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1. (黄石中考改编)阅读材料,解答问题:
已知实数$m,n$满足$m^2 - m - 1 = 0,n^2 - n - 1 = 0$,且$m≠n$,显然$m,n$是方程$x^2 - x - 1 = 0$的两个不相等的实数根,由根与系数的关系可知$m + n = 1,mn = -1$.
(1)已知实数$a,b$满足$a^2 - 15a = 5,b^2 = 15b + 5$,则$\frac{a}{b}+\frac{b}{a}$的值为
(2)已知实数$m,n$满足$m^2 - 2m = 1,n^2 + 2n = 1$且$mn≠1$,求$\frac{mn + n + 1}{n}$的值;
(3)已知实数$m,n$满足$\frac{1}{m^4}+\frac{1}{m^2}=7,n^2 - n = 7$且$n＞0$,求$\frac{1}{m^4}+n^2$的值.
已知实数$m,n$满足$m^2 - m - 1 = 0,n^2 - n - 1 = 0$,且$m≠n$,显然$m,n$是方程$x^2 - x - 1 = 0$的两个不相等的实数根,由根与系数的关系可知$m + n = 1,mn = -1$.
(1)已知实数$a,b$满足$a^2 - 15a = 5,b^2 = 15b + 5$,则$\frac{a}{b}+\frac{b}{a}$的值为
2或-47
;(2)已知实数$m,n$满足$m^2 - 2m = 1,n^2 + 2n = 1$且$mn≠1$,求$\frac{mn + n + 1}{n}$的值;
(3)已知实数$m,n$满足$\frac{1}{m^4}+\frac{1}{m^2}=7,n^2 - n = 7$且$n＞0$,求$\frac{1}{m^4}+n^2$的值.
答案
(1)2或-47 解析:当$a=b$时,$\frac{a}{b}+\frac{b}{a}=2$;当$a≠b$时,$\because a,b$满足$a^2-15a-5=0,b^2-15b-5=0$,$\therefore a,b$是方程$x^2-15x-5=0$的两个根.$\because \Delta＞0$,$\therefore a+b=15,ab=-5$,$\therefore \frac{a}{b}+\frac{b}{a}=\frac{a^2+b^2}{ab}=\frac{(a+b)^2-2ab}{ab}=\frac{15^2-2×(-5)}{-5}=-47$.
(2)解法一:由$n^2+2n-1=0$可知$n≠0$,$\therefore 1+\frac{2}{n}-\frac{1}{n^2}=0$,即$\frac{1}{n^2}-\frac{2}{n}-1=0$.又$m^2-2m-1=0$且$mn≠1$,即$m≠\frac{1}{n}$,$\therefore m,\frac{1}{n}$是方程$x^2-2x-1=0$的两个不相等的实数根,$\therefore m+\frac{1}{n}=2$,$\therefore \frac{mn+n+1}{n}=m+1+\frac{1}{n}=2+1=3$.
解法二:$\because m^2-2m-1=0,n^2+2n-1=0$,$\therefore (-m)^2+2×(-m)-1=0$,$\therefore -m$和$n$都是方程$x^2+2x-1=0$的根,$\Delta＞0$.如果$-m$和$n$是两个不相等的实数根,那么由根与系数的关系可知$-mn=-1$,与题干$mn≠1$矛盾,$\therefore -m$和$n$是同一个实数根,即$-m=n$,$\therefore \frac{mn+n+1}{n}=\frac{-n^2+n+1}{n}=\frac{2n-1+n+1}{n}=3$.
(3)令$\frac{1}{m^2}=a,-n=b$,则$a^2+a-7=0,b^2+b-7=0$.$\because n＞0$,$\therefore \frac{1}{m^2}≠-n$,即$a≠b$,$\therefore a,b$是方程$x^2+x-7=0$的两个不相等的实数根,$\therefore a+b=-1,ab=-7$,故$\frac{1}{m^4}+n^2=a^2+b^2=(a+b)^2-2ab=15$.
(2)解法一:由$n^2+2n-1=0$可知$n≠0$,$\therefore 1+\frac{2}{n}-\frac{1}{n^2}=0$,即$\frac{1}{n^2}-\frac{2}{n}-1=0$.又$m^2-2m-1=0$且$mn≠1$,即$m≠\frac{1}{n}$,$\therefore m,\frac{1}{n}$是方程$x^2-2x-1=0$的两个不相等的实数根,$\therefore m+\frac{1}{n}=2$,$\therefore \frac{mn+n+1}{n}=m+1+\frac{1}{n}=2+1=3$.
解法二:$\because m^2-2m-1=0,n^2+2n-1=0$,$\therefore (-m)^2+2×(-m)-1=0$,$\therefore -m$和$n$都是方程$x^2+2x-1=0$的根,$\Delta＞0$.如果$-m$和$n$是两个不相等的实数根,那么由根与系数的关系可知$-mn=-1$,与题干$mn≠1$矛盾,$\therefore -m$和$n$是同一个实数根,即$-m=n$,$\therefore \frac{mn+n+1}{n}=\frac{-n^2+n+1}{n}=\frac{2n-1+n+1}{n}=3$.
(3)令$\frac{1}{m^2}=a,-n=b$,则$a^2+a-7=0,b^2+b-7=0$.$\because n＞0$,$\therefore \frac{1}{m^2}≠-n$,即$a≠b$,$\therefore a,b$是方程$x^2+x-7=0$的两个不相等的实数根,$\therefore a+b=-1,ab=-7$,故$\frac{1}{m^4}+n^2=a^2+b^2=(a+b)^2-2ab=15$.
2. (1) $( \dfrac{3+\sqrt{3^2 -4\sqrt{2}}}{2\sqrt{2}} )^2 + ( \dfrac{3-\sqrt{3^2 -4\sqrt{2}}}{2\sqrt{2}} )^2 = \_\_\_\_\_\_$
(2) 代数式 $( \dfrac{1+\sqrt{1-4a}}{2} )^3 - ( \dfrac{1+\sqrt{1-4a}}{2} )^2 + a( \dfrac{1+\sqrt{1-4a}}{2} ) -2$ 的值为 ______
(2) 代数式 $( \dfrac{1+\sqrt{1-4a}}{2} )^3 - ( \dfrac{1+\sqrt{1-4a}}{2} )^2 + a( \dfrac{1+\sqrt{1-4a}}{2} ) -2$ 的值为 ______
答案
(1)$\frac{9}{2}-\sqrt{2}$ 解析:对比求根公式$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$可知$a=\sqrt{2}$,$b=-3,c=1$,构造一元二次方程$\sqrt{2}x^2-3x+1=0$,设两个根为$x_1,x_2$,$x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2=(\frac{3}{\sqrt{2}})^2-2×\frac{1}{\sqrt{2}}=\frac{9}{2}-\sqrt{2}$.
(2)$-2$ 解析:设$m=\frac{1+\sqrt{1-4a}}{2}$,$\therefore (\frac{1+\sqrt{1-4a}}{2})^3-(\frac{1+\sqrt{1-4a}}{2})^2+a(\frac{1+\sqrt{1-4a}}{2})-2=m^3-m^2+am-2$.$\because m=\frac{1+\sqrt{1-4a}}{2}=\frac{-(-1)+\sqrt{(-1)^2-4×1· a}}{2×1}$,$\therefore m$是方程$x^2-x+a=0$的根,$\therefore m^2-m=-a$,$\therefore m^3-m^2+am-2=m(m^2-m)+am-2=-am+am-2=-2$.
(2)$-2$ 解析:设$m=\frac{1+\sqrt{1-4a}}{2}$,$\therefore (\frac{1+\sqrt{1-4a}}{2})^3-(\frac{1+\sqrt{1-4a}}{2})^2+a(\frac{1+\sqrt{1-4a}}{2})-2=m^3-m^2+am-2$.$\because m=\frac{1+\sqrt{1-4a}}{2}=\frac{-(-1)+\sqrt{(-1)^2-4×1· a}}{2×1}$,$\therefore m$是方程$x^2-x+a=0$的根,$\therefore m^2-m=-a$,$\therefore m^3-m^2+am-2=m(m^2-m)+am-2=-am+am-2=-2$.
3. 改编题 阅读材料,解答问题:
如果关于$x$的方程$x^2+px+q=0$的两个根是$x_1,x_2$,那么$x_1+x_2=-p$,$x_1 · x_2=q$,这种根与系数的关系被称为韦达定理.韦达定理的逆定理同样成立.
(1)已知某一元二次方程的二次项系数为2,两个根分别为$\sqrt{2}+1,\sqrt{2}-1$,则该方程的一般形式为
(2)①已知实数$x,y,z$满足$x=6-y,z^2=xy-9$,则$z=$
②已知$a,b,c$满足$a+b+c=0,4abc=64$,求正数$c$的最小值.
(3)解方程组$\begin{cases}(m+1)(3m^2+5mn)=144, \\ m^2+4m+5n=24.\end{cases}$
如果关于$x$的方程$x^2+px+q=0$的两个根是$x_1,x_2$,那么$x_1+x_2=-p$,$x_1 · x_2=q$,这种根与系数的关系被称为韦达定理.韦达定理的逆定理同样成立.
(1)已知某一元二次方程的二次项系数为2,两个根分别为$\sqrt{2}+1,\sqrt{2}-1$,则该方程的一般形式为
2x²-4√2x+2=0
.(2)①已知实数$x,y,z$满足$x=6-y,z^2=xy-9$,则$z=$
0
;②已知$a,b,c$满足$a+b+c=0,4abc=64$,求正数$c$的最小值.
(3)解方程组$\begin{cases}(m+1)(3m^2+5mn)=144, \\ m^2+4m+5n=24.\end{cases}$
答案
(1)$2x^2-4\sqrt{2}x+2=0$ 解析:由题意设一元二次方程为$2x^2+bx+c=0$,则$\sqrt{2}+1+\sqrt{2}-1=-\frac{b}{2}$,$(\sqrt{2}+1)(\sqrt{2}-1)=\frac{c}{2}$,解得$b=-4\sqrt{2},c=2$,$\therefore$该方程为$2x^2-4\sqrt{2}x+2=0$.
(2)①$0$ 解析:$\because x=6-y,z^2=xy-9$,$\therefore x+y=6,xy=z^2+9$.设两根为$x,y$的一元二次方程为$a^2-6a+z^2+9=0$.$\because$方程有两个根,$\therefore \Delta=(-6)^2-4(z^2+9)=36-4z^2-36=-4z^2≥0$,$\therefore z^2=0$,$\therefore z=0$.
②$\because a+b+c=0,4abc=64$,$\therefore a+b=-c$,$ab=\frac{16}{c}$,$\therefore a,b$是方程$x^2+cx+\frac{16}{c}=0$的解,$\therefore \Delta=c^2-4·\frac{16}{c}≥0$,即$c^2-\frac{64}{c}≥0$.$\because c$是正数,$\therefore c^3-4^3≥0$,即$c^3≥4^3$,解得$c≥4$,$\therefore$正数$c$的最小值是$4$.
(3)原方程组整理得$\begin{cases}m(m+1)(3m+5n)=144, \\ m^2+m+3m+5n=24.\end{cases}$令$a=m^2+m,b=3m+5n$,则原方程组为$\begin{cases}ab=144, \\ a+b=24.\end{cases}$可得以$a,b$为两根的一元二次方程为$y^2-24y+144=0$,解得$y_1=y_2=12$,即$a=12,b=12$,$\therefore m^2+m=12,3m+5n=12$,$\therefore m_1=3,n_1=\frac{3}{5}$或$m_2=-4,n_2=\frac{24}{5}$.
(2)①$0$ 解析:$\because x=6-y,z^2=xy-9$,$\therefore x+y=6,xy=z^2+9$.设两根为$x,y$的一元二次方程为$a^2-6a+z^2+9=0$.$\because$方程有两个根,$\therefore \Delta=(-6)^2-4(z^2+9)=36-4z^2-36=-4z^2≥0$,$\therefore z^2=0$,$\therefore z=0$.
②$\because a+b+c=0,4abc=64$,$\therefore a+b=-c$,$ab=\frac{16}{c}$,$\therefore a,b$是方程$x^2+cx+\frac{16}{c}=0$的解,$\therefore \Delta=c^2-4·\frac{16}{c}≥0$,即$c^2-\frac{64}{c}≥0$.$\because c$是正数,$\therefore c^3-4^3≥0$,即$c^3≥4^3$,解得$c≥4$,$\therefore$正数$c$的最小值是$4$.
(3)原方程组整理得$\begin{cases}m(m+1)(3m+5n)=144, \\ m^2+m+3m+5n=24.\end{cases}$令$a=m^2+m,b=3m+5n$,则原方程组为$\begin{cases}ab=144, \\ a+b=24.\end{cases}$可得以$a,b$为两根的一元二次方程为$y^2-24y+144=0$,解得$y_1=y_2=12$,即$a=12,b=12$,$\therefore m^2+m=12,3m+5n=12$,$\therefore m_1=3,n_1=\frac{3}{5}$或$m_2=-4,n_2=\frac{24}{5}$.
4. 一题多解 (1)(连云港中考)若$W=5x^2 -4xy + y^2 -2y +8x +3$（$x,y$为实数），则$W$的最小值为________。
(2)已知$m^2 +n^2 +mn +m -n +1=0$，则$\frac{1}{m} + \frac{1}{n} = \_\_\_\_\_\_$。
(2)已知$m^2 +n^2 +mn +m -n +1=0$，则$\frac{1}{m} + \frac{1}{n} = \_\_\_\_\_\_$。
答案
(1)$-2$ 解析:解法一:由题意知$5x^2+(8-4y)x+(y^2-2y+3-W)=0$.$\because x$为实数,$\therefore (8-4y)^2-20(y^2-2y+3-W)≥0$,即$5W≥(y+3)^2-10≥-10$,$\therefore W≥-2$,$\therefore W$的最小值为$-2$.
解法二:$W=5x^2-4xy+y^2-2y+8x+3=x^2+4x^2-4xy+y^2-2y+8x+3=(4x^2-4xy+y^2)-2y+x^2+8x+3=(2x-y)^2+4x-2y+x^2+4x+3=[(2x-y)^2+2(2x-y)+1]+(x^2+4x+4)-2=(2x-y+1)^2+(x+2)^2-2$.$\because x,y$均为实数,$\therefore (2x-y+1)^2≥0,(x+2)^2≥0$,$\therefore W≥-2$,即$W$的最小值为$-2$.
(2)$0$ 解析:解法一:把$m^2+n^2+mn+m-n+1=0$看作关于$m$的一元二次方程,整理得$m^2+(n+1)m+n^2-n+1=0$,由题意可知$m$必然有实数根,$\therefore \Delta=(n+1)^2-4(n^2-n+1)=-3(n-1)^2≥0$,$\therefore n=1$,代入方程可解得$m=-1$,$\therefore \frac{1}{m}+\frac{1}{n}=-1+1=0$.
解法二:将$m^2+n^2+mn+m-n+1=0$变形,得$2m^2+2n^2+2mn+2m-2n+2=0$,整理得$(m^2+2m+1)+(n^2-2n+1)+(m^2+2mn+n^2)=0$,即$(m+1)^2+(n-1)^2+(m+n)^2=0$,$\therefore m+1=0,n-1=0$,解得$m=-1,n=1$.$\therefore \frac{1}{m}+\frac{1}{n}=-1+1=0$.
解法二:$W=5x^2-4xy+y^2-2y+8x+3=x^2+4x^2-4xy+y^2-2y+8x+3=(4x^2-4xy+y^2)-2y+x^2+8x+3=(2x-y)^2+4x-2y+x^2+4x+3=[(2x-y)^2+2(2x-y)+1]+(x^2+4x+4)-2=(2x-y+1)^2+(x+2)^2-2$.$\because x,y$均为实数,$\therefore (2x-y+1)^2≥0,(x+2)^2≥0$,$\therefore W≥-2$,即$W$的最小值为$-2$.
(2)$0$ 解析:解法一:把$m^2+n^2+mn+m-n+1=0$看作关于$m$的一元二次方程,整理得$m^2+(n+1)m+n^2-n+1=0$,由题意可知$m$必然有实数根,$\therefore \Delta=(n+1)^2-4(n^2-n+1)=-3(n-1)^2≥0$,$\therefore n=1$,代入方程可解得$m=-1$,$\therefore \frac{1}{m}+\frac{1}{n}=-1+1=0$.
解法二:将$m^2+n^2+mn+m-n+1=0$变形,得$2m^2+2n^2+2mn+2m-2n+2=0$,整理得$(m^2+2m+1)+(n^2-2n+1)+(m^2+2mn+n^2)=0$,即$(m+1)^2+(n-1)^2+(m+n)^2=0$,$\therefore m+1=0,n-1=0$,解得$m=-1,n=1$.$\therefore \frac{1}{m}+\frac{1}{n}=-1+1=0$.
