Question

Chứng tỏ rằng:

\(7^6+7^5-7^4⋮55\)

\(3^{n+2}-2^{n+2}+3^n-2^n⋮10\) ( với mọi số nguyên dương n )

\(43^{43}-17^{17}⋮10\)

\(8^7-2^{18}⋮14\)

\(23^{401}+38^{202}-2^{433}⋮5\)

Answer 1

\(a,7^6+7^5-7^4=7^4\left(7^2+7-1\right)\\ =7^4\cdot55\\ \Rightarrow7^6+7^5-7^4⋮55\)

\(b,3^{n+2}-2^{n+2}+3^n-2^n\\ =3^n\cdot3^2+3^n-2^n\cdot2^2-2^n\\ =3^n\left(3^2+1\right)-2^n\left(2^2+1\right)\\ =3^n\cdot10-2^{n-1}\cdot2\cdot5\\ =10\cdot\left(3^n-2^{n-1}\right)\\ \Rightarrow3^{n+2}-2^{n+2}+3^n-2^n⋮10\)

\(c,8^7-2^{18}=8^7-\left(2^3\right)^6\\ =8^7-8^6\\ =8^6\cdot\left(8-1\right)\\ =8^5\cdot8\cdot7\\ =8^5\cdot4\cdot14\\ \Rightarrow8^7-2^{18}⋮14\)