Chọn A

A nhé

Ta có:\(a^3+b^3+c^3=3abc\)

\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)

\(\Leftrightarrow\left(a-b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)

\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right).\left[\left(a+b\right)^2-\left(a+b\right).c+c^2\right]-3ab\left(a+b+c\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-bc-ac=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)

\(\Leftrightarrow dpcm\)

Ta có:

\(a^3+b^3+c^3=3abc\)

\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)+3abc=3abc\)

\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-ac-bc=0\end{matrix}\right.\)

Ta có:

\(a^2+b^2+c^2-ab-ac-bc=0\)

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

Vì \(\left(a-b\right)^2\ge0\)

\(\left(b-c\right)^2\ge0\)

\(\left(c-a\right)^2\ge0\)

Mà \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\)

\(\Rightarrow a=b=c\)

a, a+b+c=0 => a+b=-c

=>(a+b)³=(-c)³ 

=>a³+3ab(a+b)+b³=-c³

=>a³-3abc+b³=-c³

=>a³+b³+c³=3abc

b, a²+b²+c²=ab+bc+ca

<=>2(a²+b²+c²)=2(ab+bc+ca)

<=>2a²+2b²+2c²-2ab-2bc-2ca=0

<=>(a²-2ab+b²)+(b²-2bc+c²)+(c²-2ca+a²)=0

<=>(a-b)²+(b-c)²+(c-a)²=0

Mà \(\left(a-b\right)^2\ge0;\left(b-c\right)^2\ge0;\left(c-a\right)^2\ge0\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)

\(\Rightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Rightarrow a=b=c}\)

Áp dụng BĐT AM-GM ta có:

\(VT=\frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c}\)

\(\ge3\sqrt[3]{\frac{abc}{\left(b+c-a\right)\left(a+c-b\right)\left(a+b-c\right)}}\)

Cần chứng minh \(3\sqrt[3]{\frac{abc}{\left(b+c-a\right)\left(a+c-b\right)\left(a+b-c\right)}}\ge3\)

\(\Leftrightarrow\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\le abc\)

Ta có: \(\left(a+b-c\right)\left(b+c-a\right)\le b^2\)

Tương tự nhân theo vế ta có DPCM

câu này vừa thi hsg huyện thiệu hóa xong
a+b+c=1 =>a^3+b^3+c^3+3(a+b)(b+c)*c+a)=1 ....

 a+b+c+d=0 
=>a+b=-(c+d) 
=> (a+b)^3=-(c+d)^3 
=> a^3+b^3+3ab(a+b)=-c^3-d^3-3cd(c+d) 
=> a^3+b^3+c^3+d^3=-3ab(a+b)-3cd(c+d) 
=> a^3+b^3+c^3+d^3=3ab(c+d)-3cd(c+d) ( vi a+b = - (c+d)) 
==> a^3 +b^^3+c^3+d^3==3(c+d)(ab-cd) (đpcm)

hey you, còn câu b,c?

Ta có: \(b^2=a\cdot c\)

=>\(\frac{a}{b}=\frac{b}{c}\) (1)

Ta có: \(c^2=bd\)

=>\(\frac{b}{c}=\frac{c}{d}\) (2)

Từ (1),(2) suy ra \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)

Đặt \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k\)

=>\(\begin{cases}c=dk\\ b=ck=dk\cdot k=dk^2\\ a=bk=dk^2\cdot k=dk^3\end{cases}\)

a: \(\frac{a^3+b^3-c^3}{b^3+c^3-d^3}=\frac{\left(dk^3\right)^3+\left(dk^2\right)^3-\left(dk\right)^3}{\left(dk^2\right)^3+\left(dk\right)^3-d^3}=\frac{d^3k^3\left(k^6+k^3-1\right)}{d^3\left(k^6+k^3-1\right)}=k^3\)

\(\left(\frac{a+b-c}{b+c-d}\right)^3=\left(\frac{dk^3+dk^2-dk}{dk^2+dk-d}\right)^3\)

\(=\left\lbrack\frac{dk\left(k^2+k-1\right)}{d\left(k^2+k-1\right)}\right\rbrack^3=k^3\)

Do đó: \(\frac{a^3+b^3-c^3}{b^3+c^3-d^3}=\left(\frac{a+b-c}{b+c-d}\right)^3\)

b: \(\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\)

\(=\frac{\left(dk^3\right)^3+\left(dk^2\right)^3+\left(dk\right)^3}{\left(dk^2\right)^3+\left(dk\right)^3+d^3}=\frac{d^3k^3\left(k^6+k^3+1\right)}{d^3\left(k^6+k^3+1\right)}=k^3\)

\(\frac{a}{d}=\frac{dk^3}{d}=k^3\)

Do đó: \(\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{a}{d}\)

Lê Minh Tuấn bn tham khảo nha:

 a+b+c+d=0 
=>a+b=-(c+d) 
=> (a+b)^3=-(c+d)^3 
=> a^3+b^3+3ab(a+b)=-c^3-d^3-3cd(c+d) 
=> a^3+b^3+c^3+d^3=-3ab(a+b)-3cd(c+d) 
=> a^3+b^3+c^3+d^3=3ab(c+d)-3cd(c+d) ( vi a+b = - (c+d)) 
==> a^3 +b^^3+c^3+d^3==3(c+d)(ab-cd) (dpcm)

cảm ơn OoO Ledegill2 OoO