a)  3n = 35 => n = 5

b) 2n = 28 => n = 8

--thodagbun--

( h tớ lm b đt nè :( )

a) \(3^n=243\)

\(=>3^n=3^5\)

\(=>n=5\)

b) \(2^n=256\)

\(=>2^n=2^8\)

\(=>n=8\)

#Nothings -_-

Hài

a) 3ⁿ = 243

=> 3ⁿ = 3⁵

=> n = 5

b) 2ⁿ = 256 

=> 2ⁿ = 2⁸

=> n = 8

1:

2n^2+5n-1 chia hết cho 2n-1

=>2n^2-n+6n-3+2 chia hết cho 2n-1

=>2n-1 thuộc {1;-1;2;-2}

mà n nguyên

nên n=1 hoặc n=0

2:

a: A=n(n+1)(n+2)

Vì n;n+1;n+2 là 3 số liên tiếp

nên A=n(n+1)(n+2) chia hết cho 3!=6

b: B=(2n-1)[(2n-1)^2-1]

=(2n-1)(2n-2)*2n

=4n(n-1)(2n-1)

Vì n;n-1 là hai số nguyên liên tiếp

nên n(n-1) chia hết cho 2

=>B chia hết cho 8

c: C=n^2+14n+49-n^2+10n-25=24n+24=24(n+1) chia hết cho 24

\(a,\Rightarrow3^n=27=3^3\Rightarrow n=3\\ b,\Rightarrow13^n=13=13^1\Rightarrow n=1\\ c,\Rightarrow2^4< 2^{n-1}< 2^6\\ \Rightarrow n-1=5\Rightarrow n=6\\ d,\Rightarrow5^n=134-9=125=5^3\Rightarrow n=3\)

\(e,\Rightarrow2^n=4^{10}:8^4=2^{20}:2^{12}=2^8\\ \Rightarrow n=8\)

\(f,n\in\varnothing\)

1,N=3

2,N=6

\(2n-1+5n-2=\frac{7}{32}\)

                   \(7n-3=\frac{7}{32}\)

                           \(7n=\frac{7}{32}+3\)

                           \(7n=\frac{103}{32}\)

                              \(n=\frac{103}{32}:7\)

                              \(n=\frac{103}{224}\)

a) \(5^{n+2}+26.5^n+8^{2n+1}=25.5^n+26.6^n+8.8^{2n}\)

\(=5^n.51+8.64^n\)

Có \(64\equiv5\) (mod 59)

\(\Rightarrow64^n\equiv5^n\) (mod 59)

\(\Rightarrow8.64^n\equiv8.5^n\) (mod 59)

\(\Rightarrow5^n.51+8.64^n\equiv8.5^n+5^n.51\) (mod 59)

mà \(8.5^n+5^n.51=59.5^n\)\(\equiv0\) (mod 59)

\(\Rightarrow5^n.51+8.64^n\equiv8.5^n+5^n.51\equiv0\) (mod 59) 

\(\Rightarrow5^{n+2}+26.5^n+8^{2n+1}⋮59\)

b) \(4^{2n}-3^{2n}-7=16^n-9^n-7\)

Có \(16^n-9^n-7=\left(16-9\right)\left(16^{n-1}+...+9^{n-1}\right)-7=7\left(16^{n-1}+...+9^{n-1}\right)-7⋮\)\(7\) (I)

Có \(16\equiv1\) (mod 3) \(\Rightarrow16^n\equiv1\) (mod 3) mà \(7\equiv1\) (mod 3)

\(\Rightarrow16^n-7\equiv0\) (mod 3) mà \(9^n\equiv0\) (mod 3)

\(\Rightarrow16^n-9^n-7⋮3\) (II)

Có \(9^n\equiv1\) (mod 8)\(\Rightarrow9^n+7\equiv8\) (mod 8) 

\(\Rightarrow9^n+7⋮8\)  mà \(16^n=2^n.8^n⋮8\) 

\(\Rightarrow16^n-9^n-7⋮8\) (III)

Do \(\left(3;7;8\right)=1\)\(,3.7.8=168\)

Từ (I) (II) (III) \(\Rightarrow16^n-9^n-7⋮168\) 

\(\Rightarrow\) Đpcm

a) 5n+2+26.5n+82n+1=25.5n+26.6n+8.82n5n+2+26.5n+82n+1=25.5n+26.6n+8.82n

Có  (mod 59)

 (mod 59)

 (mod 59)

 (mod 59)

mà  (mod 59)

 (mod 59) 

cho e hỏi là 3 dấu gạch ngang là gì vậy ạ

a,\(lim\dfrac{1-2n^2}{5n+5}=lim\dfrac{\left(1-n\sqrt{2}\right)\left(1+n\sqrt{2}\right)}{5n+5}=lim\dfrac{\left(\dfrac{1}{n}-\sqrt{2}\right)\left(\dfrac{1}{n}+\sqrt{2}\right)}{5+\dfrac{5}{n}}=\dfrac{-2}{5}\)

b,\(lim\dfrac{1-2n}{5n+5n^2}=lim\dfrac{\dfrac{1}{n^2}-\dfrac{2}{n}}{\dfrac{5}{n}+5}=\dfrac{0}{5}=0\)

a,\(lim\dfrac{n^2-2n}{5n+3n^2}=lim\dfrac{1-\dfrac{2}{n}}{\dfrac{5}{n}+3}=\dfrac{1}{3}\)

b,\(lim\dfrac{n^2-2}{5n+3n^2}=lim\dfrac{1-\dfrac{2}{n^2}}{\dfrac{5}{n}+3}=\dfrac{1}{3}\)

c,\(lim\dfrac{1-2n}{5n+3n^2}=lim\dfrac{1-2n}{n\left(5+3n\right)}=lim\dfrac{\dfrac{1}{n}-2}{1\left(\dfrac{5}{n}+3\right)}=-\dfrac{2}{3}\)

d,\(lim\dfrac{1-2n^2}{5n+5}=lim\dfrac{\left(1-n\sqrt{2}\right)\left(1+n\sqrt{2}\right)}{5n+5}=lim\dfrac{\left(\dfrac{1}{n}-\sqrt{2}\right)\left(\dfrac{1}{n}+\sqrt{2}\right)}{5+\dfrac{5}{n}}=\dfrac{-2}{5}\)

a) \(\Rightarrow2\left(n+3\right)-38⋮\left(n+3\right)\)

Do \(n\in N\)

\(\Rightarrow\left(n+3\right)\inƯ\left(38\right)=\left\{19;38\right\}\)

\(\Rightarrow n\in\left\{16;35\right\}\)

b) \(\Rightarrow5\left(n+5\right)-74⋮\left(n+5\right)\)

Do \(n\in N\)

\(\Rightarrow\left(n+5\right)\inƯ\left(74\right)=\left\{37;74\right\}\)

\(\Rightarrow N\in\left\{32;69\right\}\)