求(x 1)=x2x 1\n’;’\t’;’\a’各起什么作用f(x2)-f(x1)=loga(x2^2-ax2)-loga(x1^2-ax1)f(x)=x^5/5-ax^3/3 (a 3)x a^2
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![求(x 1)=x2x 1\n’;’\t’;’\a’各起什么作用f(x2)-f(x1)=loga(x2^2-ax2)-loga(x1^2-ax1)f(x)=x^5/5-ax^3/3 (a 3)x a^2](/uploads/image/z/13950565-61-5.jpg?t=%E6%B1%82%28x+1%29%3Dx2x+1%5Cn%E2%80%99%EF%BC%9B%E2%80%99%5Ct%E2%80%99%EF%BC%9B%E2%80%99%5Ca%E2%80%99%E5%90%84%E8%B5%B7%E4%BB%80%E4%B9%88%E4%BD%9C%E7%94%A8f%28x2%29-f%28x1%29%3Dloga%28x2%5E2-ax2%29-loga%28x1%5E2-ax1%29f%28x%29%3Dx%5E5%2F5-ax%5E3%2F3+%28a+3%29x+a%5E2)
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求(x 1)=x2x 1\n’;’\t’;’\a’各起什么作用f(x2)-f(x1)=loga(x2^2-ax2)-loga(x1^2-ax1)f(x)=x^5/5-ax^3/3 (a 3)x a^2
求(x 1)=x2x 1\n’;’\t’;’\a’各起什么作用
f(x2)-f(x1)=loga(x2^2-ax2)-loga(x1^2-ax1)f(x)=x^5/5-ax^3/3 (a 3)x a^2
求(x 1)=x2x 1\n’;’\t’;’\a’各起什么作用f(x2)-f(x1)=loga(x2^2-ax2)-loga(x1^2-ax1)f(x)=x^5/5-ax^3/3 (a 3)x a^2
f(2x 1)比方D所以|x|=0,|y-2/1|=0比方Y=(A' B)(A B')C (BC)4y=28,则y=7