已知定义在R上的函数f(x)的图像关于原点对称,且当x>0时,f(x)=x²-2x+2,求f(x)的解析式.证明x ³+x在R上是增函数。解要设x1<x2∈R,假如x1=a x2=b 我算到了:(a-b)×(a²+ab+b²)+(a-b) 改成如
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![已知定义在R上的函数f(x)的图像关于原点对称,且当x>0时,f(x)=x²-2x+2,求f(x)的解析式.证明x ³+x在R上是增函数。解要设x1<x2∈R,假如x1=a x2=b 我算到了:(a-b)×(a²+ab+b²)+(a-b) 改成如](/uploads/image/z/8317635-51-5.jpg?t=%E5%B7%B2%E7%9F%A5%E5%AE%9A%E4%B9%89%E5%9C%A8R%E4%B8%8A%E7%9A%84%E5%87%BD%E6%95%B0f%28x%29%E7%9A%84%E5%9B%BE%E5%83%8F%E5%85%B3%E4%BA%8E%E5%8E%9F%E7%82%B9%E5%AF%B9%E7%A7%B0%2C%E4%B8%94%E5%BD%93x%EF%BC%9E0%E6%97%B6%2Cf%28x%29%3Dx%26sup2%3B-2x%2B2%2C%E6%B1%82f%28x%29%E7%9A%84%E8%A7%A3%E6%9E%90%E5%BC%8F.%E8%AF%81%E6%98%8Ex+%26sup3%3B%2Bx%E5%9C%A8R%E4%B8%8A%E6%98%AF%E5%A2%9E%E5%87%BD%E6%95%B0%E3%80%82%E8%A7%A3%E8%A6%81%E8%AE%BEx1%EF%BC%9Cx2%E2%88%88R%EF%BC%8C%E5%81%87%E5%A6%82x1%3Da+x2%3Db+%E6%88%91%E7%AE%97%E5%88%B0%E4%BA%86%EF%BC%9A%28a-b%29%C3%97%28a%26sup2%3B%2Bab%2Bb%26sup2%3B%29%2B%28a-b%29+%E6%94%B9%E6%88%90%E5%A6%82)
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