Question

[Mộc Lung Hoa]

Bài 1: cho x+y+z=0 và x^2+y^2+z^2=14 .Tính S=x^4+y^4+z^4

Bài 2: cho x>y>0 và a+b+c=0.Tính S= \(\dfrac{1}{a^2+b^2-c^2}\)+\(\dfrac{1}{b^2+c^2-a^2}\)+\(\dfrac{1}{c^2+a^2-b^2}\)

bài 3: cho a^2 +4b +4=0

b^2 +4c+4=0

c^2 +4a+4=0 .Tính S=a^18+b^18+c^18

Answer 1

[Phương Trâm]

1,

\(x^2+y^2+y^2=14\)

\(\Rightarrow\left(x+y+z\right)^2-2xy-2yz-2zx=14\)

\(\Rightarrow-2\left(xy+yz+zx\right)=14\)

\(\Rightarrow xy+yz+zx=-7\)

\(\Rightarrow\left(xy+yz+zx\right)^2=49\)

\(\Leftrightarrow x^2y^2+y^2z^2+z^2x^2+2x^2yz+2xy^2z+2xyz^2=49\)

\(\Leftrightarrow x^2y^2+y^2z^2+z^2x^2+2xyz\left(x+y+z\right)=49\)

\(\Leftrightarrow x^2y^2+y^2z^2+z^2x^2=49\)

Ta có: \(x^4+y^4+z^4\)

\(=\left(x^2+y^2+z^2\right)^2-2x^2y^2-2y^2z^2-2z^2x^2\)

\(=14^2-2\left(x^2y^2+y^2z^2+z^2x^2\right)\)

\(=14^2-2.49\)

\(=196-98\)

\(=98\)