1．$f(x),g(x)$定义在$\mathbf R$上，下列说法正确的是( )
A．若$f(f(x))=f(x)$，则$f(x)=x$
B．若$f(f(x))=x$，则$f(x)=x$
C．若$f(g(x))=x$，则$g(f(x))=x$
D．若$f(g(x))=x$，则$g(x_1)=g(x_2)$等价于$x_1=x_2$
2．设$A,B$为抛物线$y^2=2px$($p>0$)上不同的两点，$O$为坐标原点，则$\left|\overrightarrow{OA}+\overrightarrow{OB}\right|^2-\left|\overrightarrow{AB}\right|^2$的最小值为______．
3．已知$A,B$分别为椭圆$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$($a>b>0$)的右顶点和上顶点，直线$y=kx$($k>0$)与椭圆交于$C,D$两点．若四边形$ACBD$的面积的最大值为$2c^2$，则椭圆的离心率为______．
4．已知实数$x,y,z$满足$\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}=1$，则$\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}$的值是______．
5．已知数列$\{x_n\}$满足$$x_{n+1}=\left(\dfrac 2{n^2}+\dfrac 3n+1\right)x_n+n+1,n\in\mathbf N^*,$$且$x_1=3$，求数列$\{x_n\}$的通项公式．
6．已知正整数数列$\{a_n\}$满足$a_1=2$，$a_{n+1}=a_n^2-a_n+1$($n\in\mathbf N^*$)，求证：数列$\{a_n\}$中的任意两项都互质．
7．已知$a,b>0$，$n\in\mathbf N^*$，求证：$\dfrac{1}{a+b}+\dfrac{1}{a+2b}+\cdots +\dfrac{1}{a+nb}<\dfrac{n}{\sqrt{\left(a+\dfrac 12b\right)\left(a+\dfrac{n+1}2b\right)}}$．
参考答案
1．D．
选项A的反例：$f(x)=|x|$；
选项B的反例：$f(x)=-x$；
选项C的反例：$f(x)=\begin{cases} \tan x,&x\neq \dfrac{\pi}2+k\pi,\\1,&x=\dfrac{\pi}2+k\pi,\end{cases} $其中$k\in\mathbf Z$，$g(x)=\arctan x$．
2．$-4p^2$．
由极化恒等式，所求代数式即$4\overrightarrow {OA}\cdot \overrightarrow{OB}$．设$A(2pm^2,2pm),B(2pn^2,2pn)$，则$$4\overrightarrow{OA}\cdot\overrightarrow{OB}=16p^2\left[\left(mn+\dfrac 12\right)^2-\dfrac 14\right]\geqslant -4p^2.$$当且仅当$mn=-\dfrac 12$时取等号．
3．$\dfrac{\sqrt 2}2$．
仿射为圆，可得$\left(\dfrac 12\cdot \sqrt 2a\cdot 2a\right)\cdot \dfrac ba=2c^2$．
4．$0$．
为了出现所求的代数式，在题中条件左右两边同乘以$(x+y+z)$，有$$(x+y+z)\cdot\left(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}\right)=x+y+z,$$于是所求代数式的值为$$\sum_{cyc}\left[(x+y)\cdot \dfrac{z}{x+y}\right]-(x+y+z)=0.$$
5．根据题意，有$$x_{n+1}=\dfrac{(n+1)(n+2)}{n^2}x_n+n+1,$$于是$$\dfrac{x_{n+1}}{(n+1)^2(n+2)}=\dfrac{x_n}{n^2(n+1)}+\dfrac{1}{(n+1)(n+2)},$$进而可得$$\dfrac{x_{n+1}}{(n+1)^2(n+2)}+\dfrac{1}{n+2}=\dfrac{x_n}{n^2(n+1)}+\dfrac{1}{n+1},$$因此$$\dfrac{x_n}{n^2(n+1)}+\dfrac{1}{n+1}=\dfrac{x_{n-1}}{(n-1)^2\cdot n}+\dfrac{1}{n}=\cdots =\dfrac{x_1}{2}+\dfrac 12=2,$$所以$x_n=n^2(2n+1),n\in\mathbf N^*$．
6．根据题意，有$$a_{n+1}-1=a_n\left(a_n-1\right)=a_n\cdots a_1,$$于是对任意$p,q\in\mathbf N^*$且$p<q$，都有$ a_p \mid \left( a_q-1\right)$，因此$\left(a_p,a_q\right)=1$，因此数列$\{a_n\}$中任意两项都互质．
7．柯西加裂项 根据题意，有\[\begin{split} LHS&<\sqrt n\cdot \sqrt{\dfrac{1}{(a+b)^2}+\dfrac{1}{(a+2b)^2}+\cdots +\dfrac{1}{(a+nb)^2}}\\&<\sqrt n \cdot \sqrt{\dfrac{1}{\left(a+\dfrac 12b\right)\left(a+\dfrac 32b\right)}+\dfrac{1}{\left(a+\dfrac 32b\right)\left(a+\dfrac 52b\right)}+\cdots +\dfrac{1}{\left(a+\dfrac{n-1}2b\right)\left(a+\dfrac{n+1}2\right)}}\\&=\sqrt n\cdot \sqrt{\dfrac 1b\left(\dfrac 1{a+\dfrac 12b}-\dfrac{1}{a+\dfrac 32b}+\dfrac{1}{a+\dfrac 32b}-\dfrac 1{a+\dfrac 52b}+\cdots +\dfrac{1}{a+\dfrac{2n-1}2b}-\dfrac{1}{a+\dfrac{2n+1}2b}\right)}\\&=RHS,\end{split}\]因此原不等式得证．
A-L-G不等式 根据题意，有\[\begin{split} LHS&=\sum_{i=1}^n\dfrac{2}{\left(a+\dfrac{2i+1}2b\right)+\left(a+\dfrac{2i-1}2b\right)}\\&<\sum_{i=1}^n\dfrac{\ln\left(a+\dfrac{2i+1}2b\right)-\ln\left(a+\dfrac{2i-1}2b\right)}{b}\\&=\dfrac{\ln\left(a+\dfrac {2n+1}2b\right)-\ln\left(a+\dfrac 1b\right)}{b}\\&=n\cdot \dfrac{\ln\left(a+\dfrac {2n+1}2b\right)-\ln\left(a+\dfrac 1b\right)}{\left(a+\dfrac{2n+1}2b\right)-\left(a+\dfrac 12b\right)}\\&<\dfrac{n}{\sqrt{\left(a+\dfrac 12b\right)\left(a+\dfrac{2n+1}2b\right)}}=RHS,\end{split}\]因此原不等式得证．