\(3^{15}+3^{14}+3^{13}\)

\(=3^{13}\left(3^2+3+1\right)=3^{13}\cdot13⋮13\)

\(=3^{13}\left(3^2+3+1\right)=3^{13}\cdot13⋮13\)

\(3^{15}+3^{14}+3^{13}\)

\(=3^{13}\left(3^2+3+1\right)\)

\(=3^{13}\left(9+3+1\right)\)

\(=3^{13}.13\)⋮13

⇒\(3^{15}+3^{14}+3^{13}\)⋮13

           \(\frac{19}{13}+\frac{14}{6}+\frac{1}{9}+\frac{4}{6}+\frac{7}{13}+\frac{17}{9}\)

\(=\left(\frac{19}{13}+\frac{7}{13}\right)+\left(\frac{14}{6}+\frac{4}{6}\right)+\left(\frac{1}{9}+\frac{17}{9}\right)\)

\(=\frac{26}{13}+\frac{18}{6}+\frac{18}{9}\)

\(=2+3+2\)

\(=7\)

              \(\frac{315}{316}\times\frac{313}{314}\times\frac{316}{315}\times\frac{317}{314}\)

\(=\left(\frac{315}{316}\times\frac{316}{315}\right)\times\left(\frac{313}{314}\times\frac{317}{314}\right)\)

\(=1\times1,006339\)

\(=1,006339\)

#Chúc bạn học tốt !

#k mình nhé ?

Đơn giản như đang giỡn :

=315.313.316.317/315.314.316.317

=313/314

hok tốt k nhé bạn .

    \(\frac{315}{316}\cdot\frac{313}{314}\cdot\frac{316}{315}\cdot\frac{317}{314}\)

= \(\frac{315\cdot313\cdot316\cdot317}{316\cdot314\cdot315\cdot314}\)

=\(\frac{1\cdot313\cdot1\cdot317}{1\cdot314\cdot1\cdot314}\)(Bước này là bước rút gọn)

= \(\frac{99221}{98596}\)

                                     #Kiều

\(\frac{315}{316}.\frac{313}{314}.\frac{316}{315}.\frac{317}{314}\)\(=\frac{315.313.316.317}{316.314.315.314}\)

Sau khi rút gọn:

\(=\frac{313.317}{314.314}\)

\(\frac{315}{316}\cdot\frac{316}{314}\cdot\frac{316}{315}\cdot\frac{317}{313}=\frac{315.316.316.317}{316.314.315.313}=\frac{1.1.316.317}{1.314.1.313}=\frac{100172}{98282}\)

Ta có : \(\frac{3}{2}\times\frac{4}{5}\times\frac{2}{3}\)

\(=\frac{3}{2}\times\frac{2}{3}\times\frac{4}{5}=1\times\frac{4}{5}\)

\(=\frac{4}{5}\)

\(\frac{4}{5}\) nhé bn

315/316*313/314*316/315*317/313

=315*313*316*317/316*314*315*313

=317/314

Tick cho mik nha

\(f\left(x\right)=x^{312}-x^{313}+x^{314}-x^{315}+x^{446}\)

\(=\left(x^{312}-1\right)-\left(x^{313}-x\right)+\left(x^{314}-1\right)-\left(x^{315}-x\right)+\left(x^{446}-1\right)-2x+3\)

\(=\left[\left(x^2\right)^{156}-1\right]-x\left[\left(x^2\right)^{156}-1\right]+\left[\left(x^2\right)^{157}-1\right]-x\left[\left(x^2\right)^{157}-1\right]+\left[\left(x^2\right)^{223}-1\right]-2x+3\)

\(=\left(x^2-1\right)A_{\left(x\right)}-x\left(x^2-1\right)B_{\left(x\right)}+\left(x^2-1\right)C_{\left(x\right)}-x\left(x^2-1\right)D_{\left(x\right)}+\left(x^2-1\right)E_{\left(x\right)}-2x+3\)

Vậy số dư là -2x+3

\(C=1+3+3^2+3^3+...+3^{11}\\ a,C=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+\left(3^6+3^7+3^8\right)+\left(3^9+3^{10}+3^{11}\right)\\ =13+3^3.\left(1+3+3^2\right)+3^6.\left(1+3+3^2\right)+3^9.\left(1+3+3^2\right)\\ =13+3^3.13+3^6.13+3^9.13\\ =13.\left(1+3^3+3^6+3^9\right)⋮13\)

Ý a phải chia hết cho 13 chứ em?

b: C=(1+3+3^2+3^3)+...+3^8(1+3+3^2+3^3)

=40(1+...+3^8) chia hết cho 40

a: C ko chia hết cho 15 nha bạn

\(b,C=\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)+\left(3^8+3^9+3^{10}+3^{11}\right)\\ =40+3^4.\left(1+3+3^2+3^3\right)+3^8.\left(1+3+3^2+3^3\right)\\ =40.\left(1+3^4+3^8\right)⋮40\)

A=1+3+3^2+3^3+...+3^98+3^99+3^100

A=(1+3+ 3^2)+(3^3+3^4+3^5)+...+(3^98+3^99+3^100)

A=(1+3+3^2)+3^3x(1+3+3^2)+...+3^98x(1+3+3^2)

A=13x3^3x13+...+3^98x13

=> 13x(1+3+3^3+...+3^98)chia hết cho 13

Vậy A chia hết cho 13

câu b đâu bạn ?

 

 

cho A = 1 + 3 + 3² + 3³ + ... +  3¹¹

 

^b) chứng minh A chia hết cho 40

\(a,C=5+5^2+5^3+5^4+\cdot\cdot\cdot+5^{20}\)

\(=5\left(1+5+5^2+\cdot\cdot\cdot+5^{19}\right)\)

Ta thấy: \(5\left(1+5+5^2+\cdot\cdot\cdot+5^{19}\right)⋮5\)

nên \(C⋮5\)

\(b,C=5+5^2+5^3+5^4\cdot\cdot\cdot+5^{20}\)

\(=\left(5+5^2\right)+\left(5^3+5^4\right)+\cdot\cdot\cdot+\left(5^{19}+5^{20}\right)\)

\(=5\left(1+5\right)+5^3\left(1+5\right)+\cdot\cdot\cdot+5^{19}\left(1+5\right)\)

\(=5\cdot6+5^3\cdot6+\cdot\cdot\cdot+5^{19}\cdot6\)

\(=6\cdot\left(5+5^3+\cdot\cdot\cdot+5^{19}\right)\)

Ta thấy: \(6\cdot\left(5+5^3+\cdot\cdot\cdot+5^{19}\right)⋮6\)

nên \(C⋮6\)

\(c,C=5+5^2+5^3+5^4+\cdot\cdot\cdot+5^{20}\)

\(=\left(5+5^3\right)+\left(5^2+5^4\right)+\cdot\cdot\cdot+\left(5^{17}+5^{19}\right)+\left(5^{18}+5^{20}\right)\)

\(=5\left(1+5^2\right)+5^2\left(1+5^2\right)+\cdot\cdot\cdot+5^{17}\cdot\left(1+5^2\right)+5^{18}\left(1+5^2\right)\)

\(=5\cdot26+5^2\cdot26+\cdot\cdot\cdot+5^{17}\cdot26+5^{18}\cdot26\)

\(=26\cdot\left(5+5^2+\cdot\cdot\cdot+5^{17}+5^{18}\right)\)

Ta thấy: \(26\cdot\left(5+5^2+\cdot\cdot\cdot+5^{17}+5^{18}\right)⋮13\)

nên \(C⋮13\)

#\(Toru\)

a, ta có
C = 5 + 5^2 + 5^3 + 5^4 + ... + 5^20
=> C = 5 . ( 1 + 5 + 5^2 + 5^3 + ... + 5^19 )
=> C chia hết cho 5
b,
C = 5 + 5^2 + 5^3 + 5^4 + ... + 5^20
=> C = 5 . ( 1 + 5 ) + 5^3 . ( 1 + 5 ) + ... + 5^19 . ( 1 + 5 )
=> C = 5 . 6 + 5^3 . 6 + ... + 5^19 . 6
=> C = 6 . ( 5 + 5^3 + ... + 5^19 )
=> C chia hết cho 6
c,
C = 5 + 5^2 + 5^3 + ... + 5^20
=> C = (5 + 5^2 + 5^3 + 5^4 ) + ... + (5^17 + 5^18 + 5^19 + 5^20 )
=> C = 5 . ( 1 + 5 + 5^2 + 5^3 ) + ... + 5^17 . ( 1+ 5 + 5^2 +5^3)
=> C = 5 . 156 + 5^5 . 156 + ...+ 5^17 . 156
=> C = 5 . 12 . 13 + 5^5 . 12 . 13 + ... + 5^17 . 12 . 13
=> C = 13 . ( 5 . 12 + 5^5 . 12 + ... + 5^17 . 12 )
=> C chia hết cho 13