计算:(2012^3-2×2012^2+1)/(2012^3-3×2012^2+2012+2)
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(2012^3-2×2012^2+1)/(2012^3-3×2012^2+2012+2)把2012=x,则:
=(x³-2x²+1)/(x³-3x²+x+2)
=(x³-2x²+x-x+1)/(x³-2x²+x+1-x²+1)
= [x(x²-2x+1)-(x-1)]/[x(x²-2x+1)-(x²-1)]
= [x(x-1)²-(x-1)]/[x(x-1)²-(x-1)(x+1)]
= [(x-1)(x²-x-1)]/[(x-1)(x²-x-x-1)]
= [(x-1)(x²-x-1)]/[(x-1)(x²-2x-1)]
= (x²-x-1)/(x²-2x-1) 把2012=x,代入
= (2012²-2012-1)/(2012²-2X2012-1)
=4046131/4044199
≈1
=(x³-2x²+1)/(x³-3x²+x+2)
=(x³-2x²+x-x+1)/(x³-2x²+x+1-x²+1)
= [x(x²-2x+1)-(x-1)]/[x(x²-2x+1)-(x²-1)]
= [x(x-1)²-(x-1)]/[x(x-1)²-(x-1)(x+1)]
= [(x-1)(x²-x-1)]/[(x-1)(x²-x-x-1)]
= [(x-1)(x²-x-1)]/[(x-1)(x²-2x-1)]
= (x²-x-1)/(x²-2x-1) 把2012=x,代入
= (2012²-2012-1)/(2012²-2X2012-1)
=4046131/4044199
≈1
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(2012^3-2×2012^2+1)/(2012^3-3×2012^2+2012+2)
=[(2012^3-2×2012^2+2012)-(2012-1)]/[(2012^3-4×2012^2+4×2012)+(2012^2-3×2012+2)]
=(2012-1)[2012(2012-1)-1]/[2012(2012-2)^2+(2012-1)(2012-2)]
=2011×(2011×2012-1)/[(2012-2)(2010×2012+2011)]
=2011×(2011×2012-1)/[2010×(2011×2012-1)]
=2011/2010
=1又1/2010
=[(2012^3-2×2012^2+2012)-(2012-1)]/[(2012^3-4×2012^2+4×2012)+(2012^2-3×2012+2)]
=(2012-1)[2012(2012-1)-1]/[2012(2012-2)^2+(2012-1)(2012-2)]
=2011×(2011×2012-1)/[(2012-2)(2010×2012+2011)]
=2011×(2011×2012-1)/[2010×(2011×2012-1)]
=2011/2010
=1又1/2010
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lz你如果真想要答案,试试计算机啊。
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