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1. 若$x_{1}$,$x_{2}是关于x的一元二次方程2x^{2}+4x+m= 0$的两个实数根,且满足$x_{1}^{2}+x_{2}^{2}+5x_{1}x_{2}-x_{1}^{2}x_{2}^{2}= 0$,则$m$的值为
-2
.
答案：
1. -2 点拨:由题意得$x_{1}+x_{2}=-\frac{4}{2}=-2$,$x_{1}\cdot x_{2}=\frac{m}{2}$,且$\Delta\geq0$,即$4^{2}-4×2m\geq0$,$\therefore m\leq2$.$\because x_{1}^{2}+x_{2}^{2}+5x_{1}x_{2}-x_{1}^{2}x_{2}^{2}=0$,$\therefore (x_{1}+x_{2})^{2}+3x_{1}x_{2}-(x_{1}x_{2})^{2}=0$,$\therefore (-2)^{2}+3\cdot \frac{m}{2}-(\frac{m}{2})^{2}=0$,$\therefore m=-2$或$m=8$,又$\because m\leq2$,$\therefore m=-2$.
2. 已知关于$x的方程a(2x+a)= x(1-x)的两个实数根为x_{1}$,$x_{2}$,$S= \sqrt{x_{1}}+\sqrt{x_{2}}$.
(1)当$a$取什么整数时,$S的值为1$?
(2)是否存在负数$a$,使$S^{2}的值不小于25$? 若存在,请求出$a$的取值范围;若不存在,请说明理由.
(1)当$a$取什么整数时,$S的值为1$?
(2)是否存在负数$a$,使$S^{2}的值不小于25$? 若存在,请求出$a$的取值范围;若不存在,请说明理由.
答案：
2. 解:
(1)$S=\sqrt{x_{1}}+\sqrt{x_{2}}$,$S^{2}=x_{1}+x_{2}+2\sqrt{x_{1}x_{2}}$.由$a(2x+a)=x(1-x)$,得$x^{2}+(2a-1)x+a^{2}=0$.$\because$方程有两个实数根,$\therefore \Delta\geq0$,$\therefore (2a-1)^{2}-4a^{2}\geq0$,$\therefore a\leq0.25$.$\because x_{1},x_{2}$均为非负数,$\therefore x_{1}+x_{2}=1-2a\geq0$,$x_{1}x_{2}=a^{2}\geq0$,$\therefore a\leq0.5$.综上,$a\leq0.25$.$\therefore S^{2}=x_{1}+x_{2}+2\sqrt{x_{1}x_{2}}=1-2a+2|a|=1$.当$0\leq a\leq0.25$时,$1-2a+2a=1$,有$1=1$,满足题意;当$a＜0$时,$1-2a-2a=1$,有$a=0$,不符合设定,舍去.$\therefore$当$0\leq a\leq0.25$时,$S$的值为1.$\because a$为整数,$\therefore a=0$时,$S$的值为1.
(2)存在.要使$S^{2}=x_{1}+x_{2}+2\sqrt{x_{1}x_{2}}\geq25$,只要$1-2a+2|a|\geq25$即可.$\because a＜0$,$\therefore 1-2a-2a\geq25$,$\therefore a\leq-6$.$\therefore$使$S^{2}$的值不小于25的$a$的取值范围为$a\leq-6$.
(1)$S=\sqrt{x_{1}}+\sqrt{x_{2}}$,$S^{2}=x_{1}+x_{2}+2\sqrt{x_{1}x_{2}}$.由$a(2x+a)=x(1-x)$,得$x^{2}+(2a-1)x+a^{2}=0$.$\because$方程有两个实数根,$\therefore \Delta\geq0$,$\therefore (2a-1)^{2}-4a^{2}\geq0$,$\therefore a\leq0.25$.$\because x_{1},x_{2}$均为非负数,$\therefore x_{1}+x_{2}=1-2a\geq0$,$x_{1}x_{2}=a^{2}\geq0$,$\therefore a\leq0.5$.综上,$a\leq0.25$.$\therefore S^{2}=x_{1}+x_{2}+2\sqrt{x_{1}x_{2}}=1-2a+2|a|=1$.当$0\leq a\leq0.25$时,$1-2a+2a=1$,有$1=1$,满足题意;当$a＜0$时,$1-2a-2a=1$,有$a=0$,不符合设定,舍去.$\therefore$当$0\leq a\leq0.25$时,$S$的值为1.$\because a$为整数,$\therefore a=0$时,$S$的值为1.
(2)存在.要使$S^{2}=x_{1}+x_{2}+2\sqrt{x_{1}x_{2}}\geq25$,只要$1-2a+2|a|\geq25$即可.$\because a＜0$,$\therefore 1-2a-2a\geq25$,$\therefore a\leq-6$.$\therefore$使$S^{2}$的值不小于25的$a$的取值范围为$a\leq-6$.
3. 已知在关于$x的分式方程\frac{k - 1}{x - 1}= 2$①和一元二次方程$(2 - k)x^{2}+3mx+(3 - k)n= 0$②中,$k$,$m$,$n$均为实数,方程①的根为非负数.
(1)求$k$的取值范围;
(2)当方程②有两个整数根$x_{1}$,$x_{2}$,$k$为整数,且$k = m + 2$,$n = 1$时,求$m$的值;
(3)当方程②有两个实数根$x_{1}$,$x_{2}$,满足$x_{1}(x_{1}-k)+x_{2}(x_{2}-k)= (x_{1}-k)(x_{2}-k)$,且$k$为负整数时,试判断$|m|\leqslant 2$是否成立,请说明理由.
(1)求$k$的取值范围;
(2)当方程②有两个整数根$x_{1}$,$x_{2}$,$k$为整数,且$k = m + 2$,$n = 1$时,求$m$的值;
(3)当方程②有两个实数根$x_{1}$,$x_{2}$,满足$x_{1}(x_{1}-k)+x_{2}(x_{2}-k)= (x_{1}-k)(x_{2}-k)$,且$k$为负整数时,试判断$|m|\leqslant 2$是否成立,请说明理由.
答案：
3. 解:
(1)$\because$关于$x$的分式方程$\frac{k-1}{x-1}=2$的根为非负数,$\therefore x\geq0$且$x\neq1$,即$x=\frac{k+1}{2}\geq0$,且$\frac{k+1}{2}\neq1$,解得$k\geq-1$且$k\neq1$.又$\because$一元二次方程$(2-k)x^{2}+3mx+(3-k)n=0$中$2-k\neq0$,$\therefore k\neq2$,$\therefore k\geq-1$且$k\neq1$且$k\neq2$.
(2)把$k=m+2$,$n=1$代入原方程,得$-mx^{2}+3mx+(1-m)=0$,$\therefore mx^{2}-3mx+m-1=0$,由原方程有两个整数根,得$m\neq0$且$\Delta\geq0$,$\therefore \Delta=9m^{2}-4m(m-1)=m(5m+4)\geq0$,$\therefore m＞0$或$m\leq-\frac{4}{5}$.$\because x_{1}+x_{2}=3$,$x_{1}\cdot x_{2}=\frac{m-1}{m}=1-\frac{1}{m}$,$x_{1},x_{2}$是整数,$\therefore 1-\frac{1}{m}$为整数,$\therefore m=1$或$m=-1$.由
(1)知$k\geq-1$且$k\neq1$且$k\neq2$,则$m+2\geq-1$且$m+2\neq1$且$m+2\neq2$,即$m\geq-3$且$m\neq-1$且$m\neq0$,$\therefore m=1$.
(3)$|m|\leq2$成立.理由如下:由
(1)知$k\geq-1$且$k\neq1$且$k\neq2$,$\because k$为负整数,$\therefore k=-1$.$\because (2-k)x^{2}+3mx+(3-k)n=0$有两个实数根$x_{1},x_{2}$,$\therefore x_{1}+x_{2}=-\frac{3m}{2-k}=\frac{3m}{k-2}=-m$,$x_{1}x_{2}=\frac{(3-k)n}{2-k}=\frac{4}{3}n$.$\because x_{1}(x_{1}-k)+x_{2}(x_{2}-k)=(x_{1}-k)(x_{2}-k)$,$\therefore x_{1}^{2}-x_{1}k+x_{2}^{2}-x_{2}k=x_{1}x_{2}-x_{1}k-x_{2}k+k^{2}$,$\therefore x_{1}^{2}+x_{2}^{2}=x_{1}x_{2}+k^{2}$,$\therefore (x_{1}+x_{2})^{2}-2x_{1}x_{2}-x_{1}x_{2}=k^{2}$,$\therefore (x_{1}+x_{2})^{2}-3x_{1}x_{2}=k^{2}$,$\therefore (-m)^{2}-3×\frac{4}{3}n=(-1)^{2}$,$\therefore m^{2}-4n=1$,$\therefore n=\frac{m^{2}-1}{4}$.$\because \Delta=(3m)^{2}-4(2-k)(3-k)n=9m^{2}-48n\geq0$,$\therefore 9m^{2}-48×\frac{m^{2}-1}{4}\geq0$,即$m^{2}\leq4$,$\therefore |m|\leq2$成立.
(1)$\because$关于$x$的分式方程$\frac{k-1}{x-1}=2$的根为非负数,$\therefore x\geq0$且$x\neq1$,即$x=\frac{k+1}{2}\geq0$,且$\frac{k+1}{2}\neq1$,解得$k\geq-1$且$k\neq1$.又$\because$一元二次方程$(2-k)x^{2}+3mx+(3-k)n=0$中$2-k\neq0$,$\therefore k\neq2$,$\therefore k\geq-1$且$k\neq1$且$k\neq2$.
(2)把$k=m+2$,$n=1$代入原方程,得$-mx^{2}+3mx+(1-m)=0$,$\therefore mx^{2}-3mx+m-1=0$,由原方程有两个整数根,得$m\neq0$且$\Delta\geq0$,$\therefore \Delta=9m^{2}-4m(m-1)=m(5m+4)\geq0$,$\therefore m＞0$或$m\leq-\frac{4}{5}$.$\because x_{1}+x_{2}=3$,$x_{1}\cdot x_{2}=\frac{m-1}{m}=1-\frac{1}{m}$,$x_{1},x_{2}$是整数,$\therefore 1-\frac{1}{m}$为整数,$\therefore m=1$或$m=-1$.由
(1)知$k\geq-1$且$k\neq1$且$k\neq2$,则$m+2\geq-1$且$m+2\neq1$且$m+2\neq2$,即$m\geq-3$且$m\neq-1$且$m\neq0$,$\therefore m=1$.
(3)$|m|\leq2$成立.理由如下:由
(1)知$k\geq-1$且$k\neq1$且$k\neq2$,$\because k$为负整数,$\therefore k=-1$.$\because (2-k)x^{2}+3mx+(3-k)n=0$有两个实数根$x_{1},x_{2}$,$\therefore x_{1}+x_{2}=-\frac{3m}{2-k}=\frac{3m}{k-2}=-m$,$x_{1}x_{2}=\frac{(3-k)n}{2-k}=\frac{4}{3}n$.$\because x_{1}(x_{1}-k)+x_{2}(x_{2}-k)=(x_{1}-k)(x_{2}-k)$,$\therefore x_{1}^{2}-x_{1}k+x_{2}^{2}-x_{2}k=x_{1}x_{2}-x_{1}k-x_{2}k+k^{2}$,$\therefore x_{1}^{2}+x_{2}^{2}=x_{1}x_{2}+k^{2}$,$\therefore (x_{1}+x_{2})^{2}-2x_{1}x_{2}-x_{1}x_{2}=k^{2}$,$\therefore (x_{1}+x_{2})^{2}-3x_{1}x_{2}=k^{2}$,$\therefore (-m)^{2}-3×\frac{4}{3}n=(-1)^{2}$,$\therefore m^{2}-4n=1$,$\therefore n=\frac{m^{2}-1}{4}$.$\because \Delta=(3m)^{2}-4(2-k)(3-k)n=9m^{2}-48n\geq0$,$\therefore 9m^{2}-48×\frac{m^{2}-1}{4}\geq0$,即$m^{2}\leq4$,$\therefore |m|\leq2$成立.
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