Question

1) Cho 3 số a,b,c thỏa mãn \(\left\{{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2=2009\end{matrix}\right.\)

Tính \(A=a^4+b^4+c^4\)

2) Cho 3 số x,y,z thỏa mãn \(x+y+z=3\)

Tìm giá trị lớn nhất của \(B=xy+yz+zx\)

Answer 1

2)Đầu tiên ta cm bđt:\(xy+yz+zx\le\dfrac{\left(x+y+z\right)^2}{3}\)

\(\Leftrightarrow3\left(xy+yz+zx\right)\le x^2+y^2+z^2+2\left(xy+yz+zx\right)\)

\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx\ge0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)(luôn đúng)

\(\Rightarrow xy+yz+zx\le3\)

"="<=>x=y=z=1

Answer 2

Ta có:\(\left\{{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2=2009\end{matrix}\right.\)

\(\Rightarrow ab+bc+ca=\dfrac{-2009}{2}\)

\(\Rightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=1009020,25\)

\(\Rightarrow2\left(a^2b^2+b^2c^2+c^2a^2\right)=2018040,5\)

\(\Rightarrow a^4+b^4+c^4=\left(a^2+b^2+c^2\right)^2-2\left(a^2b^2+b^2c^2+c^2a^2\right)=\)2018040,5