第一题：已知y^2+3y-1=0,求y^4/(y^8-3y^4+1)的值.第二题：计算（2007^3-2*2007^2+1）/(2007^3+2007^2-3*2007-2)第二题中:(x^3–2x^2+1)/(x^3+x^2–3x–2)=x^3–x^2–x^2+1/(x^3–x^2-x)这个分母是怎么变

Question

第一题：已知y^2+3y-1=0,求y^4/(y^8-3y^4+1)的值.
第二题：计算（2007^3-2*2007^2+1）/(2007^3+2007^2-3*2007-2)
第二题中:
(x^3 – 2x^2 + 1 ) / (x^3 + x^2 – 3x – 2)
= x^3 – x^2 – x^2 + 1 / (x^3 – x^2 -x)
这个分母是怎么变成那个样子的呢?
说了我就给你加分!

Answer 1

第一题：已知y^2+3y-1=0,求y^4/(y^8-3y^4+1)的值.
y^2+3y-1=0,两边除以y,得 y+3-1/y = 0 ,y – 1/y = -3
两边 平方,y ^2 + 1/y^2 -2 = 9,y ^2 + 1/y^2 = 11,
再次平方,y^4 + 1/y^4 + 2 = 121,y^4 + 1/y^4 = 119,
y^4/(y^8-3y^4+1) = 1/(y^4 – 3 + 1/y^4) = 1/(119 - 3) = 1 / 116
第二题：计算（2007^3-2*2007^2+1）/(2007^3+2007^2-3*2007-2)
(x^3 – 2x^2 + 1 ) / (x^3 + x^2 – 3x – 2)
= x^3 – x^2 – x^2 + 1 / (x^3 – x^2 -x)
= (x^2 – x – 1)(x-1) / (x+2) (x^2 - x – 1)
= (x-1)/(x+2)
（2007^3-2*2007^2+1）/(2007^3+2007^2-3*2007-2)
= (2007-1)/(2007+2)
= 2006/2009