分式的基本性质 (9 13:33:3)1,(2x-1)/xy=____2,若a=2/3,则(a2-2a-3)/(a2-7a+12)的值等于____3,已知x2-4x+1=0,求x4+1/x4
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![分式的基本性质 (9 13:33:3)1,(2x-1)/xy=____2,若a=2/3,则(a2-2a-3)/(a2-7a+12)的值等于____3,已知x2-4x+1=0,求x4+1/x4](/uploads/image/z/5630706-18-6.jpg?t=%E5%88%86%E5%BC%8F%E7%9A%84%E5%9F%BA%E6%9C%AC%E6%80%A7%E8%B4%A8+%289+13%3A33%3A3%291%2C%282x-1%29%2Fxy%3D____2%2C%E8%8B%A5a%3D2%2F3%2C%E5%88%99%28a2-2a-3%29%2F%28a2-7a%2B12%29%E7%9A%84%E5%80%BC%E7%AD%89%E4%BA%8E____3%2C%E5%B7%B2%E7%9F%A5x2-4x%2B1%3D0%2C%E6%B1%82x4%2B1%2Fx4)
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分式的基本性质 (9 13:33:3)1,(2x-1)/xy=____2,若a=2/3,则(a2-2a-3)/(a2-7a+12)的值等于____3,已知x2-4x+1=0,求x4+1/x4
分式的基本性质 (9 13:33:3)
1,(2x-1)/xy=____
2,若a=2/3,则(a2-2a-3)/(a2-7a+12)的值等于____
3,已知x2-4x+1=0,求x4+1/x4
分式的基本性质 (9 13:33:3)1,(2x-1)/xy=____2,若a=2/3,则(a2-2a-3)/(a2-7a+12)的值等于____3,已知x2-4x+1=0,求x4+1/x4
1.分解为:2/y-1/xy
2.(a2-2a-3)/(a2-7a+12)
=(a-3)(a+1)/(a-3)(a-4)
=(a+1)/(a-4)
将a=2/3代入得到:
=(a+1)/(a-4)=5/-10=-1/2
3.解方程:x2-4x+1=0 得到:x=2± √3
代入x4+1/x4 =97+56√3+1/(97+56√3)
兄弟,你的题好生奇怪啊。
第一题没看懂
第二题:(a2-2a-3)/(a2-7a+12)=(a-3)(a+1)/(a-3)(a-4),约分后得:
(a+1)/(a-4),带入a=2/3得:结果为1/2
第三题没看懂
1、 (2x-1)/xy= 2x/xy - 1/xy =2/y - 1/xy
2,若a=2/3,则(a2-2a-3)/(a2-7a+12)的值等于
(a2-2a-3) = (a-3)(a+1)
(a2-7a+12)= (a-3)(a-4)
答案等于 (a+1)/(a-4)= - 1/2