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

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

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

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

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

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

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

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

\(\text{Chắc bn ghi thiếu đề :}\)

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

\(Tính\)\(a^4+b^4+c^4\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Ta có: \(a+b+c=0\)

\(\Leftrightarrow a+b=-c\)

\(\Leftrightarrow a^2+2ab+b^2=c^2\)

\(\Leftrightarrow a^2+b^2-c^2=-2ab\)

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

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

Ta lại có: \(a^2+b^2+c^2=2009\)

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

\(\Leftrightarrow2\left(a^4+b^4+c^4\right)=2009^2\)

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

thanhs

Ta có:

\(\left(a+b+c+d\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)\ge\left(a+b+c+d\right).\frac{16}{\left(a+b+c+d\right)}=16\)

\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\ge4\)

Dấu = xảy ra khi \(a=b=c=d=1\)

Áp dụng bđt bu nhi a ta có 

\(\left(a+b\right)^2\le2\left(a^2+b^2\right)\Rightarrow\left(-2-c\right)^2\le2\left(2-c^2\right)\)

=> \(c^2+4c+4\le4-2c^2\)

=> \(3c^2+4c\le0\Rightarrow c\left(3c+4\right)\le0\Rightarrow-\frac{4}{3}\le c\le0\)

\(\frac{d}{b^2}\) hay \(\frac{b^2}{d}\)hả bạn?

Ta có: \(\frac{a^4}{c}+\frac{b^4}{d}\ge\frac{\left(a^2+b^2\right)^2}{c+d}=\frac{1}{c+d}\)

Dấu = xảy ra khi \(\frac{a^2}{c}=\frac{b^2}{d}\)

Do đó: \(VT=\frac{a^2}{c}+\frac{b}{d^2}=\frac{d^2}{b}+\frac{b}{d^2}\ge2\sqrt{\frac{d^2}{b}.\frac{b}{d^2}}=2\)