Question

Cho x,y,z\(\ne\)0 \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=2\)và\(\dfrac{2}{xy}-\dfrac{1}{z^2}=4\) Tính (x+y+z)²⁰¹⁸

Answer 1

\(\)\(\left(\dfrac{1}{x};\dfrac{1}{y};\dfrac{1}{z}\right)\rightarrow\left(a;b;c\right)\)

Viết lại đề: \(\left\{{}\begin{matrix}a+b+c=2\\2ab-c^2=4\end{matrix}\right.\) . Tính \(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^{2018}\)

\(\Leftrightarrow\left(a+b+c\right)^2-2ab+c^2=0\)

\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac-2ab+c^2=0\)

\(\Leftrightarrow a^2+b^2+2c^2+2bc+2ac=0\)

\(\Leftrightarrow\left(a^2+c^2+2ac\right)+\left(b^2+c^2+2bc\right)=0\)

\(\Leftrightarrow\left(a+c\right)^2+\left(b+c\right)^2=0\)

\(\Leftrightarrow....\)