\(2^3.17+70.2^3+2^3.13\)

\(=2^3.\left(17+70+13\right)\)

\(=2^3.100\)

\(=8.100\)

\(=800.\)

2³. 17 + 70 . 2³ + 2³ . 13

= 8 . 17 + 70 . 8 + 8 . 13

= 8 . ( 17 + 70 + 13 )

= 8 . 100

= 800

a) Ta áp dụng đẳng thức sau: \(a^{2k+1}+b^{2k+1}⋮a+b\)

\(A=2^{70}+b^{70}=4^{35}+9^{35}⋮4+9=13\)

\(\Rightarrowđpcm\)

b) Ta có: \(17\equiv-1\left(mod18\right)\Rightarrow17^{19}\equiv-1\left(mod18\right)\)

\(19\equiv1\left(mod18\right)\Rightarrow19^{17}\equiv1\left(mod18\right)\)

\(\Rightarrow17^{19}+19^{17}⋮18\left(đpcm\right)\)

b) Ta có: 2^70+3^70= 4^35+9^35 chia hết cho 4+9=13

đpcm

:)? Đề bài là ntn bạn nhỉ??

đề bài ở trên ý !

Giúp mk vs

a) -15 - (- 7) + 13 = -15 - -7 + 13

= -8 + 13

= 5

b) 47 - (-15) + (-13) = 47 - -15 + -13

= 62 + -13

= 49

c) -115 + (-20) - (-11) = -115 + -20 - -11

= -135 - -11

= 124

d) -2 . (-4) . (-5) . 6 = -2 . -4 . -5 . 6

= -8 . -5 . 6

= 40 . 6

= 240

e) 11 + (-113) - (-17) + 78 = 11 + -113 - -17 + 78

= -102 - -17 + 78

= -85 + 78

= -7

g) 2³ . (-5)² . (-70)² = 8 . -25 . -4900

= -200 . -4900 

= 980000

a)-15-(-7)+13=-15+8-13=-7+(-13)=-20

b)47-(-15)+(-13)=47+15+(-13)=62+(-13)=49

c)-115+(-20)-(-11)=-135+11=-124

d)-2*(-4)*(-5)*6=8*(-5)*6=(-40)*6=-240

e)11+(-113)-(-17)+78=-102+17+78=-85+78=-7

g)2^3*(-5)^2*(-70)^2=8*25*4900=200*4900=980000

\(\left(a\right)\frac{3}{7}-\left(\frac{-2}{7}\right)=\frac{3}{7}+\frac{2}{7}=\frac{5}{7}\)

\(\left(b\right)\frac{-13}{11}+\frac{6}{22}-\frac{36}{33}=\frac{-39}{33}+\frac{9}{33}-\frac{36}{33}=\frac{-66}{33}=-2\)

\(a,\frac{3}{7}-\frac{-2}{7}=\frac{3-\left(-2\right)}{7}=\frac{3+2}{7}=\frac{5}{7}\)

\(b,-\frac{13}{11}+\frac{6}{22}-\frac{36}{33}=-\frac{13}{11}+\frac{3}{11}-\frac{12}{11}=\frac{-13+3-12}{11}=-\frac{22}{11}=-2\)

\(c,-\frac{6}{17}-\frac{5}{-12}+\frac{3}{81}=-\frac{6}{17}+\frac{5}{12}+\frac{1}{27}=\frac{-6.27.12+5.17.27+17.12}{17.12.27}=\frac{185}{1836}\)

\(\frac{2}{70}-\frac{9}{-24}+\frac{-5}{42}=\frac{1}{35}+\frac{3}{8}-\frac{5}{42}=\frac{1.24+105.3-5.20}{840}=\frac{239}{840}\)

a) Có: \(2^3=8\equiv1\left(mod7\right)\Rightarrow2^{51}\equiv1\left(mod7\right)\)

\(\Rightarrow2^{51}-1⋮7\left(đpcm\right)\)

b) 2⁷⁰ + 3⁷⁰ = (2²)³⁵ + (3²)³⁵ = 4³⁵ + 9³⁵

\(=\left(4+9\right).\left(4^{34}-4^{33}.9+....-4.9^{33}+9^{34}\right)\)

\(=13.\left(4^{34}-4^{33}.9+...-4.9^{33}+9^{34}\right)⋮13\left(đpcm\right)\)

t chỉ lm 2 câu đại diện, c` lại tương tự

phần a sai đề nha bạn 

b,Ta có

      \(2\equiv2\left(mod13\right)\)

\(\Rightarrow2^{12}\equiv1\left(mod13\right)\)

\(\Rightarrow2^{12.5}.2^{10}\equiv1.2^{10}\left(mod13\right)\)

\(\Rightarrow2^{60}.2^{10}\equiv1024\left(mod13\right)\)

\(\Rightarrow2^{70}\equiv10\left(mod13\right)\)\(\left(1\right)\)

Lại có:

\(3\equiv3\left(mod13\right)\)

\(\Rightarrow3^6\equiv1\left(mod13\right)\)

\(\Rightarrow3^{6.11}.3^4\equiv1.3^4\left(mod13\right)\)

\(\Rightarrow3^{66}.3^4\equiv81\left(mod13\right)\)

\(\Rightarrow3^{70}\equiv3\left(mod13\right)\)\(\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\Rightarrow2^{70}+3^{70}\equiv13\equiv0\left(mod13\right)\)

c, Ta có

\(17\equiv-1\left(mod18\right)\)

\(\Rightarrow17^{19}\equiv-1\left(mod18\right)\)\(\left(1\right)\)

Lại có

\(19\equiv1\left(mod18\right)\)

\(\Rightarrow19^{17}\equiv1\left(mod18\right)\)\(\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\Rightarrow17^{19}+19^{17}\equiv0\left(mod18\right)\)

\(\Rightarrow17^{19}+19^{17}⋮18\)

a) \(\dfrac{\dfrac{3}{14}-\dfrac{3}{17}+\dfrac{3}{19}}{\dfrac{5}{19}+\dfrac{5}{14}-\dfrac{5}{17}}=\dfrac{3\left(\dfrac{1}{14}-\dfrac{1}{17}+\dfrac{1}{19}\right)}{5\left(\dfrac{1}{14}-\dfrac{1}{17}+\dfrac{1}{19}\right)}=\dfrac{3}{5}\)

c) \(\dfrac{\dfrac{2}{7}+\dfrac{2}{5}+\dfrac{2}{17}-\dfrac{2}{193}}{\dfrac{3}{7}+\dfrac{3}{5}+\dfrac{3}{17}-\dfrac{3}{293}}=\dfrac{2\left(\dfrac{1}{7}+\dfrac{1}{5}+\dfrac{1}{17}-\dfrac{1}{193}\right)}{3\left(\dfrac{1}{7}+\dfrac{1}{5}+\dfrac{1}{17}-\dfrac{1}{193}\right)}=\dfrac{2}{3}\)

d)