Cho x, y, z khác 0 thỏa mãn đồng thời 1/x +1/y + 1/z =2 và 2/xy - 1/z^2 = 4 Tính giá trị của biểu thức p=(x+2y+z)^2018

Đáp án: $P=1$

Giải thích các bước giải:

Ta có:
$\dfrac1x+\dfrac1y+\dfrac1z=2$

$\to(\dfrac1x+\dfrac1y+\dfrac1z)^2=2^2$

$\to \dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+\dfrac{2}{xy}+\dfrac{2}{yz}+\dfrac{2}{zx}=4$

$\to \dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+\dfrac{2}{xy}+\dfrac{2}{yz}+\dfrac{2}{zx}=\dfrac{2}{xy}-\dfrac{1}{z^2}$

$\to \dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{2}{z^2}+\dfrac{2}{yz}+\dfrac{2}{zx}=0$

$\to (\dfrac{1}{x^2}+\dfrac{2}{xz}+\dfrac{1}{z^2})+(\dfrac{1}{y^2}+\dfrac{2}{yz}+\dfrac{1}{z^2})=0$

$\to (\dfrac1x+\dfrac1z)^2+(\dfrac1y+\dfrac1z)^2=0$
$\to \dfrac1x+\dfrac1z=\dfrac1y+\dfrac1z=0$ vì $(\dfrac1x+\dfrac1z)^2,(\dfrac1y+\dfrac1z)^2\ge 0$

$\to x=y=-z$

$\to \dfrac1{-z}+\dfrac1{-z}+\dfrac1z=2\to -\dfrac1z=2\to z=-\dfrac12$

$\to x=y=\dfrac12$

$\to x+2y+z=\dfrac12+2.\dfrac12-\dfrac12=1$

$\to P=1$