…………………………… Question …………………………… in progress 0 Môn Toán Nguyệt Ánh 10 months 2021-04-22T05:51:57+00:00 2021-04-22T05:51:57+00:00 1 Answers 11 views 0
Answers ( )
Giải thích các bước giải:
Ta có:
$\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac1{x+y+z}=\dfrac{(y+z+1)+(x+z+2)+(x+y-3)}{x+y+z}=\dfrac{2(x+y+z)}{x+y+z}=2$
$\to x+y+z=2, y+z+1=2x, x+z+2=2y, x+y-3=2z$
Ta có $y+z+1=2x\to y+z=2x-1$
Mà $x+y+z=2\to x+(2x-1)=2\to 3x-1=2\to 3x=3\to x=1$
Ta có: $x+z+2=2y$
$\to (x+y+z)+2=3y$
$\to 2+2=3y$
$\to y=\dfrac43$
Ta có $x+y-3=2z$
$\to x+y+z-3=3z$
$\to 2-3=3z$
$\to z=-\dfrac13$
$\to A=2016\cdot 1+(\dfrac43)^{2017}+(-\dfrac13)^{2017}$