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20. (5 分)先化简,再求值:$2a(a+4b)+(3b+a)(3b-a)-(a+3b)^{2}$,其中$a=-2,b=\frac{1}{2}$.
答案
【解析】$\begin{aligned}&2a(a + 4b) + (3b + a)(3b - a) - (a + 3b)^2 \\=&2a^2 + 8ab + 9b^2 - a^2 - (a^2 + 6ab + 9b^2) \\=&2a^2 + 8ab + 9b^2 - a^2 - a^2 - 6ab - 9b^2 \\=&2ab,\end{aligned}$
当$a = -2$,$b = \dfrac{1}{2}$时,原式$= 2 × (-2) × \dfrac{1}{2} = -2$.
当$a = -2$,$b = \dfrac{1}{2}$时,原式$= 2 × (-2) × \dfrac{1}{2} = -2$.
21. (6分)完全平方公式经过适当的变形,可以解决很多数学问题.
例如:若$a + b = 3,ab = 1$,求$a^2 + b^2$的值.
解:因为$a + b = 3,ab = 1$,所以$(a + b)^2 = 9,2ab = 2$,所以$a^2 + b^2 + 2ab = 9$,所以$a^2 + b^2 = 7$.
根据上面的解题思路与方法解决下列问题:
(1)若$a - b = -5,ab = 3$,则$a^2 + b^2 =$
(2)若$(a + b)^2 = 17,(a - b)^2 = 13$,求$a^2 + b^2$的值;
(3)已知$x - \frac{1}{x} = 3$,求$x^2 + \frac{1}{x^2}$的值.
例如:若$a + b = 3,ab = 1$,求$a^2 + b^2$的值.
解:因为$a + b = 3,ab = 1$,所以$(a + b)^2 = 9,2ab = 2$,所以$a^2 + b^2 + 2ab = 9$,所以$a^2 + b^2 = 7$.
根据上面的解题思路与方法解决下列问题:
(1)若$a - b = -5,ab = 3$,则$a^2 + b^2 =$
31
;(2)若$(a + b)^2 = 17,(a - b)^2 = 13$,求$a^2 + b^2$的值;
(3)已知$x - \frac{1}{x} = 3$,求$x^2 + \frac{1}{x^2}$的值.
答案
【点拨】本题考查完全平方公式变形求值,掌握完全平方公式是解题的关键.
【解析】(1)因为$a - b = -5$,$ab = 3$,所以$(a - b)^2 = 25$,$2ab = 6$,所以$a^2 - 2ab + b^2 = 25$,即$a^2 - 6 + b^2 = 25$,所以$a^2 + b^2 = 31$. 故答案为 31.
(2)因为$(a + b)^2 = 17$,$(a - b)^2 = 13$,所以$(a + b)^2 + (a - b)^2 = 30$,$a^2 + 2ab + b^2 + a^2 - 2ab + b^2 = 30$,$2a^2 + 2b^2 = 30$,所以$a^2 + b^2 = 15$.
(3)因为$x - \dfrac{1}{x} = 3$,所以$(x - \dfrac{1}{x})^2 = 3^2 = 9$,所以$x^2 + \dfrac{1}{x^2} - 2 = 9$,所以$x^2 + \dfrac{1}{x^2} = 11$.
【解析】(1)因为$a - b = -5$,$ab = 3$,所以$(a - b)^2 = 25$,$2ab = 6$,所以$a^2 - 2ab + b^2 = 25$,即$a^2 - 6 + b^2 = 25$,所以$a^2 + b^2 = 31$. 故答案为 31.
(2)因为$(a + b)^2 = 17$,$(a - b)^2 = 13$,所以$(a + b)^2 + (a - b)^2 = 30$,$a^2 + 2ab + b^2 + a^2 - 2ab + b^2 = 30$,$2a^2 + 2b^2 = 30$,所以$a^2 + b^2 = 15$.
(3)因为$x - \dfrac{1}{x} = 3$,所以$(x - \dfrac{1}{x})^2 = 3^2 = 9$,所以$x^2 + \dfrac{1}{x^2} - 2 = 9$,所以$x^2 + \dfrac{1}{x^2} = 11$.
22. (6 分)我们知道,同底数幂的乘法法则为$a^m · a^n = a^{m+n}$(其中$a≠0,m,n$为正整数),类似地,我们规定关于任意正整数$m,n$的一种新运算:$f(m) · f(n) = f(m+n)$(其中$m,n$为正整数).
例如:若$f(3)=2$,则$f(6)=f(3+3)=f(3) · f(3)=2 × 2=4$.
$f(9)=f(3+3+3)=f(3) · f(3) · f(3)=2 × 2 × 2=8$.
(1)若$f(2)=5$,①填空:$f(6)=$
(2)若$f(a)=3$,化简:$f(a) · f(2a) · f(3a) · \dots · f(10a)$.
例如:若$f(3)=2$,则$f(6)=f(3+3)=f(3) · f(3)=2 × 2=4$.
$f(9)=f(3+3+3)=f(3) · f(3) · f(3)=2 × 2 × 2=8$.
(1)若$f(2)=5$,①填空:$f(6)=$
125
;②当$f(2n)=25$,求$n$的值;(2)若$f(a)=3$,化简:$f(a) · f(2a) · f(3a) · \dots · f(10a)$.
答案
【点拨】本题考查同底数幂的乘法,数字的变化规律,解题的关键是理解清楚所给的新运算.
【解析】(1)①因为$f(2) = 5$,所以$f(6) = f(2 + 2 + 2) = f(2) · f(2) · f(2) = 5 × 5 × 5 = 125$,故答案为 125;
②因为$25 = 5 × 5 = f(2) · f(2) = f(2 + 2)$,$f(2n) = 25$,所以$f(2n) = f(2 + 2)$,所以$2n = 4$,所以$n = 2$.
(2)因为$f(2a) = f(a + a) = f(a) · f(a) = 3 × 3 = 3^{1+1} = 3^2$,$f(3a) = f(a + a + a) = f(a) · f(a) · f(a) = 3 × 3 × 3 = 3^{1+1+1} = 3^3$,$\dots$,$f(10a) = 3^{10}$,所以$f(a) · f(2a) · f(3a) · \dots · f(10a) = 3 × 3^2 × 3^3 × \dots × 3^{10} = 3^{1+2+3+\dots+10} = 3^{55}$.
【解析】(1)①因为$f(2) = 5$,所以$f(6) = f(2 + 2 + 2) = f(2) · f(2) · f(2) = 5 × 5 × 5 = 125$,故答案为 125;
②因为$25 = 5 × 5 = f(2) · f(2) = f(2 + 2)$,$f(2n) = 25$,所以$f(2n) = f(2 + 2)$,所以$2n = 4$,所以$n = 2$.
(2)因为$f(2a) = f(a + a) = f(a) · f(a) = 3 × 3 = 3^{1+1} = 3^2$,$f(3a) = f(a + a + a) = f(a) · f(a) · f(a) = 3 × 3 × 3 = 3^{1+1+1} = 3^3$,$\dots$,$f(10a) = 3^{10}$,所以$f(a) · f(2a) · f(3a) · \dots · f(10a) = 3 × 3^2 × 3^3 × \dots × 3^{10} = 3^{1+2+3+\dots+10} = 3^{55}$.
