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\(a+b+c=0\)
\(\Leftrightarrow\left(a+b+c\right)^2=0\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac=0\)
\(\Leftrightarrow a^2+b^2+c^2=-2\left(ab+bc+ac\right)\)
\(\Leftrightarrow\left(a^2+b^2+c^2\right)^2=\left[-2\left(ab+bc+ac\right)\right]^2\)
\(\Leftrightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2a^2c^2=4\left(ab+bc+ac\right)^2\)
\(\Leftrightarrow a^4+b^4+c^4=4\left(ab+bc+ac\right)^2-2a^2b^2-2b^2c^2-2a^2c^2\)
Mà \(\left(ab+bc+ac\right)^2=a^2b^2+b^2c^2+a^2c^2+abc\left(a+b+c\right)\)
\(=a^2b^2+b^2c^2+a^2c^2\)
nên \(a^4+b^4+c^4=4\left(ab+bc+ac\right)^2-2\left(ab+bc+ac\right)^2\)
\(a^4+b^4+c^4=2\left(ab+bc+ac\right)^2\left(đpcm\right)\)
\(a+b+c=0\)
\(\Leftrightarrow\left(a+b+c\right)^2=0\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac=0\)
\(\Leftrightarrow a^2+b^2+c^2=-2\left(ab+bc+ac\right)\)
\(\Leftrightarrow\left(a^2+b^2+c^2\right)^2=4\left(ab+bc+ac\right)^2\)
\(\Leftrightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2a^2c^2\)
\(=4\left(a^2b^2+b^2c^2+a^2c^2+2a^2bc+2ab^2c+2abc^2\right)\)
\(\Leftrightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2a^2c^2\)
\(=4\left(a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)\right)\)
\(\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)\)
\(=4\left(a^2b^2+b^2c^2+a^2c^2\right)\)
\(\Leftrightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+a^2c^2\right)\)
\(\Leftrightarrow2\left(a^4+b^4+c^4\right)=4\left(a^2b^2+b^2c^2+a^2c^2+2a^2bc+2ab^2c+2abc^2\right)\)
\(\Leftrightarrow2\left(a^4+b^4+c^4\right)=4\left(ab+bc+ac\right)^2\)
\(\Leftrightarrow\left(a^4+b^4+c^4\right)=2\left(ab+bc+ac\right)^2\)
Trả lời
Theo đề ra ta có:
\(a+b+c=0\)
\(\Leftrightarrow\left(a+b+c\right)^2=0\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=0\)
\(\Leftrightarrow\left(a^2+b^2+c^2\right)^2=4\left(ab+bc+ca\right)^2\)
\(\Leftrightarrow a^4+b^4+c^4+2\left(a^2\cdot b^2+b^2\cdot c^2+c^2\cdot a^2\right)=4\left(ab+bc+ca\right)^2\)(1)
Lại có:
\(\left(ab+bc+ca\right)^2\)
\(=a^2\cdot b^2+b^2\cdot c^2+c^2\cdot a^2+2bc^2\cdot c+2abc^2+2a^2bc\)
\(=a^2b^2+b^2c^2+c^2a^2=2abc\left(a+b+c\right)\)
\(=a^2b^2+b^2c^2+c^2a^2=2abc\cdot0\)(Do a+b+c=0)
\(=a^2b^2+b^2c^2+c^2a^2\)
Thay \(\left(ab+bc+ca\right)^2=a^2b^2+b^2c^2+c^2a^2\)vào (1); ta có:
\(a^4+b^4+c^4+2\left(ab+bc+ca\right)^2=4\left(ab+bc+ca\right)^2\)
\(\Leftrightarrow a^4+b^4+c^4=2\left(ab+bc+ca\right)^2\)
Vậy \(a,b,c\inℕ\), a+b+c=0 thì \(a^4+b^4+c^4=2\left(ab+bc+ca\right)^2\)(đpcm)
P/s; có gì sai thì góp ý nhé!
Sai đề nha bạn. Không tồn tại 3 số a, b, c > 0 thỏa mãn a + b + c = 0