\(\text{Chắc bn ghi thiếu đề :}\)

\(\hept{\begin{cases}a+b+c=0\\a^2+b^2+c^2=1\end{cases}}\)

\(Tính\)\(a^4+b^4+c^4\)

\(Giải:\)\(\text{Đặt}\)\(M=a^4+b^4+c^4\)

\(\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2\)

\(1=M=\left(2a^2b^2+2b^2c^2+2c^2a^2\right)\)

\(M=1-\left(2a^2b^2+2b^2c^2+2c^2a^2\right)=1-2\left(a^2b^2+b^2c^2+c^2a^2\right)\)

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

\(0=1+2ab+2ac+2bc\)

\(2\left(ab+ac+bc\right)=-1\Rightarrow ab+ac+bc=-\frac{1}{2}\)

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

\(\frac{1}{4}=^2b^2+a^2c^2+b^2c^2+2abc\left(a+b+c\right)\)

\(\Rightarrow^2b^2+a^2c^2+b^2c^2=\frac{1}{4}.0\left(vì\right)a+b+c=0\)

\(M=1-2.\frac{1}{4}=\frac{1}{2}\)

thiếu đề

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

\(\Leftrightarrow ab+bc+ca=-\frac{2009}{2}.\)

\(\Rightarrow\left(ab+bc+ca\right)^2=\frac{2009^2}{4}.\)

\(a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=\frac{2009^2}{4}.\)

\(a^2b^2+b^2c^2+c^2a^2=\frac{2009^2}{4}.\)

Ta có \(\left(a^2+b^2+c^2\right)^2=2009^2\)

\(a^4+b^4+c^4=2009^2-2\left(a^2b^2+b^2c^2+c^2a^2\right)\)

                          \(=2009^2-2.\frac{2009^2}{4}=\frac{2009^2}{2}.\)

\(a^2+b^2+c^2+2\left(ab+bc+ac\right)=0\Rightarrow ab+bc+ac=-\frac{2009}{2}\)

\(\left(ab+bc+ac\right)^2=a^2b^2+a^2c^2+b^2c^2+2abc\left(a+c+b\right)=a^2b^2+a^2c^2+b^2c^2\)\(\Rightarrow a^2b^2+a^2c^2+b^2c^2=\frac{2009^2}{4}\)

\(\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)\)

\(\Rightarrow2009^2=a^4+b^4+c^4+\frac{2009^2}{4}\cdot2\)

\(\Rightarrow a^4+b^4+c^4=\frac{2009^2}{2}\)

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

\(a^2b^2+b^2c^2+c^2a^2=\left(ab+bc+ca\right)^2-2abc\left(a+b+c\right)=\left(\frac{a^2+b^2+c^2}{2}\right)^2=\frac{2009^2}{4}\)

\(A=a^4+b^4+c^4=\left(a^2+b^2+c^2\right)^2-2\left(a^2b^2+b^2c^2+c^2a^2\right)=\frac{2009^2}{2}\)

Ko mất tính tổng quát giả sử \(a\ge b\ge c\),

Let \(f\left(x\right)=x^2\) là hàm lồi trên \(\left(0;2\right)\) và \(\left(2,1,0\right)›\left(a,b,c\right)\)

Áp dụng Bđt Karamata ta có:

\(5=2^2+1^2+0^2\ge a^2+b^2+c^2\)

Hay ta có ĐPCM

Áp dụng tính chất của dãy tỉ số bằng nhau,ta có :

\(\frac{b+c}{a}=\frac{c+a}{b}=\frac{a+b}{c}=\frac{\left(b+c\right)+\left(c+a\right)+\left(a+b\right)}{a+b+c}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)

\(\frac{b+c}{a}=2\Rightarrow b+c=2a\)( 1 )

\(\frac{c+a}{b}=2\Rightarrow c+a=2b\)( 2 )

\(\frac{a+b}{c}=2\Rightarrow a+b=2c\)( 3 )

Từ ( 1 ),(2) và ( 3 ) \(\Rightarrow a=b=c\)

Bạn nào làm hộ mình , mình TK cho 10 TK nhé 

OLM giúp em với !