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1[2025湖南衡阳期中,中]已知$(a+b)^2=9,(a-b)^2=25$,求$a^2+b^2$与$ab$的值.
答案
【解】$\because (a+b)^2=a^2+b^2+2ab=9$,①
$(a-b)^2=a^2+b^2-2ab=25$,②
$\therefore$ ①+②得$2(a^2+b^2)=34$,即$a^2+b^2=17$,①-②得$4ab=-16$,即$ab=-4.$
$(a-b)^2=a^2+b^2-2ab=25$,②
$\therefore$ ①+②得$2(a^2+b^2)=34$,即$a^2+b^2=17$,①-②得$4ab=-16$,即$ab=-4.$
2[2025广东河源校级期中,中]已知$x-y=5$，$xy=2$，求$x^2+y^2$及$(x+y)^2 -6$的值.
答案
【解】$\because x-y=5,xy=2,\therefore x^2+y^2=(x-y)^2+2xy=5^2+2×2=29,(x+y)^2-6=(x-y)^2+4xy-6=5^2+4×2-6=25+8-6=27.$
3[2025江西吉安质检,中]已知两个不相等的实数$m,n$满足$m^2 + n^2 = 40$,$m + n = -4$.
(1)求$mn$的值；
(2)求$m - n$的值.
(1)求$mn$的值；
(2)求$m - n$的值.
答案
【解】(1)$\because m^2+n^2=40,m+n=-4,\therefore (m+n)^2=m^2+2mn+n^2=40+2mn=16,\therefore mn=-12.$
(2)由(1)得$mn=-12. \because m^2+n^2=40,\therefore (m-n)^2=m^2-2mn+n^2=40+24=64,\therefore m-n=8$ 或 $m-n=-8.$
(2)由(1)得$mn=-12. \because m^2+n^2=40,\therefore (m-n)^2=m^2-2mn+n^2=40+24=64,\therefore m-n=8$ 或 $m-n=-8.$
1[2025四川成都期末,中]若$m-\frac{1}{m}=5$，则$m^2+\frac{1}{m^2}$的值为
27
。答案
27 【解析】$\because m-\frac{1}{m}=5,\therefore m^2-2× m×\frac{1}{m}+\frac{1}{m^2}=25,\therefore m^2-2+\frac{1}{m^2}=25,\therefore m^2+\frac{1}{m^2}=25+2=27,$ 故$m^2+\frac{1}{m^2}$的值为 27.
2[2026湖南郴州期末,中]已知$a^2 - 3a + 1 = 0$，则$a^4 + \frac{1}{a^4} =$
47
答案
47 【解析】$\because a^2-3a+1=0,\therefore a≠0,\therefore a-3+\frac{1}{a}=0,$ 即$a+\frac{1}{a}=3$,两边平方得,$a^2+2+\frac{1}{a^2}=9,$
$\therefore a^2+\frac{1}{a^2}=7,$ 再平方得,$a^4+2+\frac{1}{a^4}=49,\therefore a^4+\frac{1}{a^4}=47.$ 故答案为 47.
$\therefore a^2+\frac{1}{a^2}=7,$ 再平方得,$a^4+2+\frac{1}{a^4}=49,\therefore a^4+\frac{1}{a^4}=47.$ 故答案为 47.
