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Ta có: \(a+b+c=0\)
\(\Leftrightarrow\left(a+b+c\right)^2=0\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)
\(\Leftrightarrow a^2+b^2+c^2=-2\left(ab+bc+ca\right)\)
\(\Leftrightarrow\left(a^2+b^2+c^2\right)^2=\left[-2\left(ab+bc+ca\right)\right]^2\)
\(\Leftrightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2=4\left[a^2b^2+b^2c^2+2abc\left(a+b+c+\right)\right]\)
\(\Leftrightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2=4a^2b^2+4b^2c^2+4c^2a^2\) (a + b + c = 0)
\(\Leftrightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2\right)\) (1)
Mà \(\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2\) (2)
Từ (1) và (2) \(\Rightarrow\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+a^4+b^4+c^4=2\left(a^4+b^4+c^4\right)\)
=> đpcm
Ta có :
\(a+b+c=0\)
\(\Rightarrow a=-\left(b+c\right)\)
\(\Rightarrow a^2-b^2-c^2=2bc\)
\(\Rightarrow a^4+b^4+c^4=2\left(a^2b^2+a^2c^2+b^2c^2\right)\)
Cộng \(a^4+b^4+c^4\)vào \(2\left(a^2b^2 +a^2c^2+b^2c^2\right)\)
=> Đpcm
#)Giải :
a) Để C/m a và b là hai số đối nhau => a + b = 0
Ta có : \(2\left(a^2+b^2\right)=\left(a-b\right)^2\)
\(\Leftrightarrow2a^2+2b^2=a^2-2ab+b^2\)
\(\Leftrightarrow2a^2+2b^2-a^2-2ab+b^2=0\)
\(\Leftrightarrow a^2+2ab+b^2=0\)
\(\Leftrightarrow\left(a+b\right)^2=0a\Leftrightarrow a+b=0\)
\(\Rightarrowđpcm\)
Ta có:
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=4\left(a^2+b^2+c^2-ac-bc-ca\right)\)
⇔ \(a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2=4a^2+4b^2+4c^2-4ac-4bc-4ca\)
⇔ \(2a^2+2b^2+2c^2-2ac-2bc-2ca=4a^2+4b^2+4c^2-4ac-4bc-4ca\)
⇔ \(2a^2+2b^2+2c^2-2ac-2bc-2ca=0\)
⇔ \(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
⇔ \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Do \(\left\{{}\begin{matrix}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(c-a\right)^2\ge0\end{matrix}\right.\) \(\forall a,b,c\)
⇒ \(\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{matrix}\right.\) ⇔ \(\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\)
⇔ \(a=b=c\)
⇒ \(ĐPCM\)
Ta có:
\(a+b+c=0\)
\(\Leftrightarrow a+b=-c\)
\(\Leftrightarrow\left(a+b\right)^2=\left(-c\right)^2\)
\(\Leftrightarrow c^2-a^2-b^2=2ab\)
\(\Leftrightarrow\left(c^2-a^2-b^2\right)^2=\left(2ab\right)^2\)
\(\Leftrightarrow a^4+b^4+c^4-2c^2a^2-2b^2c^2+2a^2b^2=4a^2b^2\)
\(\Leftrightarrow a^4+b^4+c^4=2c^2a^2+2b^2c^2+2a^2b^2\)
\(\Leftrightarrow2\left(a^4+b^4+c^4\right)=a^4+b^4+c^4+2c^2a^2+2b^2c^2+2a^2b^2\)
\(\Leftrightarrow2\left(a^4+b^4+c^4\right)=\left(a^2+b^2+c^2\right)^2\)
Ta có: \(a+b+c=0\Rightarrow\left(a+b+c\right)^2=0\)
\(\Leftrightarrow a^2+b^2+c^2+2.\left(ab+ac+bc\right)=0\)
\(\Rightarrow a^2+b^2+c^2=-2\left(ab+bc+ac\right)\)
Thiếu dữ kiện
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=4\left(a^2+b^2+c^2-ab-ac-bc\right)\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc=4\left(a^2+b^2+c^2-ab-ac-bc\right)\)
\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-ac-bc\right)=4\left(a^2+b^2+c^2-ab-ac-bc\right)\)
\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Vì \(\hept{\begin{cases}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(c-a\right)^2\ge0\end{cases}\forall a;b;c}\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Rightarrow}\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Rightarrow}a=b=c}\)
Vậy \(a=b=c\)
Ta có: \(a+b+c=0\Leftrightarrow\left(a+b+c\right)^2=0\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)
\(\Leftrightarrow a^2+b^2+c^2=-2\left(ab+bc+ca\right)\)
\(\Leftrightarrow\left(a^2+b^2+c^2\right)^2=\left[-2\left(ab+bc+ca\right)\right]^2\)
\(\Leftrightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2=4\left[a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)\right]\)
\(\Leftrightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2=4a^2b^2+4b^2c^2+4c^2a^2\) (vì a + b + c = 0)
\(\Leftrightarrow a^4+b^4+c^4=2a^2b^2+2b^2c^2+2c^2a^2\) (1)
Lại có: \(\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)\) (2)
Thay (1) vào (2) ta được:
\(\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+a^4+b^4+c^4=2\left(a^4+b^4+c^4\right)\left(đpcm\right)\)
Ta có: \(a+b+c=0\Rightarrow\left(a+b+c\right)^2=0\)
\(\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=0\)
\(\Rightarrow a^2+b^2+c^2=-2\left(ab+bc+ac\right)\) (1)
Cần chứng minh: \(\left(a^2+b^2+c^2\right)^2=2\left(a^4+b^4+c^4\right)\)
\(\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)=2\left(a^4+b^4+c^4\right)\)
\(\Leftrightarrow2\left(a^2b^2+b^2c^2+a^2c^2\right)=a^4+b^4+c^4\)
\(\Leftrightarrow4\left(a^2b^2+b^2c^2+a^2c^2\right)=\left(a^2+b^2+c^2\right)^2\) (Cộng hai vế cho 2(a2b2+b2c2+a2c2)
\(\Leftrightarrow4\left(a^2b^2+b^2c^2+a^2c^2\right)=\left[-2\left(ab+bc+ac\right)\right]^2\) (vì (1))
\(\Leftrightarrow4\left(a^2b^2+b^2c^2+a^2c^2\right)=4\left(a^2b^2+b^2c^2+a^2c^2\right)+8\left(ab^2c+abc^2+a^2bc\right)\)
\(\Leftrightarrow8\left(ab^2c+abc^2+a^2bc\right)=0\)
<=> 8abc (a+b+c) = 0
<=> 0 = 0 (Vì a+b+c = 0 ) (luôn luôn đúng)
Vậy => đpcm