Câu trả lời hay nhất: Do a+b+c=0 =>a+b= -c
Ta có (a+b)^5=c^5
<=>a^5+5a^4b+10a^3b^2+10a^2b^3 + 5ab^4 + b^5=-c^5
<=>a^5+b^5+c^5= -5ab(a^3+2a^2b+2ab^2+b^3)
<=>a^5+b^5+c^5= -5ab( a^2(a+b)+ab(a+b)+b^2(a+b))
<=>a^5+b^5+c^5= -5ab(-c)(a^2+ab+b^2) Vì a+b= -c
<=>2(a^5+b^5+c^5)=5abc2(a^2+ab+b^2)
<=>2(a^5+b^5+c^5)=5abc(a^2+b^2+(a+b)^2)
<=>2(a^5+b^5+c^5)=5abc(a^2+b^2+(-c)^2)
<=>2(a^5+b^5+c^5)=5abc(a^2+b^2+c^2) (đpcm)

Câu hỏi của Ngô Đức Duy - Toán lớp 8 - Học toán với OnlineMath ...

Giải:

Ta có:

\(a+b+c=0\Leftrightarrow a+b=-c\Leftrightarrow\left(a+b\right)^5=\left(-c\right)^5\)

\(\Leftrightarrow a^5+b^5+5ab\left(a^3+2a^2b+2ab^2+b^3\right)=\left(-c\right)^5\)

\(\Leftrightarrow a^5+b^5+c^5=-5ab\left[\left(a+b\right)\left(a^2+b^2-ab\right)+2ab\left(a+b\right)\right]\)

\(=-5ab\left(a+b\right)\left(a^2+b^2+ab\right)\)

\(\Leftrightarrow2\left(a^5+b^5+c^5\right)=5abc\left(2a^2+2b^2+2ab\right)\)

\(=5abc\left[a^2+b^2+\left(a+b\right)^2\right]=5abc\left(a^2+b^2+c^2\right)\) (Đpcm)

Ta có:

\(a+b+c=0\)

\(\Rightarrow a+b=-c\)

\(\Rightarrow\left(a+b\right)^5=-c^5\)

\(\Rightarrow a^5+5a^4b+10a^3b+10a^2b^3+5ab^4+b^5=-c^5\)

\(\Rightarrow a^5+b^5+c^5=5ab\left(a^3+b^3+2a^2b+2ab^2\right)\)

\(\Rightarrow a^5+b^5+c^5=5ab\left[\left(a^3+b^3\right)+2ab\left(a+b\right)\right]\)

\(\Rightarrow a^5+b^5+c^5=5ab\left[\left(a+b\right)\left(a^2-ab+b^2\right)+2ab\left(a+b\right)\right]\)

\(\Rightarrow a^5+b^5+c^5=5ab\left(a+b\right)\left(a^2+ab+b^2\right)\)

\(\Rightarrow a^5+b^5+c^5=-5abc\left(a^2+ab+b^2\right)\)

\(\Rightarrowđpcm\)

a, Vì a,b,c dương nên : \(a+b+c\ge3\sqrt[3]{abc}\)      (1)

 \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}\)    (2)

Nhân vế theo vế 1 và 2 ta có : \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\sqrt[3]{\frac{abc}{abc}}=9\)

Mà a+b+c=1 nên \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge9\)

còn câu b nữa giúp với

\(a+b+c=0\Leftrightarrow a+b=-c\)

\(\Rightarrow\left(a+b\right)^5=-c^5\)

\(\Leftrightarrow a^5+b^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4=-c^2\)

\(\Leftrightarrow a^5+b^5+c^5=-5ab\left(a^3+2a^2b+2ab^2+b^3\right)\)

\(\Leftrightarrow a^5+b^5+c^5=-5ab\left(a+b\right)\left(a^2+ab+b^2\right)\)

\(\Leftrightarrow a^5+b^5+c^5=5abc\left(a^2+ab+b^2\right)\)

\(\Leftrightarrow2\left(a^5+b^5+c^5\right)=5abc\left(2a^2+2ab+2b^2\right)\)

\(\Leftrightarrow2\left(a^5+b^5+c^5\right)=5abc\left(a^2+b^2+\left(a+b\right)^2\right)=5abc\left(a^2+b^2+c^2\right)\)