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1. 已知$13x^{2}-6xy+y^{2}-4x+1= 0$,则$x+y$的值为(
A.0
B.1
C.2
D.3
C
)A.0
B.1
C.2
D.3
答案：
C 点拨:$13x^{2}-6xy+y^{2}-4x+1=0,9x^{2}+4x^{2}-6xy+y^{2}-4x+1=0$,即$(9x^{2}-6xy+y^{2})+(4x^{2}-4x+1)=0$,$\therefore (3x-y)^{2}+(2x-1)^{2}=0$,$\therefore 3x-y=0,2x-1=0$,$\therefore x=\frac {1}{2},y=\frac {3}{2}$,$\therefore x+y=2.$
2. 已知实数$x$,$y满足x^{2}+3x+y-3= 0$,求$x+y$的最大值。
答案：
解:由题意,得$y=-x^{2}-3x+3$,$\therefore x+y=-x^{2}-2x+3=-(x+1)^{2}+4≤4$,$\therefore x+y$的最大值为 4.
3. 已知实数$m$,$n满足m^{2}+n^{2}= 2+mn$,求$(2m-3n)^{2}+(m+2n)(m-2n)$的最大值和最小值。
答案：
解:$\because m^{2}+n^{2}=2+mn$,$\therefore (2m-3n)^{2}+(m+2n)(m-2n)=4m^{2}+9n^{2}-12mn+m^{2}-4n^{2}=5m^{2}+5n^{2}-12mn=5(mn+2)-12mn=10-7mn$.$\because m^{2}+n^{2}=2+mn$,$\therefore (m+n)^{2}=2+3mn≥0$,$\therefore mn≥-\frac {2}{3}$,$\because m^{2}+n^{2}=2+mn$,$\therefore (m-n)^{2}=2-mn≥0$,$\therefore mn≤2$,$\therefore -\frac {2}{3}≤mn≤2$,$\therefore -14≤-7mn≤\frac {14}{3}$,$\therefore -4≤10-7mn≤\frac {44}{3}$.$\therefore (2m-3n)^{2}+(m+2n)(m-2n)$的最大值为$\frac {44}{3}$,最小值为-4.
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