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Nhận xét:\(\left(a+b\right)^3=a^3+b^3+3a^2b+3ab^2\)
=> \(a^3+b^3=\left(a+b\right)^3-3a^2b-3ab^2\)
ta có \(a^3+b^3+c^3-3abc\)
Thay vào biểu thức trên ta có:
\(\left(a+b\right)^3+c^3-3a^2b-3ab^2-3abc\)
= \(\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
=\(\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)
= \(\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
=\(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)
Vay \(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ac-bc-ab\right)\)
Do \(a^3+b^3+c^3=3abc\)và theo đầu bài \(a+b+c\ne0\)nen \(a^2+b^2+c^2-ac-bc-ab=0\)
=> \(a=b=c\)
Vay N = \(\frac{3a^2}{\left(3a\right)^2}=\frac{1}{3}\)
\(B=\dfrac{a^3+c^3+3ac\left(a+c\right)-b^3-3ac\left(a+c\right)+3abc}{\left(a+b\right)^2+\left(b+c\right)^2+\left(c-a\right)^2}\)
\(=\dfrac{\left(a+c\right)^3-b^3-3ac\left(a+c-b\right)}{\left(a+b\right)^2+\left(b+c\right)^2+\left(c-a\right)^2}\)
\(=\dfrac{\left(a+c-b\right)\left[\left(a+c\right)^2+b\left(a+c\right)+b^2\right]-3ac\left(a+c-b\right)}{\left(a+b\right)^2+\left(b+c\right)^2+\left(c-a\right)^2}\)
\(=\dfrac{\left(a+c-b\right)\left(a^2+b^2+c^2+ab+bc-ac\right)}{\left(a+b\right)^2+\left(b+c\right)^2+\left(c-a\right)^2}\)
\(=\dfrac{-2\left(2a^2+2b^2+2c^2+2ab+2bc-2ca\right)}{\left(a+b\right)^2+\left(b+c\right)^2+\left(c-a\right)^2}\)
\(=\dfrac{-2\left[\left(a+b\right)^2+\left(b+c\right)^2+\left(c-a\right)^2\right]}{\left(a+b\right)^2+\left(b+c\right)^2+\left(c-a\right)^2}=-2\)
Ta có:
\(a^3+b^3+c^3=3abc=>a^3+b^3+c^3-3abc=0\)
\(=>\left(a+b\right)^3-3a^2b-3ab^2+c^3-3abc=0\)
\(=>\left[\left(a+b\right)^3+c^3\right]-3a^2b-3ab^2-3abc=0\)
\(=>\left[\left(a+b\right)^3+c^3\right]-3ab\left(a+b+c\right)=0\)
\(=>\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=0\)
\(=>\left(a+b+c\right)\left(a^2+2ab+b^2-ca-bc+c^2-3ab\right)=0\)
\(=>\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
Vì a3+b3+c3=3abc và a+b+c khác 0
=>\(a^2+b^2+c^2-ab-bc-ca=0\)
\(=>2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(=>\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(=>\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Tổng 3 số không âm = 0 <=> chúng đều = 0
\(< =>\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}< =>a=b=c}\)
Vậy \(\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\frac{3a^2}{\left(3a\right)^2}=\frac{1}{3}\)
\(\)
Ta có ; \(a^3+b^3+c^3=3abc\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-bc-ac\right)-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow\frac{a+b+c}{2}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)
Vì \(a+b+c\ne0\) nên ta có \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\Leftrightarrow a=b=c\)
a) Thay a = b = c vào biểu thức được : \(\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\frac{3a^2}{\left(3a\right)^2}=\frac{3a^2}{9a^2}=\frac{1}{3}\)
b) Thay a = b = c vào P : \(P=\frac{2}{a}.\frac{2}{b}\frac{2}{c}=\frac{8}{abc}\)
Câu 1:
a: \(\left(a+b\right)^3-3ab\left(a+b\right)\)
\(=a^3+3a^2b+3ab^2+b^3-3a^2b-3ab^2\)
\(=a^3+b^3\)
b: \(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)
Ta có:\(A^3+B^3+C^3-3ABC=A^3+3A^2B+3AB^2+B^3+C^3-3AB\left(A+B+C\right)\)
\(=\left(A+B\right)^3+C^3-3AB\left(A+B+C\right)\)\(=\left(A+B+C\right)\left(A^2+B^2+C^2-AB-BC-CA\right)\)
Mặt khác:\(\left(A-B\right)^2+\left(B-C\right)^2+\left(C-A\right)^2=A^2-2AB+B^2+B^2-2BC+C^2+C^2-2CA+A^2\)
\(=2\left(A^2+B^2+C^2-AB-BC-CA\right)\)
Nên giá trị của phân thức là:\(\frac{A+B+C}{2}\)
a. \(\dfrac{a^3+b^3+c^3-3abc}{a^2+b^2+c^2-ab-bc-ac}=\dfrac{\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc}{a^2+b^2+c^2-ab-bc-ac}\)
\(=\dfrac{\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)}{a^2+b^2+c^2-ab-bc-ac}\)
\(=\dfrac{\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)}{a^2+b^2+c^2-ab-bc-ac}\)
\(=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)}{a^2+b^2+c^2-ab-bc-ac}=a+b+c\)
b. \(\dfrac{x^3-y^3+z^3+3xyz}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}=\dfrac{\left(x-y\right)^3+z^3+3xy\left(x-y\right)+3xyz}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)
\(=\dfrac{\left(x-y+z\right)\left[\left(x-y\right)^2-\left(x-y\right)z+z^2\right]+3xy\left(x-y+z\right)}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)
\(=\dfrac{\left(x-y+z\right)\left(x^2-2xy+y^2-xz+yz+z^2+3xy\right)}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)
\(=\dfrac{\left(x-y+z\right)\left(x^2+y^2+z^2+xy+yz-xz\right)}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)
\(=\dfrac{2\left(x-y+z\right)\left(x^2+y^2+z^2+xy+yz-xz\right)}{2\left[\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2\right]}\)
\(=\dfrac{\left(x+y-z\right)\left(2x^2+2y^2+2z^2+2xy+2yz-2zx\right)}{2\left[\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2\right]}\)
\(=\dfrac{\left(x-y+z\right)\left[\left(x^2+2xy+y^2\right)+\left(y^2+2yz+z^2\right)+\left(z^2-2zx+x^2\right)\right]}{2\left[\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2\right]}\)
\(=\dfrac{\left(x-y+z\right)\left[\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2\right]}{2\left[\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2\right]}=\dfrac{x-y+z}{2}\)
ta có: \(a^3+b^3+c^3=3abc\)
=>\(a^3+b^3+c^3-3abc=0\)
=>\(\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)
=>\(\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)
=>\(\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc-3ab+c^2\right)=0\)
=>\(\left(a+b+c\right)\left(a^2+b^2+c^2-ac-bc-ab\right)=0\)
=>\(\left(a+b+c\right)2\left(a^2+b^2+c^2-ac-bc-ab\right)=0\)
=>(a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ac+a^2)=0
=>(a-b)^2+(b-c)^2+(c-a)^2=0
=>a=b=c
N=\(\dfrac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\dfrac{3a^2}{9a^2}=\dfrac{1}{3}\)