22. (4分)如果aᶜ = b,那么我们规定(a,b) = c,例如:因为2³ = 8,所以(2,8) = 3.
(1)根据上述规定,填空:
(3,27) = ,(4,1) = (2,0.25) = .
(2)记(3,5) = a,(3,6) = b,(3,30) = c.求证:a + b = c.
解:∵(3,5) = a,(3,6) = b,(3,30) = c,
∴3ᵃ = 5,3ᵇ = 6,3ᶜ = 30,∴3ᵃ×3ᵇ = 30,
∴3ᵃ×3ᵇ = 3ᶜ,∴3ᵃ⁺ᵇ = 3ᶜ,∴a + b = c.
(1)根据上述规定,填空:
(3,27) = ,(4,1) = (2,0.25) = .
(2)记(3,5) = a,(3,6) = b,(3,30) = c.求证:a + b = c.
解:∵(3,5) = a,(3,6) = b,(3,30) = c,
∴3ᵃ = 5,3ᵇ = 6,3ᶜ = 30,∴3ᵃ×3ᵇ = 30,
∴3ᵃ×3ᵇ = 3ᶜ,∴3ᵃ⁺ᵇ = 3ᶜ,∴a + b = c.
答案:
(1)3 0 -2
(2)$\because(3,5)=a,(3,6)=b,(3,30)=c$,$\therefore 3^{a}=5,3^{b}=6,3^{c}=30,\therefore 3^{a}×3^{b}=30$,$\therefore 3^{a}×3^{b}=3^{c},\therefore 3^{a + b}=3^{c},\therefore a + b = c$.
(1)3 0 -2
(2)$\because(3,5)=a,(3,6)=b,(3,30)=c$,$\therefore 3^{a}=5,3^{b}=6,3^{c}=30,\therefore 3^{a}×3^{b}=30$,$\therefore 3^{a}×3^{b}=3^{c},\therefore 3^{a + b}=3^{c},\therefore a + b = c$.
23. (6分)中考新考法 证明代数结论 你能比较两个数2022²⁰²³和2023²⁰²²的大小吗? 为了解决这个问题,先把问题一般化,即比较nⁿ⁺¹和(n + 1)ⁿ的大小(n≥1且n为整数),然后从分析n = 1,n = 2,n = 3,…,这些简单的情形入手,从中发现规律,经过归纳、总结,最后猜想出结论.
(1)通过计算,比较下列各组数的大小(在横线处填上“>”“=”或“<”);
①1² 2¹;②2³ 3²;③3⁴ 4³;
④4⁵ 5⁴;⑤5⁶ 6⁵;⑥6⁷ 7⁶;…
(2)由第(1)小题的结果归纳、猜想nⁿ⁺¹与(n + 1)ⁿ的大小关系;
(3)根据第(2)小题得到的一般结论,可以得到2022²⁰²³ 2023²⁰²².(填“>”“<”或“=”)
解:(2)由(1)可知,当n = 1,2时,nⁿ⁺¹ < (n + 1)ⁿ;
当n≥3时,nⁿ⁺¹ > (n + 1)ⁿ.
(1)通过计算,比较下列各组数的大小(在横线处填上“>”“=”或“<”);
①1² 2¹;②2³ 3²;③3⁴ 4³;
④4⁵ 5⁴;⑤5⁶ 6⁵;⑥6⁷ 7⁶;…
(2)由第(1)小题的结果归纳、猜想nⁿ⁺¹与(n + 1)ⁿ的大小关系;
(3)根据第(2)小题得到的一般结论,可以得到2022²⁰²³ 2023²⁰²².(填“>”“<”或“=”)
解:(2)由(1)可知,当n = 1,2时,nⁿ⁺¹ < (n + 1)ⁿ;
当n≥3时,nⁿ⁺¹ > (n + 1)ⁿ.
答案:
(1)< < > > > >
(2)由
(1)可知,当$n = 1,2$时,$n^{n + 1}<(n + 1)^{n}$;当$n\geqslant3$时,$n^{n + 1}>(n + 1)^{n}$.
(3)>
(1)< < > > > >
(2)由
(1)可知,当$n = 1,2$时,$n^{n + 1}<(n + 1)^{n}$;当$n\geqslant3$时,$n^{n + 1}>(n + 1)^{n}$.
(3)>
查看更多完整答案,请扫码查看