x^2n+(2x)^n+4^(n-1) 因式分解
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x^2n+(2x)^n+4^(n-1)因式分解x^2n+(2x)^n+4^(n-1)因式分解x^2n+(2x)^n+4^(n-1)因式分解x^2n+(2x)^n+4^(n-1)=(x^n)^2+2*2
x^2n+(2x)^n+4^(n-1) 因式分解
x^2n+(2x)^n+4^(n-1) 因式分解
x^2n+(2x)^n+4^(n-1) 因式分解
x^2n+(2x)^n+4^(n-1)
=(x^n)^2+2*2^(n-1)*x^n+[2^(n-1)]^2
=[x^n+2^(n-1)]^2
=(xn- 1_22n)2
x^2n+(2x)^n+4^(n-1)
=(x^n)^2+2^nx^n+2^2(n-1)
=(x^n)^2+2*2^(n-1)x^n+[2^(n-1)]^2
=[x^n+2^(n-1)]^2
f(x)=e^x-x 求证(1/n)^n+(2/n)^n+...+(n/n)^n
因式分解：x^(2n)+(2x)^(n)+4^(n-1)
x^2n+(2x)^n+4^(n-1) 因式分解
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