\(B=2+2^2+2^3+...+2^{30}\)

\(B=\left(2+2^2+2^3+2^4+2^5+2^6\right)+...+\left(2^{25}+2^{26}+2^{27}+2^{28}+2^{29}+2^{30}\right)⋮21\)\(B=2.\left(1+2+2^2+2^3+2^4+2^5\right)+...+2^{25}.\left(1+2+2^2+2^3+2^4+2^5\right)⋮21\)\(B=2.\left(1+2+4+8+16+32\right)+2^{25}.\left(1+2+4+8+16+32\right)⋮21\)

\(B=\left(2+...+2^{25}\right).63⋮21\)

\(B⋮21\)

\(A=5+5^2+5^3+5^4+...+5^{40}\)

Ta có:

\(A=\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{39}+5^{40}\right)⋮2\)

\(A=5.\left(1+5\right)+5^3.\left(1+5\right)+...+5^{39}.\left(1+5\right)⋮2\)

\(A=\left(1+5\right).\left(5+5^3+...+5^{39}\right)⋮2\)

\(A=6.\left(5+5^3+...+5^{39}\right)⋮2\)

\(A⋮2\)

Ta có:

\(A=\left(5+5^2+5^3+5^4\right)+...+\left(5^{37}+5^{38}+5^{39}+5^{40}\right)⋮3\)

\(A=5.\left(1+5+5^2+5^3\right)+...+5^{37}.\left(1+5+5^2+5^3\right)⋮3\)

\(A=\left(1+5+25+125\right).\left(5+...+5^{37}\right)⋮3\)

\(A=156.\left(5+...+5^{37}\right)⋮3\)

\(A⋮3\)

Chia nhỏ ra cậu ơi

0:0 = (pích cồ bôn) ko xác định

Hành vi của `A` thuộc loại tệ nạn ma túy.

Bài đúng nhé cậu!

Đề sai không bạn?

\(x^3+5x^2+26x+2^4?\) hay \(x^3+5x^2+2^{6x}+2^4?\)

:< = 😞

`1`

\(5,6\times4+5,6\times3+5,6\times2+5,6\)

\(=5,6\times\left(4+3+2+1\right)\)

\(=5,6\times10=56\)

`2`

\(a.1,83+0,38+0,1+4,62+2,17+4,9\)

\(=\left(1,83+2,17\right)+\left(0,38+4,62\right)+\left(0,1+4,9\right)\)

\(=4+5+5=14\)

\(b.2,9+1,7+1,71+0,29+2,1+1,3\)

\(=\left(2,9+2,1\right)+\left(1,7+1,3\right)+\left(1,71+0,29\right)\)

\(=5+3+2\)

\(=8+2=10\)