Question

[Đặng Xuân Quyền A6]

Giúp mình nha

Answer 1

[Yuri Sweet[𝕿𝖊𝖆𝖒 𝕹𝖊𝖕𝖆𝖑]]

a) Ta có : \(a^3+b^3+c^3=3abc\)

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

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

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

(luôn đúng ,do \(a+b+c=0\))

b) Ta có : \(\left(a+b+c\right)^2=3\left(a^2+b^2+c^2\right)\)

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

\(\Rightarrow2\left(a^2+b^2+c^2\right)=2\left(ab+bc+ca\right)\)

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

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

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

\(\Rightarrow a=b=c\)

c) Ta có : \(\left(a+b+c\right)^2=3\left(ab+bc+ca\right)\)

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

\(\Rightarrow a^2+b^2+c^2=ab+bc+ca\)

\(\Rightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)

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

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

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

\(\Rightarrow a=b=c\)