登录
1. 已知$A=2a^2 - a + \frac{9}{4}, B=2a + 1$. 对于任意实数$a$,试比较$A$与$B$的大小.
答案
解:$A - B = 2a^2 - a + \frac{9}{4} - (2a + 1) = 2a^2 - 3a + \frac{5}{4} = 2[a^2 - \frac{3}{2}a + (\frac{3}{4})^2 - (\frac{3}{4})^2] + \frac{5}{4} = 2(a - \frac{3}{4})^2 + \frac{1}{8}.$
$\because (a - \frac{3}{4})^2 ≥ 0,\therefore 2(a - \frac{3}{4})^2 + \frac{1}{8} ＞ 0,\therefore A ＞ B.$
$\because (a - \frac{3}{4})^2 ≥ 0,\therefore 2(a - \frac{3}{4})^2 + \frac{1}{8} ＞ 0,\therefore A ＞ B.$
2. [2025 盐城经开区期中] 已知关于$x$的一元二次方程$2x^2 + (m+3)x -1 + m = 0$. 求证:无论$m$取何值,方程总有两个不相等的实数根.
答案
证明:由题意,得$a=2,b=m+3,c=-1+m$,
$b^2 - 4ac = (m + 3)^2 - 4 × 2 × (-1 + m) = m^2 - 2m + 17 = (m^2 - 2m + 1) + 16 = (m - 1)^2 + 16.$
$\because (m - 1)^2 ≥ 0,\therefore (m - 1)^2 + 16 ≥ 16 ＞ 0,$
$\therefore$ 无论$m$取何值,一元二次方程$2x^2 + (m + 3)x - 1 + m = 0$总有两个不相等的实数根.
$b^2 - 4ac = (m + 3)^2 - 4 × 2 × (-1 + m) = m^2 - 2m + 17 = (m^2 - 2m + 1) + 16 = (m - 1)^2 + 16.$
$\because (m - 1)^2 ≥ 0,\therefore (m - 1)^2 + 16 ≥ 16 ＞ 0,$
$\therefore$ 无论$m$取何值,一元二次方程$2x^2 + (m + 3)x - 1 + m = 0$总有两个不相等的实数根.
3. [2025苏州吴江青云中学月考] 阅读材料:若 $ m^2 - 2mn + 2n^2 -8n +16=0 $,求 $ m,n $ 的值.
解:∵ $ m^2 -2mn +2n^2 -8n +16=0 $,
∴ $ (m^2 -2mn +n^2)+(n^2 -8n +16)=0 $,
∴ $ (m-n)^2 + (n-4)^2=0 $,
∴ $ m-n=0, n-4=0, ∴n=4, m=4 $.
根据你的观察,探究下面的问题:
(1)已知 $ x^2 +4xy +5y^2 +6y +9=0 $,求 $ x-y $ 的值.
(2)已知 $ △ ABC $ 的三边长 $ a,b,c $ 都是正整数,且满足 $ a^2 -4a +2b^2 -4b +6=0 $,求 $ c $ 的值.
解:∵ $ m^2 -2mn +2n^2 -8n +16=0 $,
∴ $ (m^2 -2mn +n^2)+(n^2 -8n +16)=0 $,
∴ $ (m-n)^2 + (n-4)^2=0 $,
∴ $ m-n=0, n-4=0, ∴n=4, m=4 $.
根据你的观察,探究下面的问题:
(1)已知 $ x^2 +4xy +5y^2 +6y +9=0 $,求 $ x-y $ 的值.
(2)已知 $ △ ABC $ 的三边长 $ a,b,c $ 都是正整数,且满足 $ a^2 -4a +2b^2 -4b +6=0 $,求 $ c $ 的值.
答案
解:(1)$\because x^2 +4xy +5y^2 +6y +9=0,$
$\therefore (x^2 +4xy +4y^2) + (y^2 +6y +9) = 0,$
$\therefore (x + 2y)^2 + (y + 3)^2 = 0,$
$\therefore x + 2y = 0,y + 3 = 0,\therefore y = -3,x = 6,$
$\therefore x - y = 6 - (-3) = 9.$
(2)$\because a^2 -4a +2b^2 -4b +6=0,$
$\therefore (a^2 -4a +4) + (2b^2 -4b +2) = 0,$
$\therefore (a - 2)^2 + 2(b - 1)^2 = 0,$
$\therefore a - 2 = 0,b - 1 = 0,\therefore a = 2,b = 1.$
$\because 2 - 1 ＜ c ＜ 2 + 1,\therefore 1 ＜ c ＜ 3.$
$\because c$ 为正整数,$\therefore c = 2.$
$\therefore (x^2 +4xy +4y^2) + (y^2 +6y +9) = 0,$
$\therefore (x + 2y)^2 + (y + 3)^2 = 0,$
$\therefore x + 2y = 0,y + 3 = 0,\therefore y = -3,x = 6,$
$\therefore x - y = 6 - (-3) = 9.$
(2)$\because a^2 -4a +2b^2 -4b +6=0,$
$\therefore (a^2 -4a +4) + (2b^2 -4b +2) = 0,$
$\therefore (a - 2)^2 + 2(b - 1)^2 = 0,$
$\therefore a - 2 = 0,b - 1 = 0,\therefore a = 2,b = 1.$
$\because 2 - 1 ＜ c ＜ 2 + 1,\therefore 1 ＜ c ＜ 3.$
$\because c$ 为正整数,$\therefore c = 2.$
4. [2025 扬州江都区期中]利用我们学过的完全平方公式及不等式知识能解决代数式的一些问题.观察下列式子：
①$x^2 +4x +2=(x^2 +4x +4)-2=(x +2)^2 -2$.
$\because (x +2)^2 ≥0, \therefore x^2 +4x +2=(x +2)^2 -2 ≥ -2$,
$\therefore$代数式$x^2 +4x +2$有最小值$-2$.
②$-x^2 +2x +3=-(x^2 -2x +1)+4=-(x -1)^2 +4$.
$\because -(x -1)^2 ≤0, \therefore -x^2 +2x +3=-(x -1)^2 +4 ≤4, \therefore$代数式$-x^2 +2x +3$有最大值$4$.
阅读上述材料并完成下列问题：
(1)代数式$a^2 -2a +3$的最小值为
(2)求代数式$a^2 +b^2 +4b -8a +13$的最小值.
(3)如图,在四边形$ABCD$中,对角线$AC,BD$相交于点$O$,且$AC ⊥ BD$,若$AC + BD =12$,求四边形$ABCD$面积的最大值.
①$x^2 +4x +2=(x^2 +4x +4)-2=(x +2)^2 -2$.
$\because (x +2)^2 ≥0, \therefore x^2 +4x +2=(x +2)^2 -2 ≥ -2$,
$\therefore$代数式$x^2 +4x +2$有最小值$-2$.
②$-x^2 +2x +3=-(x^2 -2x +1)+4=-(x -1)^2 +4$.
$\because -(x -1)^2 ≤0, \therefore -x^2 +2x +3=-(x -1)^2 +4 ≤4, \therefore$代数式$-x^2 +2x +3$有最大值$4$.
阅读上述材料并完成下列问题：
(1)代数式$a^2 -2a +3$的最小值为
2
；代数式$-a^2 -6a +4$的最大值为13
.(2)求代数式$a^2 +b^2 +4b -8a +13$的最小值.
(3)如图,在四边形$ABCD$中,对角线$AC,BD$相交于点$O$,且$AC ⊥ BD$,若$AC + BD =12$,求四边形$ABCD$面积的最大值.
答案
解:(1)2 13
$a^2 - 2a + 3 = a^2 - 2a + 1 + 2 = (a - 1)^2 + 2.$
$\because (a - 1)^2 ≥ 0,\therefore (a - 1)^2 + 2 ≥ 2,即 a^2 - 2a + 3 ≥ 2,$
$\therefore a^2 - 2a + 3$ 的最小值为 2.
$-a^2 - 6a + 4 = -(a + 3)^2 + 13.$
$\because -(a + 3)^2 ≤ 0,\therefore -(a + 3)^2 + 13 ≤ 13,$
$\therefore -a^2 - 6a + 4$ 的最大值为 13.
(2)$a^2 + b^2 + 4b - 8a + 13 = a^2 - 8a + 16 + b^2 + 4b + 4 - 7 = (a - 4)^2 + (b + 2)^2 - 7.$
$\because (a - 4)^2 ≥ 0,(b + 2)^2 ≥ 0,$
$\therefore a^2 + b^2 + 4b - 8a + 13 = (a - 4)^2 + (b + 2)^2 - 7 ≥ -7,$
$\therefore$ 代数式 $a^2 + b^2 + 4b - 8a + 13$ 的最小值为 -7.
(3)$S_{\mathrm{四边形}ABCD} = S_{△ ABD} + S_{△ BDC} = \frac{1}{2}BD · OA + \frac{1}{2}BD · OC = \frac{1}{2}BD · (OA + OC) = \frac{1}{2}BD · AC.$
设 $BD = x$,则 $AC = 12 - x$,
$\therefore S_{\mathrm{四边形}ABCD} = \frac{1}{2} · x · (12 - x) = -\frac{1}{2}(x - 6)^2 + 18.$
$\because -\frac{1}{2}(x - 6)^2 ≤ 0,\therefore S_{\mathrm{四边形}ABCD} = -\frac{1}{2}(x - 6)^2 + 18 ≤ 18,$
$\therefore$ 四边形 $ABCD$ 面积的最大值为 18.
$a^2 - 2a + 3 = a^2 - 2a + 1 + 2 = (a - 1)^2 + 2.$
$\because (a - 1)^2 ≥ 0,\therefore (a - 1)^2 + 2 ≥ 2,即 a^2 - 2a + 3 ≥ 2,$
$\therefore a^2 - 2a + 3$ 的最小值为 2.
$-a^2 - 6a + 4 = -(a + 3)^2 + 13.$
$\because -(a + 3)^2 ≤ 0,\therefore -(a + 3)^2 + 13 ≤ 13,$
$\therefore -a^2 - 6a + 4$ 的最大值为 13.
(2)$a^2 + b^2 + 4b - 8a + 13 = a^2 - 8a + 16 + b^2 + 4b + 4 - 7 = (a - 4)^2 + (b + 2)^2 - 7.$
$\because (a - 4)^2 ≥ 0,(b + 2)^2 ≥ 0,$
$\therefore a^2 + b^2 + 4b - 8a + 13 = (a - 4)^2 + (b + 2)^2 - 7 ≥ -7,$
$\therefore$ 代数式 $a^2 + b^2 + 4b - 8a + 13$ 的最小值为 -7.
(3)$S_{\mathrm{四边形}ABCD} = S_{△ ABD} + S_{△ BDC} = \frac{1}{2}BD · OA + \frac{1}{2}BD · OC = \frac{1}{2}BD · (OA + OC) = \frac{1}{2}BD · AC.$
设 $BD = x$,则 $AC = 12 - x$,
$\therefore S_{\mathrm{四边形}ABCD} = \frac{1}{2} · x · (12 - x) = -\frac{1}{2}(x - 6)^2 + 18.$
$\because -\frac{1}{2}(x - 6)^2 ≤ 0,\therefore S_{\mathrm{四边形}ABCD} = -\frac{1}{2}(x - 6)^2 + 18 ≤ 18,$
$\therefore$ 四边形 $ABCD$ 面积的最大值为 18.
