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Ta có: \(a^3+b^3+c^3-3abc=0\) \(\Leftrightarrow a+b+c=0\) hoặc a = b = c
theo gt thi a + b + c \(\ne0\) \(\Rightarrow a=b=c\)
\(\Rightarrow N=\dfrac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\dfrac{3a^2}{9a^2}=\dfrac{1}{3}\)
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 = \(\frac{3a^2}{\left(3a\right)^2}=\frac{1}{3}\)
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 : \(P=\frac{2}{a}.\frac{2}{b}\frac{2}{c}=\frac{8}{abc}\)
Từng ý nhé !!!
\(P=\frac{a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ab}=\frac{a^3}{abc}+\frac{b^3}{abc}+\frac{c^3}{abc}=\frac{1}{abc}\left(a^3+b^3+c^3\right)\)
\(\frac{1}{abc}.3abc=3\)
\(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\frac{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}{2}\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}a+b+c=0\\a=b=c\end{cases}}\)
Xét \(a+b+c=0\) ta có :\(\hept{\begin{cases}a+b=-c\\a+c=-b\\b+c=-a\end{cases}}\)
\(Q=\frac{a^2}{\left(a-b\right)\left(a+b\right)-c^2}+\frac{b^2}{\left(b+c\right)\left(b-c\right)-a^2}+\frac{c^2}{\left(c+a\right)\left(c-a\right)-b^2}\)
\(=\frac{a^2}{-ac+bc-c^2}+\frac{b^2}{-ab+ac-a^2}+\frac{c^2}{-bc+ab-b^2}\)
\(=\frac{a^2}{-c\left(a+c\right)+bc}+\frac{b^2}{-a\left(a+b\right)+ac}+\frac{c^2}{-b\left(c+b\right)+ab}\)
\(=\frac{a^2}{bc+bc}+\frac{b^2}{ac+ac}+\frac{c^2}{ab+ab}\)
\(=\frac{a^2}{2bc}+\frac{b^2}{2ac}+\frac{c^2}{2ab}=\frac{1}{2abc}\left(a^3+b^3+c^3\right)=\frac{1}{2abc}.3abc=\frac{3}{2}\)
Xét \(a=b=c\) ta có :
\(Q=\frac{a^2}{a^2-a^2-a^2}+\frac{b^2}{b^2-b^2-b^2}+\frac{c^2}{c^2-c^2-c^2}=-1-1-1=-3\)
\(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
Hoặc \(a+b+c=0\)
Hoặc \(\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
TH1 : \(a+b+c=0\Rightarrow a=-\left(b+c\right);b=-\left(a+c\right);c=-\left(a+b\right)\)
\(\Rightarrow\)\(A=\left[1-\frac{\left(b+c\right)}{b}\right]\left[1-\frac{\left(a+c\right)}{c}\right]\left[1-\frac{\left(a+b\right)}{a}\right]\)
\(\Rightarrow\)\(A=\left(1-1-\frac{c}{b}\right)\left(1-1-\frac{a}{c}\right)\left(1-1-\frac{b}{a}\right)\)
\(\Rightarrow\)\(A=\left(\frac{-c}{b}\right)\left(\frac{-a}{c}\right)\left(\frac{-b}{a}\right)=-1\)
TH2 : \(\left(a^2+b^2+c^2-ab-bc-ac\right)=0\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Rightarrow\)\(a-b=b-c=c-a=0\)hay \(a=b=c=0\)
\(\Rightarrow\)\(A=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
1.CMR:
a) 3.\(\left(x^2+y^2+z^2\right)-\left(x-y\right)^2\) \(-\left(y-z\right)^2-\left(z-x\right)^2=\left(x+y+z\right)^2\)
a3 + b3 + c3 = 3abc
⇒ a3 + b3 + c3 - 3abc = 0
⇒ ( a3 + b3 ) + c3 - 3abc = 0
⇒ ( a + b )3 - 3ab( a + b ) + c3 - 3abc = 0
⇒ [ ( a + b )3 + c3 ] - [ 3ab( a + b ) + 3abc ] = 0
⇒ ( a + b + c )[ ( a + b )2 - ( a + b ).c + c2 ] - 3ab( a + b + c ) = 0
⇒ ( a + b + c )( a2 + b2 + c2 - ab - bc - ac ) = 0
Vì a + b + c ≠ 0
⇒ a2 + b2 + c2 - ab - bc - ac = 0
⇒ 2( a2 + b2 + c2 - ab - bc - ac ) = 0
⇒ 2a2 + 2b2 + 2c2 - 2ab - 2bc - 2ac = 0
⇒ ( a2 - 2ab + b2 ) + ( b2 - 2bc + c2 ) + ( a2 - 2ac + c2 ) = 0
⇒ ( a - b )2 + ( b - c )2 + ( a - c )2 = 0
Vì \(\hept{\begin{cases}\left(a-b\right)^2\\\left(b-c\right)^2\\\left(a-c\right)^2\end{cases}}\ge0\forall a,b,c\)⇒ ( a - b )2 + ( b - c )2 + ( a - c )2 ≥ 0 ∀ a,b,c
Dấu "=" xảy ra khi a = b = c
Khi đó \(N=\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\frac{a^2+a^2+a^2}{\left(a+a+a\right)^2}=\frac{3a^2}{\left(3a\right)^2}=\frac{3a^2}{9a^2}=\frac{1}{3}\)
Từ \(a^3+b^3+c^3=3abc\)
\(\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)^3-3\left(a+b\right).c\left(a+b+c\right)-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b+c\right)^2-3\left(a+b\right)c-3ab\right]=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2+2ab+2bc+2ca-3ab-3bc-3ca\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(2a^2+2b^2+2c^2-2ab-2bc-2ca\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)\right]=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)
Vì \(a+b+c\ne0\)\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Vì \(\left(a-b\right)^2\ge0\), \(\left(b-c\right)^2\ge0\), \(\left(c-a\right)^2\ge0\)\(\forall a,b,c\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)\(\forall a,b,c\)
Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Leftrightarrow a=b=c\)
Thay \(a=b=c\)vào N ta có: \(N=\frac{3a^2}{\left(3a\right)^2}=\frac{3a^2}{9a^2}=\frac{1}{3}\)
Vậy \(N=\frac{1}{3}\)