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1. 已知$x_{1},x_{2}是一元二次方程x^{2}-3x-1= 0$的两个根,不解方程,求下列各式的值.
(1)$\frac {1}{x_{1}}+\frac {1}{x_{2}}$; (2)$(1+x_{1})(1+x_{2})$; (3)$\frac {x_{2}}{x_{1}}+\frac {x_{1}}{x_{2}}$.
(1)$\frac {1}{x_{1}}+\frac {1}{x_{2}}$; (2)$(1+x_{1})(1+x_{2})$; (3)$\frac {x_{2}}{x_{1}}+\frac {x_{1}}{x_{2}}$.
答案：
1. 解：
∵$x_{1},x_{2}$是一元二次方程$x^{2}-3x-1=0$的两个根，
∴$x_{1}+x_{2}=3,x_{1}x_{2}=-1$.
(1)$\frac {1}{x_{1}}+\frac {1}{x_{2}}=\frac {x_{1}+x_{2}}{x_{1}x_{2}}=\frac {3}{-1}=-3$.
(2)$(1+x_{1})(1+x_{2})=1+x_{1}x_{2}+(x_{1}+x_{2})=1-1+3=3$.
(3)$\frac {x_{2}}{x_{1}}+\frac {x_{1}}{x_{2}}=\frac {(x_{1}+x_{2})^{2}-2x_{1}x_{2}}{x_{1}x_{2}}=\frac {9+2}{-1}=-11$.
∵$x_{1},x_{2}$是一元二次方程$x^{2}-3x-1=0$的两个根，
∴$x_{1}+x_{2}=3,x_{1}x_{2}=-1$.
(1)$\frac {1}{x_{1}}+\frac {1}{x_{2}}=\frac {x_{1}+x_{2}}{x_{1}x_{2}}=\frac {3}{-1}=-3$.
(2)$(1+x_{1})(1+x_{2})=1+x_{1}x_{2}+(x_{1}+x_{2})=1-1+3=3$.
(3)$\frac {x_{2}}{x_{1}}+\frac {x_{1}}{x_{2}}=\frac {(x_{1}+x_{2})^{2}-2x_{1}x_{2}}{x_{1}x_{2}}=\frac {9+2}{-1}=-11$.
2. 已知$x_{1},x_{2}是一元二次方程2x^{2}-3x-5= 0$的两个根,不解方程,求下列代数式的值.
(1)$x_{1}^{2}+x_{2}^{2}$; (2)$|x_{1}-x_{2}|$.
(1)$x_{1}^{2}+x_{2}^{2}$; (2)$|x_{1}-x_{2}|$.
答案：
2. 解：根据题意，得$x_{1}+x_{2}=\frac {3}{2},x_{1}x_{2}=-\frac {5}{2}$.
(1)$x_{1}^{2}+x_{2}^{2}=(x_{1}+x_{2})^{2}-2x_{1}x_{2}$
$=(\frac {3}{2})^{2}-2×(-\frac {5}{2})=\frac {29}{4}$.
(2)$|x_{1}-x_{2}|^{2}=(x_{1}+x_{2})^{2}-4x_{1}x_{2}$
$=(\frac {3}{2})^{2}-4×(-\frac {5}{2})=\frac {49}{4}$，则$|x_{1}-x_{2}|=\frac {7}{2}$.
(1)$x_{1}^{2}+x_{2}^{2}=(x_{1}+x_{2})^{2}-2x_{1}x_{2}$
$=(\frac {3}{2})^{2}-2×(-\frac {5}{2})=\frac {29}{4}$.
(2)$|x_{1}-x_{2}|^{2}=(x_{1}+x_{2})^{2}-4x_{1}x_{2}$
$=(\frac {3}{2})^{2}-4×(-\frac {5}{2})=\frac {49}{4}$，则$|x_{1}-x_{2}|=\frac {7}{2}$.
3. 已知$x_{1},x_{2}是一元二次方程x^{2}-2x+k+2= 0$的两个实数根,且满足$\frac {1}{x_{1}}+\frac {1}{x_{2}}= k-2$,求$k$的值.
答案：
3. 解：根据题意，得$b^{2}-4ac=(-2)^{2}-4(k+2)≥0$，
解得$k≤-1$.
∵$x_{1},x_{2}$是一元二次方程$x^{2}-2x+k+2=0$的两个实数根，
∴$x_{1}+x_{2}=2,x_{1}x_{2}=k+2$.
∵$x_{1},x_{2}$满足$\frac {1}{x_{1}}+\frac {1}{x_{2}}=k-2$，
∴$\frac {x_{1}+x_{2}}{x_{1}x_{2}}=k-2$，
即$\frac {2}{k+2}=k-2$，解得$k=\pm \sqrt {6}$.
∵$k≤-1$，
∴$k=-\sqrt {6}$.
即$k$的值为$-\sqrt {6}$.
解得$k≤-1$.
∵$x_{1},x_{2}$是一元二次方程$x^{2}-2x+k+2=0$的两个实数根，
∴$x_{1}+x_{2}=2,x_{1}x_{2}=k+2$.
∵$x_{1},x_{2}$满足$\frac {1}{x_{1}}+\frac {1}{x_{2}}=k-2$，
∴$\frac {x_{1}+x_{2}}{x_{1}x_{2}}=k-2$，
即$\frac {2}{k+2}=k-2$，解得$k=\pm \sqrt {6}$.
∵$k≤-1$，
∴$k=-\sqrt {6}$.
即$k$的值为$-\sqrt {6}$.
4. 若$m,n是一元二次方程x^{2}-3x+1= 0$的两个根,求代数式$m^{2}+n^{2}-2m-2n+2025$的值.
答案：
4. 解：
∵$m,n$是一元二次方程$x^{2}-3x+1=0$的两个根，
∴$m+n=3,mn=1$，
∴$m^{2}+n^{2}-2m-2n+2025=(m+n)^{2}-2mn-2(m+n)+2025=3^{2}-2×1-2×3+2025=2026$.
∵$m,n$是一元二次方程$x^{2}-3x+1=0$的两个根，
∴$m+n=3,mn=1$，
∴$m^{2}+n^{2}-2m-2n+2025=(m+n)^{2}-2mn-2(m+n)+2025=3^{2}-2×1-2×3+2025=2026$.
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