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Ta có: \(a+b+c=0\)
\(\Rightarrow\left(a+b+c\right)^2=0\)
\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ac=0\)
Mặt khác: \(a^2\ge0\forall a;b^2\ge0\forall b;c^2\ge0\forall c\)
\(\Rightarrow a^2+b^2+c^2\ge0\)
Suy ra: \(2ab+2bc+2ac=0\)
\(\Rightarrow2\left(ab+bc+ac\right)=0\)
\(\Rightarrow ab+bc+ac=0\Leftrightarrow2\left(ab+bc+ac\right)^2=0\) (1)
Lại có: \(a^4+b^4+c^4\)
\(=\left(a^2+b^2+c^2\right)^2-2\left[\left(ab\right)^2+\left(bc\right)^2+\left(ac\right)^2\right]\)
\(=0-2\left[\left(ab\right)^2+\left(bc\right)^2+\left(ac\right)^2+2\left(ab+bc+ac\right)-2\left(ab+bc+ac\right)\right]\)
\(=-2\left(ab+bc+ac\right)^2-4\left(ab+bc+ac\right)\)
\(=0\) (2)
Từ (1) và (2) \(\Rightarrow a^4+b^4+c^4=2\left(ab+bc+ac\right)^2=0\)
hay \(a^4+b^4+c^4=2\left(ab+ac+bc\right)^2\)
Kiểm tra hộ mình xem có đúng không ạ!
a) Ta có: \(a+b+c=0\)
\(\Rightarrow a^2+b^2+c^2+2ab+2ac+2bc=0\)
\(\Rightarrow a^2+b^2+c^2=-2\left(ab+bc+ca\right)\)
\(\Rightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2=4\left(a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2\right)\)
\(\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=4\left[a^2b^2+b^2c^2+c^2a^2+2abc\left(b+a+c\right)\right]\)
\(\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=4\left(a^2b^2+b^2c^2+c^2a^2\right)\)
\(\Rightarrow a^4+b^4+c^4=4\left(a^2b^2+b^2c^2+c^2a^2\right)-2\left(a^2b^2+b^2c^2+c^2a^2\right)\)
\(\Rightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2\right)\)
b) Ta có: \(a+b+c=0\)
\(\Rightarrow2abc\left(a+b+c\right)=0\)
\(\Rightarrow2a^2bc+2ab^2c+2abc^2=0\)
Ta lại có:
\(a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2\right)^2\)(chứng minh câu a)
\(\Rightarrow a^4+b^4+c^4=2a^2b^2+2b^2c^2+2c^2a^2+4a^2bc+4ab^2c+4abc^2\)
\(\Rightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2\right)\)
\(\Rightarrow a^4+b^4+c^4=2\left(ab+bc+ca\right)^2\)
Đặt \(\frac{a}{2002}=\frac{b}{2003}=\frac{c}{2004}=k\)
\(\Rightarrow\hept{\begin{cases}a=2002k\\b=2003k\\c=2004k\end{cases}}\)
\(VT=4\left(a-b\right)\left(b-c\right)=4\left(2002k-2003k\right)\left(2003k-2004k\right)=4\left(-1k\right)\left(-1k\right)=4k^2\)
\(VP=\left(c-a\right)^2=\left(2004k-2002k\right)^2=\left(2k\right)^2=4k^2\)
\(\Rightarrow VT=VP\)
\(\Rightarrow4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\left(đpcm\right)\)
4) Ta có :\(\frac{a+1}{2}=\frac{b-1}{3}=\frac{c+2}{4}=\frac{a+b+c+2}{2a+5}=\frac{a+b+c+1-1+2}{2+3+4}=\frac{a+b+c+2}{9}\)(1)
=> 2a + 5 = 9
=> 2a = 4
=> a = 2
Thay a vào (1) ta có :
\(\frac{b-1}{3}=\frac{c+2}{4}=\frac{3}{2}\)
=> \(\hept{\begin{cases}\frac{b-1}{3}=\frac{3}{2}\\\frac{c+2}{4}=\frac{3}{2}\end{cases}}\Rightarrow\hept{\begin{cases}2\left(b-1\right)=9\\2\left(c+2\right)=12\end{cases}}\Rightarrow\hept{\begin{cases}2b-2=9\\2c+4=12\end{cases}}\Rightarrow\hept{\begin{cases}2b=11\\2c=8\end{cases}\Rightarrow\hept{\begin{cases}b=5,5\\c=4\end{cases}}}\)
Vậy a = 2 ; b = 5,5 ; c = 4
5) Đặt \(\frac{a}{2002}=\frac{b}{2003}=\frac{c}{2004}=k\)
=> \(\hept{\begin{cases}a=2002k\\b=2003k\\c=2004k\end{cases}}\)
4(a - b)(b - c) = (c - a)2
=> 4(2002k - 2003k)(2003k - 2004k) = (2002k - 2004k)2
=> 4(-k)(-k) = (-2k)2
=> (-2)2(-k)2 = (-2k)2
=> 22k2 = (2k)2
=> (2k)2 = (2k)2
=> 4(a - b)(b - c) = (c - a)2 (đpcm)
