\(ab+ba=(10a+b)+(10b+a)\)

\(=10a+b+10b+a\)

\(=11a+11b\)

\(=11\left(a+b\right)\)

\(a+b\inℕ\Rightarrow ab+ba⋮11\)

\(A=2+2^2+2^3+\cdot\cdot\cdot+2^{2008}\)

\(\Rightarrow2A=2^2+2^3+2^4+\cdot\cdot\cdot+2^{2009}\)
\(\Rightarrow2A-A=\left(2^2+\cdot\cdot\cdot2^{2009}\right)-\left(2+\cdot\cdot\cdot+2^{2008}\right)\)

\(\Rightarrow A=2^{2009}-2\)

\(15:1=15\)

\(15:3=5\)

\(15:5=3\)

\(15:15=1\)

\((15;1);(5;3);(3;5);(1;15)\)

bạn giải rõ ra nha

\(\frac{-5}{19}\cdot\frac{8}{19}+\frac{-14}{19}\cdot\frac{8}{19}-\frac{11}{19}\)

\(=\frac{8}{19}\left(\frac{-5}{19}+\frac{-14}{19}\right)-\frac{11}{19}\)

\(=\frac{8}{19}\cdot\left(-1\right)-\frac{11}{19}\)

\(=\frac{-8}{19}-\frac{11}{19}\)

\(=-1\)

=))

cái này là hệ 3 ẩn rồi 

===================================

a, theo bài ra 

x+y=6 (1)

-y +z = - 5 (2) 

(1) + (2) <=> x+z = 6-5=1 , lại có x-z=9 

=> (x+z)+(x-z)=1+9<=> 2x=10<=> x=5 => z = -4 

Thay x=5 vào (1) => y=6-x=6-5=1 

vậy x=5 , y=1 , z = -4

:V tương tự với câu b nhé

Mk có cách khác nhé:

b) Ta có:

\(x+y-y-z-z-x=6+7+13\) 

\(-2z=26\Rightarrow z=-13\) 

\(\Rightarrow y=6;x=0\) 

Vậy .....

Đặt \(N=2^{2014}+2^{2013}+...+2+1\)

\(2N=2^{2015}+2^{2014}+...+2^2+2\)

\(=>2N-N=\left(2^{2015}+2^{2014}+...+2^2+2\right)-\left(2^{2014}+2^{2013}+...+2+1\right)\)

\(N=2^{2015}-1\)

\(=>M=2^{2015}-\left(2^{2015}-1\right)=1\)

(480 x 2 - 961 + 1) + (15 x 9 + 15 x 3 - 15 - 15)

= (960 - 961 + 1) + 15 x (9 + 3 - 1 - 1)

= (960 + 1 - 961) + 15 x 10

= (961 - 961) + 150

= 0 + 150

= 150

Rất vui được giúp bạn 

\(\left(480.2-961+1\right)+\left(12.9+15.3-15-15\right)\)

\(=\left(480.2-961+1\right)+\left(12.9+15.3-15.1-15.1\right)\)

\(=\left(960-961+1\right)+15.\left(9+3-1-1\right)\)

\(=\left(-1+1\right)+15.10\)

\(=0+150\)

\(=150\)

Rất vui vì giúp đc bạn <3

Ta có \(\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{29}+5^{30}\right)=\left(5+5^2\right)+5^2\left(5+5^2\right)+...+5^{28}\left(5+5^2\right)\)

\(=\left(5+5^2\right)\left(1+5^2+...+5^{28}\right)=30\left(1+5^2+..+5^{28}\right)⋮6\)

Ta có \((5+5^2+5^3)+...+\left(5^{28}+5^{29}+5^{30}\right)=(5+5^2+5^3)+...+5^{27}(5+5^2+5^3)\)

\(=155+...+5^{27}.155=155\left(1+...+5^{27}\right)⋮31\)

-\(5+5^2+5^3+...+5^{30}=\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{29}+5^{30}\right)\)

         \(=30+5^2\cdot30+5^4\cdot30+...+5^{28}\cdot30=30\left(1+5^2+...+^{28}\right)⋮6\)

-\(5+5^2+5^3+...+5^{30}=5\left(1+5+5^2\right)+5^4\left(1+5+5^2\right)+...+5^{28}\left(1+5+5^2\right)\)

          \(5\cdot31+5^4\cdot31+...+5^{28}\cdot31=31\left(5+5^4+...+5^{28}\right)⋮31\)

Ta có: \(80^{21}=80^{3.7}=\left(80^3\right)^7=512000^7\)

           \(28^{28}=28^{4.7}=\left(28^4\right)^7=614656\)

Vì \(512000^7< 614656^7\)\(\Rightarrow\)\(80^{21}< 28^{28}\)