\(a,A=2+2^2+2^3+....+2^{12}\)

\(=\left(2+2^2\right)+\left(2^3+2^4\right)+....+\left(2^{11}+2^{12}\right)\)

\(=2.\left(1+2\right)+2^3.\left(1+2\right)+....+2^{11}.\left(1+2\right)\)

\(=2.3+2^3.3+....+2^{11}.3\)

\(=3.\left(2+2^3+....+2^{11}\right)\)

Vì  \(3⋮3;\left(2+2^3+....+2^{11}\right)\in Z\) nên \(3.\left(2+2^3+.....+2^{11}\right)⋮3\)

Vậy  \(A⋮3\)

Câu b bn làm tương tự.

\(S=1+2+2^2+...+2^{2017}\)

\(2S=2+2^2+...+2^{2018}\)

\(2S-S=\left(2+2^2+...+2^{2018}\right)-\left(1+2+...+2^{2017}\right)\)

\(S=2^{2018}-1\)

\(S=1+2+2^2+2^3+.............+2^{2017}\)

\(\Rightarrow2S=2+2^2+2^3+2^4+........+2^{2018}\)

\(\Rightarrow2S-S=\left(2+2^2+2^3+...+2^{2018}\right)-\left(1+2+2^2+...+2^{2017}\right)\)

\(\Rightarrow S=2^{2018}-1\)

GTTD cua x <5 =>GTTD cua x la 1 ;2;3;4

cac so do co the la : 1;2;3;4;(-1);(-2);(-3);(-4)

tong cac so do la : 1+2+3+4+(-1)+(-2)+(-3)+(-4) = 0 ( vi do la tong cac so doi )

a) x={4;3;2;1;-1;-2;-3;-4}

b            tổng =0

x=1,bạn nhé

\(x^{15}=x-x\)

\(x^{15}=0\)

\(x\left(x+1\right)\left(x-1\right)\left(x^6-x^5+x^4-x^3+x^2-x+1\right)\left(x^6+x^5+x^4+x^3+x^2+x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)

. ghi sai đề à không hiểu từ chỗ biết rằng á

Ở chỗ hoặc hai chú thì mình sửa thành hai chục nhé.

Gọi số truyện là x

Theo đề bài, ta có: \(x⋮10,x⋮12,x⋮20\)

\(x\in\left(10;12;20\right)\)

10 = 2.5

\(12=2^2.3\)

\(20=2^2.5\)

\(\Rightarrow BCNN\left(10;12;20\right)=2^2.3.5=60\)

\(BC\left(60\right)=\left(0;60;120;180;240;300;360;...\right)\)

Đk: 150 < x < 200

=> Số truyện quyên góp được là 180 cuốn

c)x-11=15

x=26

= 250 cc

sai rồi bạn vo hoang long oi 

\(10A=\frac{10\left(10^{11}-1\right)}{10^{12}-1}=\frac{10^{12}-10}{10^{12}-1}=\frac{10^{12}-1-9}{10^{12}-1}=1-\frac{9}{10^{12}-1}\)

\(10B=\frac{10\left(10^{12}-1\right)}{10^{13}-1}=\frac{10^{13}-10}{10^{13}-1}=\frac{10^{13}-1-9}{10^{13}-1}=1-\frac{9}{10^{13}-1}\)

Vì \(10^{13}-1>10^{12}-1\Rightarrow\frac{9}{10^{13}-1}< \frac{9}{10^{12}-1}\Rightarrow-\frac{9}{10^{13}-1}>-\frac{9}{10^{12}-1}\)

\(\Rightarrow1-\frac{9}{10^{13}-1}>1-\frac{9}{10^{12}-1}\Rightarrow10B>10A\Rightarrow B>A\)

\(A=\frac{10^{11}-1}{10^{12}-1}\Leftrightarrow10A=\frac{10^{12}-10}{10^{12}-1}=\frac{10^{12}-1-9}{10^{12}-1}=1-\frac{9}{10^{12}-1}\)

\(B=\frac{10^{12}-1}{10^{13}-1}\Leftrightarrow10B=\frac{10^{13}-10}{10^{13}-1}=\frac{10^{13}-1-9}{10^{13}-1}=1-\frac{9}{10^{13}-1}\)

\(\text{Vì }1-\frac{9}{10^{12}-1}< 1-\frac{9}{10^{13}-1}\Rightarrow10A< 10B\)

\(\Rightarrow A< B\)