Question

[Khang Nguyen Minh]

gúp mình bài 6 với

Answer 1

[Akai Haruma]

Bài 6:

a)

\(A=-0,7(43^{43}-17^{17})=\frac{7(17^{17}-43^{43})}{10}\)

Ta có: \(17\equiv 7\pmod {10}\Rightarrow 17^{17}\equiv 7^{17}\pmod {10}\)

\(43\equiv 3\pmod {10}\Rightarrow 43^{43}\equiv 3^{43}\pmod {10}\)

Do đó, \(17^{17}-43^{43}\equiv 7^{17}-3^{43}\pmod {10}\)

Lại có:

\(7^2\equiv -1\pmod {10}\Rightarrow 7^{16}\equiv 1\pmod {10}\Rightarrow 7^{17}\equiv 7\pmod {10}\)

\(3^{2}\equiv -1\pmod {10}\Rightarrow 3^{42}\equiv -1\pmod {10}\Rightarrow 3^{43}\equiv -3\pmod {10}\)

\(\Rightarrow 7^{17}-3^{43}\equiv 7-(-3)\equiv 0\pmod {10}\) hay

Do đó, \(17^{17}-43^{43}\equiv 0\pmod {10}\Rightarrow A=\frac{7(17^{17}-43^{43})}{10}\in\mathbb{Z}\)

b)

Vì \(a,b,c\leq 1\Rightarrow B=\frac{a}{bc+1}+\frac{b}{ac+1}+\frac{c}{ab+1}\leq \frac{a}{abc+1}+\frac{b}{abc+1}+\frac{c}{abc+1}\)

\(\Leftrightarrow B\leq \frac{a+b+c}{abc+1}\)\((1)\)

Ta sẽ cm \(\frac{a+b+c}{abc+1}\leq 2\Leftrightarrow a+b+c\leq 2abc+2\)

Thật vậy:

Vì \(a,b\leq 1\rightarrow (a-1)(b-1)\geq 0\Leftrightarrow ab+1\geq a+b\)

\(\Leftrightarrow ab+1+c\geq a+b+c\)

Xét \(2abc+2-(ab+1+c)=abc+(c-1)(ab-1)\)

Vì \(c,ab\leq 1\Rightarrow (c-1)(ab-1)\geq 0\), mà \(a,b,c\geq 0\rightarrow abc\geq 0\)

\(\Rightarrow abc+(c-1)(ab-1)\geq 0\Leftrightarrow 2abc+2\geq ab+1+c\)

\(\Rightarrow 2abc+2\geq a+b+c\) \(\Rightarrow \frac{a+b+c}{abc+1}\leq 2\) \((2)\)

Từ (1),(2) ta có đpcm.

Answer 2

[Khang Nguyen Minh]

mai nộp rùi gúp mình với chỉ cần câu b bài 6 thui