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\(a+b+c=0\Rightarrow a+b=-c\)
\(\Rightarrow\left(a+b\right)^5=-c^5\)
\(\Rightarrow a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5=-c^5\)
\(\Rightarrow a^5+b^5+c^5+5ab\left[a^3+2a^2b+2ab^2+b^3\right]=0\)
\(\Rightarrow a^5+b^5+c^5+5ab\left[\left(a+b\right)\left(a^2-ab+b^2\right)+2ab\left(a+b\right)\right]=0\)
\(\Rightarrow a^5+b^5+c^5+5ab\left(a+b\right)\left(a^2+ab+b^2\right)=0\)
\(\Rightarrow2\left(a^5+b^5+c^5\right)+5ab\left(-c\right)\left[2a^2+2ab+2b^2\right]=0\)
\(\Rightarrow2\left(a^5+b^5+c^5\right)-5abc\left[\left(a^2+2ab+b^2\right)+a^2+b^2\right]=0\)
\(\Rightarrow2\left(a^5+b^5+c^5\right)-5abc\left[a^2+b^2+c^2\right]=0\)
\(\Rightarrow2\left(a^5+b^5+c^5\right)=5abc\left(a^2+b^2+c^2\right)\)
Chúc bạn học tốt.
Ta có:
\(a+b+c=0\)
\(\Leftrightarrow\left(a+b\right)^5=\left(-c\right)^5\)
\(\Leftrightarrow a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5=-c^5\)
\(\Leftrightarrow a^5+b^5+c^5=-5ab\left(a^3+2a^2b+2ab^2+b^3\right)\)
\(\Leftrightarrow a^5+b^5+c^5=-5ab\left[\left(a+b\right)\left(a^2-ab+b^2\right)+2ab\left(a+b\right)\right]\)
\(\Leftrightarrow a^5+b^5+c^5=5abc\left(a^2+ab+b^2\right)\)
\(\Leftrightarrow2\left(a^5+b^5+c^5\right)=5abc\left[a^2+b^2+\left(a^2+2ab+b^2\right)\right]\)
\(\Leftrightarrow2\left(a^5+b^5+c^5\right)=5abc\left(a^2+b^2+c^2\right)\)
Câu a : Không hiểu
Câu b :
\(2x^2-x-1=0\)
\(\Leftrightarrow2x^2-2x+x-1=0\)
\(\Leftrightarrow2x\left(x-1\right)+\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\Rightarrow x=1\\2x+1=0\Rightarrow x=-\dfrac{1}{2}\end{matrix}\right.\)
a,\(\left(x+5\right)^2-\left(x+5\right)\left(x-5\right)=20\)
\(\Leftrightarrow\left(x+5\right)\left(x+5-x+5\right)=20\)
\(\Leftrightarrow10x+50=20\)\(\Leftrightarrow x=-3\)
b,\(2x^2-x-1=2x^2-2x+x-1\)
\(=2x\left(x-1\right)+\left(x-1\right)\)\(=\left(x-1\right)\left(2x+1\right)\)\(=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-1}{2}\end{matrix}\right.\)
\(a+b+c=0\Rightarrow c=-\left(a+b\right)\)
\(\Rightarrow a^3+b^3+c^3=a^3+b^3+[-\left(a+b\right)]^3=\)\(a^3+b^3-a^3-3a^2b-3ab^2-b^3\)
\(=3ab[-\left(a+b\right)]=3abc\left(đpcm\right)\)
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)