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有限数学 示例
f(x)=x3√x2-1f(x)=x3√x2−1
解题步骤 1
解题步骤 1.1
化简分母。
解题步骤 1.1.1
将 11 重写为 1212。
f(x)=x3√x2-12f(x)=x3√x2−12
解题步骤 1.1.2
因为两项都是完全平方数,所以使用平方差公式 a2-b2=(a+b)(a-b)a2−b2=(a+b)(a−b) 进行因式分解,其中 a=xa=x 和 b=1b=1。
f(x)=x3√(x+1)(x-1)f(x)=x3√(x+1)(x−1)
f(x)=x3√(x+1)(x-1)f(x)=x3√(x+1)(x−1)
解题步骤 1.2
将 x3√(x+1)(x-1)x3√(x+1)(x−1) 乘以 3√(x+1)(x-1)23√(x+1)(x-1)23√(x+1)(x−1)23√(x+1)(x−1)2。
f(x)=x3√(x+1)(x-1)⋅3√(x+1)(x-1)23√(x+1)(x-1)2f(x)=x3√(x+1)(x−1)⋅3√(x+1)(x−1)23√(x+1)(x−1)2
解题步骤 1.3
合并和化简分母。
解题步骤 1.3.1
将 x3√(x+1)(x-1)x3√(x+1)(x−1) 乘以 3√(x+1)(x-1)23√(x+1)(x-1)23√(x+1)(x−1)23√(x+1)(x−1)2。
f(x)=x3√(x+1)(x-1)23√(x+1)(x-1)3√(x+1)(x-1)2f(x)=x3√(x+1)(x−1)23√(x+1)(x−1)3√(x+1)(x−1)2
解题步骤 1.3.2
对 3√(x+1)(x-1)3√(x+1)(x−1) 进行 11 次方运算。
f(x)=x3√(x+1)(x-1)23√(x+1)(x-1)3√(x+1)(x-1)2f(x)=x3√(x+1)(x−1)23√(x+1)(x−1)3√(x+1)(x−1)2
解题步骤 1.3.3
使用幂法则 aman=am+naman=am+n 合并指数。
f(x)=x3√(x+1)(x-1)23√(x+1)(x-1)1+2f(x)=x3√(x+1)(x−1)23√(x+1)(x−1)1+2
解题步骤 1.3.4
将 11 和 22 相加。
f(x)=x3√(x+1)(x-1)23√(x+1)(x-1)3f(x)=x3√(x+1)(x−1)23√(x+1)(x−1)3
解题步骤 1.3.5
将 3√(x+1)(x-1)33√(x+1)(x−1)3 重写为 (x+1)(x-1)(x+1)(x−1)。
解题步骤 1.3.5.1
使用 n√ax=axnn√ax=axn,将3√(x+1)(x-1)3√(x+1)(x−1) 重写成 ((x+1)(x-1))13((x+1)(x−1))13。
f(x)=x3√(x+1)(x-1)2(((x+1)(x-1))13)3f(x)=x3√(x+1)(x−1)2(((x+1)(x−1))13)3
解题步骤 1.3.5.2
运用幂法则并将指数相乘,(am)n=amn(am)n=amn。
f(x)=x3√(x+1)(x-1)2((x+1)(x-1))13⋅3f(x)=x3√(x+1)(x−1)2((x+1)(x−1))13⋅3
解题步骤 1.3.5.3
组合 1313 和 33。
f(x)=x3√(x+1)(x-1)2((x+1)(x-1))33f(x)=x3√(x+1)(x−1)2((x+1)(x−1))33
解题步骤 1.3.5.4
约去 33 的公因数。
解题步骤 1.3.5.4.1
约去公因数。
f(x)=x3√(x+1)(x-1)2((x+1)(x-1))33
解题步骤 1.3.5.4.2
重写表达式。
f(x)=x3√(x+1)(x-1)2(x+1)(x-1)
f(x)=x3√(x+1)(x-1)2(x+1)(x-1)
解题步骤 1.3.5.5
化简。
f(x)=x3√(x+1)(x-1)2(x+1)(x-1)
f(x)=x3√(x+1)(x-1)2(x+1)(x-1)
f(x)=x3√(x+1)(x-1)2(x+1)(x-1)
解题步骤 1.4
化简分子。
解题步骤 1.4.1
将 3√(x+1)(x-1)2 重写为 3√((x+1)(x-1))2。
f(x)=x3√((x+1)(x-1))2(x+1)(x-1)
解题步骤 1.4.2
对 (x+1)(x-1) 运用乘积法则。
f(x)=x3√(x+1)2(x-1)2(x+1)(x-1)
f(x)=x3√(x+1)2(x-1)2(x+1)(x-1)
f(x)=x3√(x+1)2(x-1)2(x+1)(x-1)
解题步骤 2
The word linear is used for a straight line. A linear function is a function of a straight line, which means that the degree of a linear function must be 0 or 1. In this case, The degree of f(x)=x3√(x+1)2(x-1)2(x+1)(x-1) is -1, which makes the function a nonlinear function.
f(x)=x3√(x+1)2(x-1)2(x+1)(x-1) is not a linear function