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Xét $f(t)=\ln t$ khả vi và liên tục với $t\in \left[2n;2n+1 {} \right]$$+f'(t)=\frac{1}{t}$. Theo đlí Lagrange thì $\exists c\in (2n;2n+1)$:$\frac{f(2n+1)-f(2n)}{2n+1-2n}=f'(c)$$\Rightarrow \ln(2n+1)-\ln2n=\frac{1}{e}>\frac{1}{2n+1}$$\Rightarrow \ln\frac{2n+1}{2n}>\frac{1}{2n+1}$$\Rightarrow \ln(\frac{2n+1}{2n})^2>1$$\Rightarrow (\frac{2n+1}{2n})^{2n+1}>e$$\Rightarrow (\frac{2n}{2n+1})^{2n+1}<\frac{1}{e}$ (1)Xét $f(x)=x^{2n}(1-x), 0$+f'(x)=\left[ 2n-(2n+1x) {} \right]x^{2n-1}=0\Rightarrow x=\frac{2n}{2n+1}$Từ bbt$\Rightarrow x^{2n}(1-x)\leq (\frac{2n}{2n+1})^{2n}.\frac{1}{2n+1}$$\Rightarrow 2n.x^{2n}(1-x)<(\frac{2n}{2n+1})^{2n+1}$ (2)Từ $(1)(2)\Rightarrow đpcm$

Xét $f(t)=\ln t$ khả vi và liên tục với $t\in \left[2n;2n+1 {} \right]$$+f'(t)=\frac{1}{t}$. Theo đlí Lagrange thì $\exists c\in (2n;2n+1)$:$\frac{f(2n+1)-f(2n)}{2n+1-2n}=f'(c)$$\Rightarrow \ln(2n+1)-\ln2n=\frac{1}{e}>\frac{1}{2n+1}$$\Rightarrow \ln\frac{2n+1}{2n}>\frac{1}{2n+1}$$\Rightarrow \ln(\frac{2n+1}{2n})^2>1$$\Rightarrow (\frac{2n+1}{2n})^{2n+1}>e$$\Rightarrow (\frac{2n}{2n+1})^{2n+1}<\frac{1}{e}$ (1)Xét $f(x)=x^{2n}(1-x), 0<x<1$$+f'(x)=\left[ 2n-(2n+1x) {} \right]x^{2n-1}=0\Rightarrow x=\frac{2n}{2n+1}$Từ bbt$\Rightarrow x^{2n}(1-x)\leq (\frac{2n}{2n+1})^{2n}.\frac{1}{2n+1}$$\Rightarrow 2n.x^{2n}(1-x)<(\frac{2n}{2n+1})^{2n+1}$ (2)Từ $(1)(2)\Rightarrow đpcm$

Xét $f(t)=\ln t$ khả vi và liên tục với $t\in \left[2n;2n+1 {} \right]$$+f'(t)=\frac{1}{t}$. Theo đlí Lagrange thì $\exists c\in (2n;2n+1)$:$\frac{f(2n+1)-f(2n)}{2n+1-2n}=f'(c)$$\Rightarrow \ln(2n+1)-\ln2n=\frac{1}{e}>\frac{1}{2n+1}$$\Rightarrow \ln\frac{2n+1}{2n}>\frac{1}{2n+1}$$\Rightarrow \ln(\frac{2n+1}{2n})^2>1$$\Rightarrow (\frac{2n+1}{2n})^{2n+1}>e$$\Rightarrow (\frac{2n}{2n+1})^{2n+1}<\frac{1}{e}$ (1)Xét $f(x)=x^{2n}(1-x), 0<x<1$$+f'(x)=\left[ 2n-(2n+1x) {} \right]x^{2n-1}=0\Rightarrow x=\frac{2n}{2n+1}$Từ bbt$\Rightarrow x^{2n}(1-x)\leq (\frac{2n}{2n+1})^{2n}.\frac{1}{2n+1}$$\Rightarrow 2n.x^{22n}(1-x)<(\frac{2n}{2n+1})^{2n+1}$(2)Từ$(1)(2)\Rightarrow đpcm$

Xét $f(t)=\ln t$ khả vi và liên tục với $t\in \left[2n;2n+1 {} \right]$$+f'(t)=\frac{1}{t}$. Theo đlí Lagrange thì $\exists c\in (2n;2n+1)$:$\frac{f(2n+1)-f(2n)}{2n+1-2n}=f'(c)$$\Rightarrow \ln(2n+1)-\ln2n=\frac{1}{e}>\frac{1}{2n+1}$$\Rightarrow \ln\frac{2n+1}{2n}>\frac{1}{2n+1}$$\Rightarrow \ln(\frac{2n+1}{2n})^2>1$$\Rightarrow (\frac{2n+1}{2n})^{2n+1}>e$$\Rightarrow (\frac{2n}{2n+1})^{2n+1}<\frac{1}{e}$ (1)Xét $f(x)=x^{2n}(1-x), 0$+f'(x)=\left[ 2n-(2n+1x) {} \right]x^{2n-1}=0\Rightarrow x=\frac{2n}{2n+1}$Từ bbt$\Rightarrow x^{2n}(1-x)\leq (\frac{2n}{2n+1})^{2n}.\frac{1}{2n+1}$$\Rightarrow 2n.x^{2n}(1-x)<(\frac{2n}{2n+1})^{2n+1}$ (2)Từ $(1)(2)\Rightarrow đpcm$