Question

数学，八年级计算题！

Answer 1

由于 y² + 1/y² = (y+1/y)² - 2
原式左边 = (y+1/y)² - 2 + 2y + 2/y - 6 = (y+1/y)² + 2(y+1/y) - 8 = (y+1/y+4)(y+1/y-2) = 0
于是 y+1/y+4 = 0 或者 y+1/y-2 = 0
解得 y = -2±√3 或 y = 1

Answer 2

解 y²+1/y²+2y+（2-6y）/y=0
y²+（1/y）²+2×y×1/y-2×y×1/y+2y+2/y-6=0
（y+1/y）²+2（y+1/y）-8=0
（y+1/y+1）²-9=0
（ y+1/y+1-3）（y+1/y+1+3）=0
（y+1/y-2）（y+1/y+4）=0
所以y+1/y-2=0或y+1/y+4=0
解得y=1或-2±√3

Answer 3

y^2 1/y^2 2y 2/y-6=0
(y 1/y)^2 2(y 1/y)-8=0
y 1/y=-4或2
y=1或(-2加减根号3)