\(Tacó:\left(2+2^2\right)\cdot\left(2^3+2^4\right)\cdot...\cdot\left(2^{59}+2^{60}\right)\)

\(A=6\cdot\left(2^3+2^4\right)\cdot...\cdot\left(2^{59}+2^{60}\right)\)

A \(⋮\)6 do A \(\div\)6 \(\times\)6=A

-  Xét \(A⋮2\)

Ta có :\(A=2+2^2+2^3+....+2^{60}\)

\(=2.\left(1+2+2^2+.....+2^{59}\right)\)

Vì \(2⋮2;\left(1+2+2^2+....+2^{59}\right)\inℕ^∗\)

Nên \(2.\left(1+2+2^2+....+2^{59}\right)⋮2\)

Do đó : \(A⋮2\)          \(\left(1\right)\)

- Xét \(A⋮3\)

Ta có : \(A=2+2^2+2^3+.....+2^{60}\)

\(=\left(2+2^2\right)+\left(2^3+2^4\right)+\left(2^5+2^6\right)+.....+\left(2^{59}+2^{60}\right)\)

\(=2\left(1+2\right)+2^3\left(1+2\right)+2^5\left(1+2\right)+.....+2^{59}\left(1+2\right)\)

\(=2.3+2^3.3+2^5.3+.....+2^{59}.3\)

\(=3.\left(2+2^3+2^5+....+2^{59}\right)\)

Vì \(3⋮3;\left(2+2^3+2^5+....+2^{59}\right)\inℕ^∗\)

Nên \(3.\left(2+2^3+2^5+....+2^{59}\right)⋮3\)            \(\left(2\right)\)

Từ (1) và (2), kết hợp với \(2.3=6;\left(2,3\right)=1\) suy ra  \(A⋮6\)      \(\left(đpcm\right)\)

y đâu vậy bạn -_-" đúng đề chưa mình giải cho :)

Đúng đề rồi

a) \(S=2+2^2+2^3+...+2^{100}\)

\(S=\left(2+2^2+2^3+2^4\right)+...+\left(2^{97}+2^{98}+2^{99}+2^{100}\right)\)

\(S=2\left(1+2+2^2+2^3\right)+...+2^{97}\left(1+2+2^2+2^3\right)\)

\(S=2\cdot15+...+2^{97}\cdot15\)

\(S=15\cdot\left(2+...+2^{97}\right)⋮15\left(đpcm\right)\)

b) \(S=2+2^2+...+2^{100}\)

\(2S=2^2+2^3+...+2^{101}\)

\(2S-S=\left(2^2+2^3+...+2^{101}\right)-\left(2+2^2+...+2^{100}\right)\)

\(S=2^{101}-2< 2^{101}\)

s=2+2^2+2^3+......÷2^100=> 2s=2^2+2^3+2^4+.....+2^101

=>2s-s=s=2^101-2=>s<2^101

(-12)x7+(-8)x7+4x(-25)+(-6)x(-25)=(-12-8)x7+(-25)(4x-6x)=7x(-20)+(-25)(-2x)=-140x+50x=-90x

126-53+20-(53+126) = 126-53+20-179 = 73+20-179 = 93-179 = -86