Câu hỏi của Edogawa Conan

Question

chứng minhb rằng x+y+z=0 thì

$\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2=\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}$

lê thị hương giang · Accepted Answer

$\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{x}\right)^2$

$=\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+2\left(\dfrac{1}{xy}+\dfrac{1}{xz}+\dfrac{1}{yz}\right)$

$=\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+2\left(\dfrac{z}{xyz}+\dfrac{y}{xyz}+\dfrac{x}{xyz}\right)$

$=\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+2\left(\dfrac{x+y+z}{xyz}\right)$

$=\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+2\left(\dfrac{0}{xyz}\right)$

$=\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}$Câu trả lời: $\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{x}\right)^2$

$=\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+2\left(\dfrac{1}{xy}+\dfrac{1}{xz}+\dfrac{1}{yz}\right)$

$=\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+2\left(\dfrac{z}{xyz}+\dfrac{y}{xyz}+\dfrac{x}{xyz}\right)$

$=\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+2\left(\dfrac{x+y+z}{xyz}\right)$

$=\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+2\left(\dfrac{0}{xyz}\right)$

$=\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}$

Violympic toán 8