Ta có : 2 + 4 + 6 + ... + 2(n - 1) + 2n = 210

<=> 2[1 + 2 + 3 + ... + (n - 1) + n] = 210

<=> 1 + 2 + 3 + ... + n = 105

<=> [(n - 1) : 1 + 1)(n + 1) : 2 = 105

<=> n(n + 1) = 210

<=> n(n + 1) = 14.15

=> n = 14

Vậy n = 14

b) Ta có : 1 + 3 + 5 + ... + (2n - 1) = 225

<=> [(2n - 1 - 1) : 2 + 1](2n - 1 + 1) : 2 = 225

<=> n² = 225

<=> n² = 15²

<=> n = 15

Vậy n = 15

a) \(2^n=8\)

\(\Rightarrow2^n=2^3\)

\(\Rightarrow n=3\)

b) \(5^{n+1}=125\)

\(\Rightarrow5^{n+1}=5^3\)

\(\Rightarrow n+1=3\)

\(\Rightarrow n=3-1=2\)

c) Mình không rõ đề:

d) \(2\cdot7^{n-1}+3=101\)

\(\Rightarrow2\cdot7^{n-1}=101-3\)

\(\Rightarrow2\cdot7^{n-1}=98\)

\(\Rightarrow7^{n-1}=\dfrac{98}{2}\)

\(\Rightarrow7^{n-1}=49\)

\(\Rightarrow7^{n-1}=7^2\)

\(\Rightarrow n-1=2\)

\(\Rightarrow n=1+2=3\)

e) \(3\cdot5^{2n+1}-6^2=339\)

\(\Rightarrow3\cdot5^{2n+1}=339+36\)

\(\Rightarrow3\cdot5^{2n+1}=375\)

\(\Rightarrow5^{2n+1}=125\)

\(\Rightarrow5^{2n+1}=5^3\)

\(\Rightarrow2n+1=3\)

\(\Rightarrow2n=2\)

\(\Rightarrow n=\dfrac{2}{2}=1\)

Mình mẫu đầu với cuối nhé:

a)  Đặt \(ƯCLN\left(3n+4,3n+7\right)=d\)  

\(\Rightarrow\left\{{}\begin{matrix}3n+4⋮d\\3n+7⋮d\end{matrix}\right.\)

\(\Rightarrow\left(3n+7\right)-\left(3n+4\right)⋮d\)

\(\Rightarrow3⋮d\)

 \(\Rightarrow d\in\left\{1,3\right\}\)

Nhưng do \(3n+4,3n+7⋮̸3\) nên \(d\ne3\Rightarrow d=1\)

Vậy \(ƯCLN\left(3n+4,3n+7\right)=1\) hay \(3n+4,3n+7\) nguyên tố cùng nhau.

 e) \(ƯCLN\left(2n+3,3n+5\right)=d\)

 \(\Rightarrow\left\{{}\begin{matrix}2n+3⋮d\\3n+5⋮d\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}6n+9⋮d\\6n+10⋮d\end{matrix}\right.\)

\(\Rightarrow\left(6n+10\right)-\left(6n+9\right)⋮d\)

\(\Rightarrow1⋮d\) \(\Rightarrow d=1\)

Vậy \(ƯCLN\left(2n+3,3n+5\right)=1\), ta có đpcm.

a, Ta có : \(\text{n + 5 = (n - 1)+6}\)

Vì \(\text{(n-1) ⋮ n-1}\)

Nên để \(\text{n+5 ⋮ n-1}\)⋮ `n-1`

Thì \(\text{6 ⋮ n-1}\) 

\(\Rightarrow\) \(\text{n - 1 ∈ Ư(6)}\)

\(\Rightarrow\) \(\text{n - 1 ∈}\) \(\left\{\text{±1;±2;±3;±6}\right\}\)

\(\Rightarrow\) \(\text{n ∈}\) \(\left\{\text{0;-1;-2;-5;2;3;4;7}\right\}\) \(\text{( TM )}\)

\(\text{________________________________________________________}\)

b, Ta có : \(\text{2n-4 = (2n+4)- 8 = 2(n+2) - 8}\)

Vì \(\text{2(n+2) ⋮ n+2}\)

Nên để \(\text{2n-4 ⋮ n+2}\)

Thì \(\text{8 ⋮ n+2}\)

\(\Rightarrow\) \(\text{n + 2 ∈ Ư(8)}\)

\(\Rightarrow\) \(\text{n + 2 ∈}\) \(\left\{\text{±1;±2;±4;±8}\right\}\)

\(\Rightarrow\) \(\text{n ∈}\) \(\left\{\text{-3;-4;-6;-10;-1;0;2;6}\right\}\) ( TM )

\(\text{_________________________________________________________________ }\)

c, Ta có :\(\text{ 6n + 4 = (6n + 3) +1 = 3(2n+1) + 1}\)

Vì \(\text{3(2n+1) ⋮ 2n+1}\)

Nên để\(\text{ 6n+4 ⋮ 2n+1}\)

Thì \(\text{1 ⋮ 2n+1}\)

\(\Rightarrow\) \(\text{2n + 1 ∈ Ư(1)}\)

\(\Rightarrow\) \(\text{2n + 1 ∈}\) \(\left\{\text{±1}\right\}\)

\(\Rightarrow\) \(\text{2n ∈}\) \(\left\{\text{-2;0}\right\}\)

\(\Rightarrow\) \(\text{n ∈}\) \(\left\{\text{-1;0}\right\}\) ( TM )

\(\text{_______________________________________}\)

Ta có : \(\text{3 - 2n = -( 2n - 3 ) = -( 2n + 2 ) + 5 = -2( n+1)+5}\)

Vì \(\text{-2(n+1) ⋮ n+1}\)

Nên để \(\text{3-2n ⋮ n+1}\)

Thì\(\text{ 5 ⋮ n + 1}\)

\(\Rightarrow\) \(\text{n + 1 ∈}\) \(\left\{\text{±1;±5}\right\}\)

\(\Rightarrow\) \(\text{n ∈}\) \(\text{-2;-6;0;4}\) ( TM )

1 . tìm giá trị x 

\(\left(x+1\right)+\left(x+4\right)+....+\left(x+28\right)=115.\)

\(\Rightarrow\left(x+x+x+....x\right)+\left(1+4+..+28\right)=115\)

\(\Rightarrow10x+\left(28+1\right).10:2=115\)

\(\Rightarrow10x+145=115\)

\(\Rightarrow10x=115-145=-30\)

\(\Rightarrow x=-30:10=-3\)

Bai 1

so so hang la: [(28+x)-(x+1)]/3+1= 10 so hang

tong =[(x+1)+(x+28)]*10/2=(2x+29)*10/2=115

(2x+29)*5=115

2x+29=115/5=23

2x=23-29=-6

x=-3

bai 2

a)

so so hang =(2n-2)/2+1=n

tong=(2n+2).n/2=n(n+1) 

n.(n+1)=210=14*15=> n=14

b) tuong tu tong=(2n-1+1).n/2=n^2=225=25^2=> n=25

Ta có: \(\frac{2n+6}{2n-1}=\frac{2n-1+7}{2n-1}=1+\frac{7}{2n-1}\)

để \(2n-6⋮2n-1\) thì \(7⋮2n-1\)

hay 2n -1 \(\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

xét bảng 

vậy........

\(\left(\dfrac{1}{4}\right)^{2n}=\left(\dfrac{1}{8}\right)^2\)

\(\Rightarrow\left(\dfrac{1}{2}\right)^{2.2n}=\left(\dfrac{1}{2}\right)^{3.2}\)

\(\Rightarrow\left(\dfrac{1}{2}\right)^{4n}=\left(\dfrac{1}{2}\right)^6\)

\(\Rightarrow4n=6\)

\(\Rightarrow n=\dfrac{6}{4}=\dfrac{3}{2}\)

n = 3/2