两道初二奥数分式运算题.在线等,满意追加2a^5-5a^4+2a^3-8a^2 1、若a是x^2-3x+1=0的根,求----------------------(这是分数线) a^2+12、若r,y,z为实数,且(y-z)^2+(z-x)^2+(x-y)^2=(y+z-2x)^2+(z+
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![两道初二奥数分式运算题.在线等,满意追加2a^5-5a^4+2a^3-8a^2 1、若a是x^2-3x+1=0的根,求----------------------(这是分数线) a^2+12、若r,y,z为实数,且(y-z)^2+(z-x)^2+(x-y)^2=(y+z-2x)^2+(z+](/uploads/image/z/7648664-32-4.jpg?t=%E4%B8%A4%E9%81%93%E5%88%9D%E4%BA%8C%E5%A5%A5%E6%95%B0%E5%88%86%E5%BC%8F%E8%BF%90%E7%AE%97%E9%A2%98.%E5%9C%A8%E7%BA%BF%E7%AD%89%2C%E6%BB%A1%E6%84%8F%E8%BF%BD%E5%8A%A02a%5E5-5a%5E4%2B2a%5E3-8a%5E2+1%E3%80%81%E8%8B%A5a%E6%98%AFx%5E2-3x%2B1%3D0%E7%9A%84%E6%A0%B9%2C%E6%B1%82----------------------%28%E8%BF%99%E6%98%AF%E5%88%86%E6%95%B0%E7%BA%BF%EF%BC%89++++++++++++++++++++++++++++++++++a%5E2%2B12%E3%80%81%E8%8B%A5r%2Cy%2Cz%E4%B8%BA%E5%AE%9E%E6%95%B0%2C%E4%B8%94%28y-z%29%5E2%2B%28z-x%29%5E2%2B%28x-y%29%5E2%3D%28y%2Bz-2x%29%5E2%2B%28z%2B)
两道初二奥数分式运算题.在线等,满意追加2a^5-5a^4+2a^3-8a^2 1、若a是x^2-3x+1=0的根,求----------------------(这是分数线) a^2+12、若r,y,z为实数,且(y-z)^2+(z-x)^2+(x-y)^2=(y+z-2x)^2+(z+
两道初二奥数分式运算题.在线等,满意追加
2a^5-5a^4+2a^3-8a^2
1、若a是x^2-3x+1=0的根,求----------------------(这是分数线)
a^2+1
2、若r,y,z为实数,且(y-z)^2+(z-x)^2+(x-y)^2=(y+z-2x)^2+(z+x-2y)^2+(x=y-2z)^2
(yz+1)(zx+1)(xy+1)
化简 ------------------------
(x^2+1)(y^2+1)(z^2+1)
晕死.出来的格式居然是这样.第一题那第一行和最后一行分别是分子和分母.
第二题第二行和第四行分别是分子和分母
两道初二奥数分式运算题.在线等,满意追加2a^5-5a^4+2a^3-8a^2 1、若a是x^2-3x+1=0的根,求----------------------(这是分数线) a^2+12、若r,y,z为实数,且(y-z)^2+(z-x)^2+(x-y)^2=(y+z-2x)^2+(z+
1.-1
因为
2a^5-5a^4+2a^3-8a^2=2a^3(a^2-3a+1)+a^2(a^2-3a+1)+3a(a^2-3a+1)-3a
a^2+1=(a^2-3a+1)+3a
且a是x^2-3x+1=0的根,
所以
a^2-3a+1=0
所以
(a^5-5a^4+2a^3-8a^2 )/(a^2+1)=[2a^3(a^2-3a+1)+a^2(a^2-3a+1)+3a(a^2-3a+1)-3a]/[(a^2-3a+1)+3a]=(-3a)/(3a)=-1
2.
解法一:
因为(y-z)^2+(z-x)^2+(x-y)^2=(y+z-2x)^2+(z+x-2y)^2+(x+-2z)^2为轮换对称式
所以
x=y=z
所以原式=1
解法二:
由已知式得:
(x-y)^2+(y-z)^2+(x-z)^2=0
所以
x=y=z
所以
原式=1