giải jup mik mai mik đi học

A=3+3² +3³+...+3¹⁰⁰

3A=3²+3³+3⁴+...+3¹⁰¹

3A-A=3¹⁰¹-3

2A=3¹⁰¹-3

A=\(\frac{3^{101}-3}{2}\)

3 + 3² + 3³ + ... + 3¹⁰⁰

3A = 3² + 3³ + 3⁴ + ... + 3¹⁰⁰ + 3¹⁰¹

3A - A = 3¹⁰¹ - 3 

2A = 3¹⁰¹ - 3

A = ( 3¹⁰¹ - 3 ) : 2

Bạn tự tính kết quả nhé.

Học tốt.

ta có A=3+3mu2+3mu3+..+3mu100

A=3+3²+3³+...+3¹⁰⁰

3A=3²+3³+...+3¹⁰⁰+3¹⁰¹

3A-A=3²+3³+...+3¹⁰⁰+3¹⁰¹-(3+3mu2+3mu3+..+3mu100)

2A=3¹⁰¹-3

2A+3=3n

suy ra:3¹⁰¹-3+3=3n

suy ra:3¹⁰¹=3n

suy ra: n =3¹⁰⁰

đợi tí nhé

là bội của 3 vì có thừa số 3

tick nhé

Rút gọn đc

3^10 - 3 = 3(3^9  - 1) = 3.(19683-1) = 3.1514.13 chia hết cho 13 

Ta có: \(3+3^2+3^3+3^4+3^5+3^6+3^7+3^8+3^9\)

\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+\left(3^7+3^8+3^9\right)\)

\(=3\left(1+3+9\right)+3^4\left(1+3+9\right)+3^7\left(1+3+9\right)\)

\(=\left(3+3^4+3^7\right).13\)chia hết cho 13

1. ĐKXĐ: \(x\ne\pm1\)

2. \(A=\left(\dfrac{x+1}{x-1}-\dfrac{x+3}{x+1}\right)\cdot\dfrac{x+1}{2}\)

\(=\dfrac{\left(x+1\right)^2-\left(x-3\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{2}\)

\(=\dfrac{x^2+2x+1-x^2+4x-3}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{2}\)

\(=\dfrac{6x-2}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{2}\)

\(=\dfrac{2\left(x-3\right)\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{x-3}{x-1}\)

3. Tại x = 5, A có giá trị là:

\(\dfrac{5-3}{5-1}=\dfrac{1}{2}\)

4. \(A=\dfrac{x-3}{x-1}\) \(=\dfrac{x-1-3}{x-1}=1-\dfrac{3}{x-1}\)

Để A nguyên => \(3⋮\left(x-1\right)\) hay \(\left(x-1\right)\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)

\(\Rightarrow\left\{{}\begin{matrix}x-1=1\\x-1=-1\\x-1=3\\x-1=-3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\left(tmđk\right)\\x=0\left(tmđk\right)\\x=4\left(tmđk\right)\\x=-2\left(tmđk\right)\end{matrix}\right.\)

Vậy: A nguyên khi \(x=\left\{2;0;4;-2\right\}\)