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HT
1
K
1
27 tháng 11 2023
\(\left(a+b+c\right)^2=a^2+b^2+c^2\)
=>\(a^2+b^2+c^2+2\left(ab+bc+ac\right)=a^2+b^2+c^2\)
=>\(2\left(ab+bc+ac\right)=0\)
=>ab+bc+ac=0
\(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\)
=>\(\dfrac{\left(bc\right)^3+\left(ac\right)^3+\left(ab\right)^3}{\left(abc\right)^3}=\dfrac{3}{abc}\)
=>\(\left(bc\right)^3+\left(ac\right)^3+\left(ab\right)^3=3\left(abc\right)^2\)
\(\Leftrightarrow\left(ab+bc\right)^3-3\cdot ab\cdot bc\cdot\left(ab+bc\right)+\left(ac\right)^3=3\left(abc\right)^2\)
=>\(\left(-ac\right)^3-3\cdot ab\cdot bc\cdot\left(-ac\right)+\left(ac\right)^3-3\left(abc\right)^2=0\)
=>\(-a^3c^3+a^3c^3+3a^2b^2c^2-3a^2b^2c^2=0\)
=>0=0(đúng)
3 tháng 1 2022
\(\Leftrightarrow a^3+b^3+c^3-3abc>=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc>=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)>=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac>=0\)(vì a+b+c>0)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2>=0\)(luôn đúng)
3 tháng 1 2022
\(a^3+b^3+c^3\ge3abc\\ \Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\ge0\\ \Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\ge0\\ \Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\ge0\)
Vì \(a,b,c>0\Leftrightarrow a+b+c>0\)
Lại có \(a^2+b^2+c^2-ab-bc-ca=\dfrac{1}{2}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\ge0\)
Nhân vế theo vế ta được đpcm
Dấu \("="\Leftrightarrow a=b=c\)
PT
2
VT
21 tháng 12 2021
a3+b3+c3= (a+b)3-3ab(a+b)+c3
Thay a+b=-c vào, ta được:
a3 + b3 +c3 = (-c)3 -3ab(-c) +c3 = 3abc (đpcm)
PT
1
25 tháng 8 2021
a+b+c+d=0 => a+d= -b-c; (a+b)3=a3+b3+3ab(a+b) => a3+b3=(a+b)3-3ab(a+b)
a3+d3+b3+d3
=(a+d)3- 3ad(a+d)+ (b+c)3-3bc(b+c) (1)
Do a+d=-b-c nên pt (1) trở thành:
-(b+c)3-3ad(-b-c)+ (b+c)3-3bc(b+c)
=3ad(b+c)-3bc(b+c)
=3(b+c)(ad-bc) <đccm>
NH
2
29 tháng 10 2021
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)
29 tháng 10 2021
\(a^3+b^3+c^3=3abc\\ \Leftrightarrow a^3+b^3+c^3-3abc=0\\ \Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\\ \Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)=0\\ \Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\left(1\right)\end{matrix}\right.\\ \left(1\right)\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\\ \Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\Leftrightarrow a=b=c\)
Vậy \(a^3+b^3+c^3=3abc\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 7 2021
1.
\(a+b+c=0\)
\(\Rightarrow\left(a+b+c\right)^2=0\)
\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ca=0\)
\(\Rightarrow a^2+b^2+c^2=-2\left(ab+bc+ca\right)\)
Ta có:
\(\dfrac{\left(a+2b\right)^2+\left(b+2c\right)^2+\left(c+2a\right)^2}{\left(a-2b\right)^2+\left(b-2c\right)^2+\left(c-2a\right)^2}\)
\(=\dfrac{a^2+4b^2+4ab+b^2+4c^2+4bc+c^2+4a^2+4ca}{a^2+4b^2-4ab+b^2+4c^2-4bc+c^2+4a^2-4ca}\)
\(=\dfrac{5\left(a^2+b^2+c^2\right)+4\left(ab+bc+ca\right)}{5\left(a^2+b^2+c^2\right)-4\left(ab+bc+ca\right)}\)
\(=\dfrac{-10\left(ab+bc+ca\right)+4\left(ab+bc+ca\right)}{-10\left(ab+bc+ca\right)-4\left(ab+bc+ca\right)}\)
\(=\dfrac{-6}{-14}=\dfrac{3}{7}\)
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 7 2021
b.
\(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+3ab\left(a+b\right)-3ab\left(a+b\right)+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(\left(a+b\right)^2-c\left(a+b\right)+c^2\right)-3abc\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\) \(\Leftrightarrow a=b=c\)
\(\Rightarrow\dfrac{ab+2bc+3ca}{3a^2+4b^2+5c^2}=\dfrac{a^2+2a^2+3a^2}{3a^2+4a^2+5a^2}=\dfrac{6}{12}=\dfrac{1}{2}\)
HF
20 tháng 10 2019
a, \(\left(a+b+c\right)^2=3\left(ab+bc+ac\right)\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac=3\left(ab+bc+ac\right)\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ac=0\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
=> a=b=c
Ta có: \(a+b+c=0\Leftrightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\end{matrix}\right.\)
Lại có: \(a^3+a^2c-abc+b^2c+b^3\)
\(=a^2\left(a+c\right)+b^2\left(c+b\right)-abc\)
\(=a^2\left(-b\right)+b^2\left(-a\right)-abc\)
\(=-ab\left(a+b+c\right)=\left(-ab\right).0=0\) (đpcm)