Question

Cứu mình bài 4 5 với mấy cái khoang tròn th nhà

Answer 1

\(x+y+z=0\Rightarrow x=-\left(y+z\right)\)\(\Rightarrow x^2=y^2+z^2+2yz\)

Thay vào

 \(P=\dfrac{1}{y^2+z^2+2yz+y^2-z^2}+\dfrac{1}{y^2+z^2-y^2-z^2-2yz}+\dfrac{1}{z^2+y^2+z^2+2yz-y^2}\)

\(=\dfrac{1}{2y^2+2yz}-\dfrac{1}{2yz}+\dfrac{1}{2z^2+2yz}\)

\(=\dfrac{1}{2y\left(y+z\right)}-\dfrac{1}{2yz}+\dfrac{1}{2z\left(y+z\right)}\)

\(=\dfrac{z-\left(y+z\right)+y}{2yz\left(y+z\right)}=\dfrac{0}{2yz\left(y+z\right)}=0\)