Question

Помогите пожалуйста решить задачу

Answer 1

$3) \; \frac{(k-6)(k-4)}{(k-4)!} = \frac{(k-6)(k-4)}{(k-4)(k-5)!}=\frac{k-6}{(k-5)!}$

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$4) \; A^3_{2n}=100A^2_n\\\\\frac{(2n)!}{(2n-3)!}=100\frac{n!}{(n-2)!} \; \; \; \; \; ODZ: n \notin (-\infty; \; 1)\\\\\frac{2n(2-1)(2n-2)(2n-3)!}{(2n-3)!}=100\frac{n(n-1)(n-2)!}{(n-2)!}\\\\2n(2-1)(2n-2)=100n(n-1)\\\\8n^3-112n^2+104n=0\\\\8n(n^2-14n+13)=0\\\\8n(n^2-n-13n+13)=0\\\\8n(n-1)(n-13)=0 \; \; \; | \div 8\\\\n(n-1)(n-13)=0\\\\\underbrace{n_1=0 \; \; n_2 =1}_{n \; \in \; (-\infty; \; 1)} \; \; \underline{n_3 =13}\\\\n=13$

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$5) \; (2+ \sqrt{2})^4 = (2+ \sqrt{2})^{2+2}=\\\\= (2+ \sqrt{2})^2 \cdot (2+ \sqrt{2})^2=\\\\=(4+4\sqrt {2}+2)(4+4\sqrt{2}+2)=\\\\=16+16\sqrt{2}+8+16\sqrt{2}+32+8\sqrt{2}+8+8\sqrt{2}+4=\\\\=68+48\sqrt{2}$