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10.(12分)(2025·南充)设$x_1,x_2$是关于$x$的方程$(x-1)(x-2)=m^2$的两根.
(1)当$x_1=-1$时,求$x_2$及$m$的值;
(2)求证:$(x_1-1)(x_2-1)≤0$.
(1)当$x_1=-1$时,求$x_2$及$m$的值;
(2)求证:$(x_1-1)(x_2-1)≤0$.
答案
10. (1)解:把$x_1=-1$代入方程$(x-1)(x-2)=m^2$,
得$m^2=6,\therefore m=\pm\sqrt{6},\therefore (x-1)(x-2)=6$,
即$x^2-3x-4=0,\therefore (x-4)(x+1)=0$,
$\therefore x_1=-1,x_2=4,\therefore x_2=4,m=\pm\sqrt{6}$.
(2)证明:方程$(x-1)(x-2)=m^2$可化为$x^2-3x+2-m^2=0$.
$\because \Delta=9-4(2-m^2)=4m^2+1＞0$,
$\therefore$方程有两个不相等的实数根.
$\because$方程$(x-1)(x-2)=m^2$,
即$x^2-3x+2-m^2=0$的两根为$x_1,x_2$,
$\therefore x_1+x_2=3,x_1· x_2=2-m^2$,
$\therefore (x_1-1)(x_2-1)=x_1· x_2-(x_1+x_2)+1=2-m^2-3+1=-m^2$.
$\because m^2≥0,\therefore -m^2≤0$,即$(x_1-1)(x_2-1)≤0$.
得$m^2=6,\therefore m=\pm\sqrt{6},\therefore (x-1)(x-2)=6$,
即$x^2-3x-4=0,\therefore (x-4)(x+1)=0$,
$\therefore x_1=-1,x_2=4,\therefore x_2=4,m=\pm\sqrt{6}$.
(2)证明:方程$(x-1)(x-2)=m^2$可化为$x^2-3x+2-m^2=0$.
$\because \Delta=9-4(2-m^2)=4m^2+1＞0$,
$\therefore$方程有两个不相等的实数根.
$\because$方程$(x-1)(x-2)=m^2$,
即$x^2-3x+2-m^2=0$的两根为$x_1,x_2$,
$\therefore x_1+x_2=3,x_1· x_2=2-m^2$,
$\therefore (x_1-1)(x_2-1)=x_1· x_2-(x_1+x_2)+1=2-m^2-3+1=-m^2$.
$\because m^2≥0,\therefore -m^2≤0$,即$(x_1-1)(x_2-1)≤0$.
11.(16分)已知关于x的一元二次方程$x^2 - (m+2)x + m -1 = 0$.
(1)求证:无论m取何值,方程都有两个不相等的实数根;
(2)如果方程的两个实数根为$x_1, x_2$,且$x_1^2 + x_2^2 - x_1x_2 = 9$,求m的值.
(1)求证:无论m取何值,方程都有两个不相等的实数根;
(2)如果方程的两个实数根为$x_1, x_2$,且$x_1^2 + x_2^2 - x_1x_2 = 9$,求m的值.
答案
11. (1)证明:方程$x^2-(m+2)x+m-1=0$中$a=1,b=-(m+2),c=m-1$.
$\because b^2-4ac=[-(m+2)]^2-4×1×(m-1)=m^2+4m+4-4m+4=m^2+8$,又$m^2≥0,\therefore m^2+8＞0$,
$\therefore$无论$m$取何值,方程都有两个不相等的实数根.
(2)解:$\because$方程$x^2-(m+2)x+m-1=0$的两个实数根为$x_1,x_2,\therefore x_1+x_2=m+2,x_1x_2=m-1$.
$\because x_1^2+x_2^2-x_1x_2=9$,即$(x_1+x_2)^2-3x_1x_2=9$,
$\therefore (m+2)^2-3(m-1)=9$,整理,得$m^2+m-2=0$,
即$(m+2)(m-1)=0$,解得$m_1=-2,m_2=1$.
$\therefore m$的值为$-2$或$1$.
$\because b^2-4ac=[-(m+2)]^2-4×1×(m-1)=m^2+4m+4-4m+4=m^2+8$,又$m^2≥0,\therefore m^2+8＞0$,
$\therefore$无论$m$取何值,方程都有两个不相等的实数根.
(2)解:$\because$方程$x^2-(m+2)x+m-1=0$的两个实数根为$x_1,x_2,\therefore x_1+x_2=m+2,x_1x_2=m-1$.
$\because x_1^2+x_2^2-x_1x_2=9$,即$(x_1+x_2)^2-3x_1x_2=9$,
$\therefore (m+2)^2-3(m-1)=9$,整理,得$m^2+m-2=0$,
即$(m+2)(m-1)=0$,解得$m_1=-2,m_2=1$.
$\therefore m$的值为$-2$或$1$.
12.(20分)阅读材料并解决问题.
材料:已知实数$m,n$满足$m^2 - m -1=0,n^2 -n -1=0$,且$m≠n$,求$\frac{n}{m}+\frac{m}{n}$的值.
解:由题意知,$m,n$是方程$x^2 -x -1=0$的两个不相等的实数根,则$m+n=1,mn=-1$,
$\therefore \frac{n}{m}+\frac{m}{n}=\frac{m^2 +n^2}{mn}=\frac{(m+n)^2 -2mn}{mn}=\frac{1+2}{-1}=-3$.
根据上述材料解决下面的问题:
(1)若一元二次方程$5x^2 +10x -1=0$的两根分别为$x_1,x_2$,则$x_1 +x_2=$
(2)已知实数$m,n$满足$3m^2 -3m -1=0,3n^2 -3n -1=0$,且$m≠n$,求$m^2n +mn^2$的值;
(3)已知实数$p,q$满足$p^2=7p -2,2q^2=7q -1$,且$p≠2q$,求$p^2 +4q^2$的值.
材料:已知实数$m,n$满足$m^2 - m -1=0,n^2 -n -1=0$,且$m≠n$,求$\frac{n}{m}+\frac{m}{n}$的值.
解:由题意知,$m,n$是方程$x^2 -x -1=0$的两个不相等的实数根,则$m+n=1,mn=-1$,
$\therefore \frac{n}{m}+\frac{m}{n}=\frac{m^2 +n^2}{mn}=\frac{(m+n)^2 -2mn}{mn}=\frac{1+2}{-1}=-3$.
根据上述材料解决下面的问题:
(1)若一元二次方程$5x^2 +10x -1=0$的两根分别为$x_1,x_2$,则$x_1 +x_2=$
-2
, $x_1x_2=$$-\frac{1}{5}$
;(2)已知实数$m,n$满足$3m^2 -3m -1=0,3n^2 -3n -1=0$,且$m≠n$,求$m^2n +mn^2$的值;
(3)已知实数$p,q$满足$p^2=7p -2,2q^2=7q -1$,且$p≠2q$,求$p^2 +4q^2$的值.
答案
12. (1)$-2$ $-\frac{1}{5}$
(2)解:$\because m,n$满足$3m^2-3m-1=0,3n^2-3n-1=0$,且$m≠ n$,
$\therefore m,n$是$3x^2-3x-1=0$的两个不相等的实数根,
$\therefore m+n=1,mn=-\frac{1}{3}$,
$\therefore m^2n+mn^2=mn(m+n)=-\frac{1}{3}×1=-\frac{1}{3}$.
(3)解:由题意知,$p$与$2q$为方程$x^2-7x+2=0$的两个不相等的实数根,
$\therefore p+2q=7,2pq=2$,
$\therefore p^2+4q^2=(p+2q)^2-4pq=7^2-2×2=45$.
(2)解:$\because m,n$满足$3m^2-3m-1=0,3n^2-3n-1=0$,且$m≠ n$,
$\therefore m,n$是$3x^2-3x-1=0$的两个不相等的实数根,
$\therefore m+n=1,mn=-\frac{1}{3}$,
$\therefore m^2n+mn^2=mn(m+n)=-\frac{1}{3}×1=-\frac{1}{3}$.
(3)解:由题意知,$p$与$2q$为方程$x^2-7x+2=0$的两个不相等的实数根,
$\therefore p+2q=7,2pq=2$,
$\therefore p^2+4q^2=(p+2q)^2-4pq=7^2-2×2=45$.
