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ta co :a + b+c=0
=>(a+b+c)^3= 0
<=> a^3 + b^3 + c^3 + 3a^2b+3a^2c + 3b^2a+3b^2c + 3c^2a+3c^2b + 6abc =0
<=>(a^3+b^3+c^3) + (3a^2b+3a^2c+3abc ) +(3b^2a+3b^c +3abc) +(3c^2a+3c^b +3abc ) - 3abc=0
<=>(a^3+b^3+c^3) + 3a(ab+ac+bc) + 3b(ab+bc+ac) + 3c(ac+bc+ab) - 3abc=0
<=>(a^3+b^3+c^3) +3(ab+bc+ac)(a+b+c) -3abc=0
<=>(a^3+b^3+c^3) +3(ab+bc+ac).0 - 3abc =0
<=> a^3+b^3+c^3 -3abc=0
=>a^3+b^3+c^3 =3abc (dpcm)
Ta co
\(a^3+b^3+c^3-3abc\)
=\(\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)
=\(\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
=\(\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2-3ab\right]\)
Ma a+b+c=3
=>\(a^3+b^3+c^3-3abc=0\)
=>\(a^3+b^3+c^3=3abc\)(\(ĐPCM\))
Ta có a3 + b3 + c3 = 3abc
<=> (a + b)3 - 3ab(a + b) + c3 = 3abc
<=> (a + b + c)[(a + b)2 - (a + b)c + c2] - 3ab(a + b + c) = 0
<=> (a + b + c)(a2 + 2ab + b2 - ac - bc + c2 - 3ab) = 0
<=> (a + b + c)(a2 + b2 + c2 - ab - ac - bc) = 0
<=> \(\orbr{\begin{cases}a+b+c=0\left(\text{tmđk}\right)\\a^2+b^2+c^2-ab-ac-bc=0\end{cases}}\)
Khi a2 + b2 + c2 - ab - ac - bc = 0
<=> 2a2 + 2b2 + 2c2 - 2ab - 2ac - 2bc = 0
<=> (a2 - 2ab + b2) + (b2 - 2bc + c2) + (a2 - 2ac + c2) = 0
<=> (a - b)2 + (b - c)2 + (c - a)2 = 0
<=> \(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Leftrightarrow a=b=c\left(\text{loại}\right)\)
Vậy a + b + c = 0
\(a^3+b^3+c^3=3abc\)\(\Leftrightarrow\)\(a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)
\(\Leftrightarrow\)\(\left[\left(a+b\right)^3+c^3\right]-\left[3ab\left(a+b\right)+3abc\right]=0\)\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-ac-bc\right)-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Leftrightarrow\frac{\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]}{2}=0\)
Vì a,b,c > 0 nên a+b+c > 0
Do đó : \(\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}\Rightarrow}a=b=c\)
1) có: a^3 + b^3 + c^3 - 3abc = 0
((a + b)3 + c^3( - 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- ac - bc - ab( = 0
Từ đây cho nhận xét:
+ Nếu a + b + c = 0 có a3 + b3 + c3 = 3abc (I)
a + b + c = 0
+ Nếu a^3 + b^3 + c^3 = 3abc thì
a = b = c
Ta có:
\(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)
\(=\frac{1}{2}\left(a+b+c\right)\left(2a^2+2b^2+2c^2-2ab-2ac-2bc\right)\)
\(=\frac{1}{2}\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\) (1)
Mà \(a+b+c=0\)
\(\left(1\right)\Rightarrow\frac{1}{2}.0.\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)
Vậy: nếu \(a+b+c=0\) thì \(a^3+b^3+c^3-3abc=0\)
Chúc bạn học tốt và tíck cho mìk vs nha bùi thị thu hương!
Xét \(a^3+b^3+c^3-3abc\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
Mà \(a+b+c=0\)
\(\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow a^3+b^3+c^3=3abc\)
thay a^3+b^3=(a+b)^3 -3ab(a+b) .Ta có :
a^3+b^3+c^3-3abc=0
<=>(a+b)^3 -3ab(a+b) +c^3 - 3abc=0
<=>[(a+b)^3 +c^3] -3ab.(a+b+c)=0
<=>(a+b+c). [(a+b)^2 -c.(a+b)+c^2] -3ab(a+b+c)=0
<=>(a+b+c).(a^2+2ab+b^2-ca-cb+c^2-3ab)...
<=>(a+b+c).(a^2+b^2+c^2-ab-bc-ca)=0
luôn đúng do a+b+c=0
Lời giải:
\(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow (a+b+c)^3-3(a+b)(b+c)(c+a)=3abc\)
\(\Leftrightarrow (a+b+c)^3-3[(a+b+c)(ab+bc+ac)-abc]=3abc\)
\(\Leftrightarrow (a+b+c)^3-3(a+b+c)(ab+bc+ac)=0\)
\(\Leftrightarrow (a+b+c)(a^2+b^2+c^2-ab-bc-ac)=0\)
Vì \(a,b,c>0\Rightarrow a+b+c>0\)
Do đó \(a^2+b^2+c^2-ab-bc-ac=0\)
\(\Leftrightarrow 2(a^2+b^2+c^2-ab-bc-ac)=0\)
\(\Leftrightarrow (a-b)^2+(b-c)^2+(c-a)^2=0\)
Ta thấy \((a-b^2;(b-c)^2;(c-a)^2\geq 0\), do đó điều trên xảy ra khi mà:
\(\left\{\begin{matrix} (a-b)^2=0\\ (b-c)^2=0\\ (c-a)^2=0\end{matrix}\right.\Leftrightarrow a=b=c\)
Ta có đpcm.
\(\text{Ta có }:a^3+b^3+c^3=3abc\\ \Leftrightarrow a^3+b^3+c^3-3abc=0\\ \Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Leftrightarrow a^2+b^2+c^2-ab-ac-bc=0\)
\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-ac-bc\right)\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(b^2-2bc+c^2\right)\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(b+c\right)^2=0\)
\(Do\left(a-b\right)^2\ge0\forall x\\ \left(a-c\right)^2\ge0\forall x\\ \left(b-c\right)^2\ge0\forall x\\ \Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(b+c\right)^2\ge0\forall x\)
\(\text{Dấu "=" xảy ra khi: }\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(a-c\right)^2=0\\\left(b-c\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a-b=0\\a-c=0\\b-c=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b\\a=c\\b=c\end{matrix}\right.\Leftrightarrow a=b=c\)
Vậy \(a=b=c\text{ }khi\text{ }a^3+b^3+c^3=3abc\)