrac{-2\left(2a^2+2b^2+2c^2+2ab+2bc-2ca\right)}{\left(a+b\right)^2+\left(b+c\right)^2+\left(c-a\right)^2}\)
\(=\dfrac{-2\left[\left(a+b\right)^2+\left(b+c\right)^2+\left(c-a\right)^2\right]}{\left(a+b\right)^2+\left(b+c\right)^2+\left(c-a\right)^2}=-2\)
Nhớ tik nha ko lần sau mk ko giúp đâu